3.308 \(\int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=150 \[ \frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (a \cos (c+d x)+b)}+\frac {2 \sqrt {a-b} \sqrt {a+b} \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 b^3 d}+\frac {x}{a^2}-\frac {2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {\tan (c+d x)}{b d (a \cos (c+d x)+b)} \]

[Out]

x/a^2-2*a*arctanh(sin(d*x+c))/b^3/d+(2*a^2-b^2)*sin(d*x+c)/a/b^2/d/(b+a*cos(d*x+c))+2*(2*a^2+b^2)*arctanh((a-b
)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/a^2/b^3/d+tan(d*x+c)/b/d/(b+a*cos(d*x+c))

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Rubi [A]  time = 0.33, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3898, 2890, 3057, 2659, 208, 3770} \[ \frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (a \cos (c+d x)+b)}+\frac {2 \sqrt {a-b} \sqrt {a+b} \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 b^3 d}+\frac {x}{a^2}-\frac {2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {\tan (c+d x)}{b d (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

x/a^2 - (2*a*ArcTanh[Sin[c + d*x]])/(b^3*d) + (2*Sqrt[a - b]*Sqrt[a + b]*(2*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Ta
n[(c + d*x)/2])/Sqrt[a + b]])/(a^2*b^3*d) + ((2*a^2 - b^2)*Sin[c + d*x])/(a*b^2*d*(b + a*Cos[c + d*x])) + Tan[
c + d*x]/(b*d*(b + a*Cos[c + d*x]))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2890

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b*d*(n + 1)*(m + 1)), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1)*Simp[a^2*(n + 1)
*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m
+ n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[((a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n
+ 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*b*d^2*f*(n + 1)*(m + 1)), x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (b+a \cos (c+d x))}+\frac {\tan (c+d x)}{b d (b+a \cos (c+d x))}+\frac {\int \frac {\left (-2 a^2-a b \cos (c+d x)+b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{b+a \cos (c+d x)} \, dx}{a b^2}\\ &=\frac {x}{a^2}+\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (b+a \cos (c+d x))}+\frac {\tan (c+d x)}{b d (b+a \cos (c+d x))}-\frac {(2 a) \int \sec (c+d x) \, dx}{b^3}-\frac {\left (-2 a^4+a^2 b^2+b^4\right ) \int \frac {1}{b+a \cos (c+d x)} \, dx}{a^2 b^3}\\ &=\frac {x}{a^2}-\frac {2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (b+a \cos (c+d x))}+\frac {\tan (c+d x)}{b d (b+a \cos (c+d x))}-\frac {\left (2 \left (-2 a^4+a^2 b^2+b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 b^3 d}\\ &=\frac {x}{a^2}-\frac {2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {2 \sqrt {a-b} \sqrt {a+b} \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 b^3 d}+\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (b+a \cos (c+d x))}+\frac {\tan (c+d x)}{b d (b+a \cos (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.58, size = 327, normalized size = 2.18 \[ \frac {\sec ^2(c+d x) (a \cos (c+d x)+b) \left (\frac {\left (a^2-b^2\right ) \sin (c+d x)}{a b^2}+\frac {(c+d x) (a \cos (c+d x)+b)}{a^2}+\frac {2 \left (-2 a^4+a^2 b^2+b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^3 \sqrt {a^2-b^2}}+\frac {2 a (a \cos (c+d x)+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^3}-\frac {2 a (a \cos (c+d x)+b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{b^3}+\frac {\sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{b^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{b^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(((c + d*x)*(b + a*Cos[c + d*x]))/a^2 + (2*(-2*a^4 + a^2*b^2 + b^4)*ArcTa
nh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x]))/(a^2*b^3*Sqrt[a^2 - b^2]) + (2*a*(b + a*
Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b^3 - (2*a*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]])/b^3 + ((b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(b^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))
+ ((b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(b^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + ((a^2 - b^2)*Sin[c + d
*x])/(a*b^2)))/(d*(a + b*Sec[c + d*x])^2)

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fricas [A]  time = 0.65, size = 584, normalized size = 3.89 \[ \left [\frac {2 \, a b^{3} d x \cos \left (d x + c\right )^{2} + 2 \, b^{4} d x \cos \left (d x + c\right ) + {\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{2} b^{2} + {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{3} b^{3} d \cos \left (d x + c\right )^{2} + a^{2} b^{4} d \cos \left (d x + c\right )\right )}}, \frac {a b^{3} d x \cos \left (d x + c\right )^{2} + b^{4} d x \cos \left (d x + c\right ) + {\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} b^{2} + {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{a^{3} b^{3} d \cos \left (d x + c\right )^{2} + a^{2} b^{4} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(2*a*b^3*d*x*cos(d*x + c)^2 + 2*b^4*d*x*cos(d*x + c) + ((2*a^3 + a*b^2)*cos(d*x + c)^2 + (2*a^2*b + b^3)*
cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*co
s(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(a^4*cos(d*x
+ c)^2 + a^3*b*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*(a^4*cos(d*x + c)^2 + a^3*b*cos(d*x + c))*log(-sin(d*x
+ c) + 1) + 2*(a^2*b^2 + (2*a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 + a^2*b^4*d*c
os(d*x + c)), (a*b^3*d*x*cos(d*x + c)^2 + b^4*d*x*cos(d*x + c) + ((2*a^3 + a*b^2)*cos(d*x + c)^2 + (2*a^2*b +
b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))
- (a^4*cos(d*x + c)^2 + a^3*b*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^4*cos(d*x + c)^2 + a^3*b*cos(d*x + c))*
log(-sin(d*x + c) + 1) + (a^2*b^2 + (2*a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 +
a^2*b^4*d*cos(d*x + c))]

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giac [B]  time = 1.55, size = 294, normalized size = 1.96 \[ \frac {\frac {d x + c}{a^{2}} - \frac {2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, {\left (2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} a b^{2}} + \frac {2 \, {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{2} b^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

((d*x + c)/a^2 - 2*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + 2*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - 2*(
2*a^2*tan(1/2*d*x + 1/2*c)^3 - a*b*tan(1/2*d*x + 1/2*c)^3 - b^2*tan(1/2*d*x + 1/2*c)^3 - 2*a^2*tan(1/2*d*x + 1
/2*c) - a*b*tan(1/2*d*x + 1/2*c) + b^2*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*
c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*a*b^2) + 2*(2*a^4 - a^2*b^2 - b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2
)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a^2 +
b^2)*a^2*b^3))/d

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maple [B]  time = 0.40, size = 353, normalized size = 2.35 \[ -\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {4 a^{2} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{3}}-\frac {1}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x)

[Out]

-2/d/b^2*a*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)+2/d/a*tan(1/2*d*x+1/2*c)/(a*
tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)+4/d/b^3*a^2/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a
-b)/((a-b)*(a+b))^(1/2))-2/d/b/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-2/d*b
/a^2/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-1/d/b^2/(tan(1/2*d*x+1/2*c)-1)+
2/d*a/b^3*ln(tan(1/2*d*x+1/2*c)-1)-1/d/b^2/(tan(1/2*d*x+1/2*c)+1)-2/d*a/b^3*ln(tan(1/2*d*x+1/2*c)+1)+2/d/a^2*a
rctan(tan(1/2*d*x+1/2*c))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.47, size = 7044, normalized size = 46.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^2,x)

[Out]

(2*atan((((((((((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 + 5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^
8))/(a^3*b^8) - (tan(c/2 + (d*x)/2)*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*
b^10)*8192i)/(a^6*b^8))*1i)/a^2 - (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^1
2 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a
^4*b^8))*1i)/a^2 - (8192*(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 +
40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8))*1i)/a^2 + (8192*t
an(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44
*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8))/a^2 - ((((((((8192*(8*
a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 + 5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) + (tan(c/2
+ (d*x)/2)*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10)*8192i)/(a^6*b^8))*1
i)/a^2 + (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10
- 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8))*1i)/a^2 - (8192*
(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6
+ 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8))*1i)/a^2 - (8192*tan(c/2 + (d*x)/2)*(16*a^1
2*b - a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 1
2*a^8*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8))/a^2)/((((((((((8192*(8*a^5*b^13 - 3*a^4*b^14 -
9*a^6*b^12 + 5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) - (tan(c/2 + (d*x)/2)*(2*a^6*b^15 -
6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10)*8192i)/(a^6*b^8))*1i)/a^2 - (8192*tan(c/2 +
(d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^
8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8))*1i)/a^2 - (8192*(3*b^14 - 5*a*b^13 + 6*a
^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 + 24*a^10*b
^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8))*1i)/a^2 + (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*a^13 +
b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*b^4 -
16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8))*1i)/a^2 + (((((((((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 + 5*a^7*b
^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) + (tan(c/2 + (d*x)/2)*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*
b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10)*8192i)/(a^6*b^8))*1i)/a^2 + (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^1
5 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28
*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8))*1i)/a^2 - (8192*(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11
- 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 2
4*a^12*b^2))/(a^3*b^8))*1i)/a^2 - (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 -
2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^1
1*b^2))/(a^4*b^8))*1i)/a^2 - (16384*(2*a*b^9 - 16*a^9*b + 16*a^10 - 2*b^10 - 16*a^2*b^8 + 24*a^3*b^7 - 18*a^4*
b^6 + 26*a^5*b^5 + 12*a^6*b^4 - 36*a^7*b^3 + 8*a^8*b^2))/(a^3*b^8))))/(a^2*d) + ((2*tan(c/2 + (d*x)/2)*(a*b +
2*a^2 - b^2))/(a*b^2) - (2*tan(c/2 + (d*x)/2)^3*(a - b)*(2*a + b))/(a*b^2))/(d*(a + b + tan(c/2 + (d*x)/2)^4*(
a - b) - 2*a*tan(c/2 + (d*x)/2)^2)) + (atan((((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((2*(a^2
- b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((2*
(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 + 5*a^7*b^11
+ 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) - (8192*tan(c/2 + (d*x)/2)*((2*(a^2 - b^2)^(1/2))/b^3 + (a^
2 - b^2)^(1/2)/(a^2*b))*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10))/(a^4*
b^8)) - (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10
- 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)) - (8192*(3*b^14 -
5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*
b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8)) + (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16
*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9
*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8))*1i - ((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((
2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))
*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 + 5*a^7
*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) + (8192*tan(c/2 + (d*x)/2)*((2*(a^2 - b^2)^(1/2))/b^3
+ (a^2 - b^2)^(1/2)/(a^2*b))*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10))
/(a^4*b^8)) + (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7
*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)) - (8192*(3*
b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 1
4*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8)) - (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^1
2 - 16*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 +
4*a^9*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8))*1i)/(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b
))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a
^2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 +
5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) - (8192*tan(c/2 + (d*x)/2)*((2*(a^2 - b^2)^(1/2
))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*
b^10))/(a^4*b^8)) - (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 +
3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)) - (81
92*(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b
^6 + 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8)) + (8192*tan(c/2 + (d*x)/2)*(16*a^12*b -
a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8
*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8)) + ((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^
2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)
/(a^2*b))*(((2*(a^2 - b^2)^(1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^1
2 + 5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) + (8192*tan(c/2 + (d*x)/2)*((2*(a^2 - b^2)^(
1/2))/b^3 + (a^2 - b^2)^(1/2)/(a^2*b))*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^
11*b^10))/(a^4*b^8)) + (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^1
1 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)) -
(8192*(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^
8*b^6 + 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8)) - (8192*tan(c/2 + (d*x)/2)*(16*a^12*
b - a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*
a^8*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8)) - (16384*(2*a*b^9 - 16*a^9*b + 16*a^10 - 2*b^10 -
16*a^2*b^8 + 24*a^3*b^7 - 18*a^4*b^6 + 26*a^5*b^5 + 12*a^6*b^4 - 36*a^7*b^3 + 8*a^8*b^2))/(a^3*b^8)))*(((a^2
- b^2)^(1/2)*4i)/b^3 + ((a^2 - b^2)^(1/2)*2i)/(a^2*b)))/d + (a*atan(((a*((2*a*((2*a*((2*a*((8192*(8*a^5*b^13 -
3*a^4*b^14 - 9*a^6*b^12 + 5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) - (16384*tan(c/2 + (d
*x)/2)*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10))/(a^3*b^11)))/b^3 - (81
92*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9
+ 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)))/b^3 - (8192*(3*b^14 - 5*a*b
^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 +
24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8)))/b^3 + (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*
a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*
b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^8))*2i)/b^3 - (a*((2*a*((2*a*((2*a*((8192*(8*a^5*b^13 - 3*a^4*b^14 -
9*a^6*b^12 + 5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) + (16384*tan(c/2 + (d*x)/2)*(2*a^6*
b^15 - 6*a^7*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10))/(a^3*b^11)))/b^3 + (8192*tan(c/2 + (
d*x)/2)*(2*a^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8
- 28*a^10*b^7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)))/b^3 - (8192*(3*b^14 - 5*a*b^13 + 6*a^2*b^
12 - 8*a^3*b^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 + 24*a^10*b^4 -
52*a^11*b^3 + 24*a^12*b^2))/(a^3*b^8)))/b^3 - (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*a^13 + b^13 +
2*a^2*b^11 - 2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*b^4 - 16*a^10*
b^3 + 16*a^11*b^2))/(a^4*b^8))*2i)/b^3)/((2*a*((2*a*((2*a*((2*a*((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 +
5*a^7*b^11 + 6*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) - (16384*tan(c/2 + (d*x)/2)*(2*a^6*b^15 - 6*a^7
*b^14 + 8*a^8*b^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10))/(a^3*b^11)))/b^3 - (8192*tan(c/2 + (d*x)/2)*(2*a
^2*b^15 - 6*a^3*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^
7 + 28*a^11*b^6 - 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)))/b^3 - (8192*(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b
^11 - 5*a^4*b^10 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3
+ 24*a^12*b^2))/(a^3*b^8)))/b^3 + (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 -
2*a^3*b^10 + 5*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^1
1*b^2))/(a^4*b^8)))/b^3 + (2*a*((2*a*((2*a*((2*a*((8192*(8*a^5*b^13 - 3*a^4*b^14 - 9*a^6*b^12 + 5*a^7*b^11 + 6
*a^8*b^10 - 13*a^9*b^9 + 6*a^10*b^8))/(a^3*b^8) + (16384*tan(c/2 + (d*x)/2)*(2*a^6*b^15 - 6*a^7*b^14 + 8*a^8*b
^13 - 8*a^9*b^12 + 6*a^10*b^11 - 2*a^11*b^10))/(a^3*b^11)))/b^3 + (8192*tan(c/2 + (d*x)/2)*(2*a^2*b^15 - 6*a^3
*b^14 + 10*a^4*b^13 - 10*a^5*b^12 + a^6*b^11 + 3*a^7*b^10 - 9*a^8*b^9 + 25*a^9*b^8 - 28*a^10*b^7 + 28*a^11*b^6
- 24*a^12*b^5 + 8*a^13*b^4))/(a^4*b^8)))/b^3 - (8192*(3*b^14 - 5*a*b^13 + 6*a^2*b^12 - 8*a^3*b^11 - 5*a^4*b^1
0 + 11*a^5*b^9 - 18*a^6*b^8 + 40*a^7*b^7 - 34*a^8*b^6 + 14*a^9*b^5 + 24*a^10*b^4 - 52*a^11*b^3 + 24*a^12*b^2))
/(a^3*b^8)))/b^3 - (8192*tan(c/2 + (d*x)/2)*(16*a^12*b - a*b^12 - 16*a^13 + b^13 + 2*a^2*b^11 - 2*a^3*b^10 + 5
*a^4*b^9 - 21*a^5*b^8 + 44*a^6*b^7 - 44*a^7*b^6 + 12*a^8*b^5 + 4*a^9*b^4 - 16*a^10*b^3 + 16*a^11*b^2))/(a^4*b^
8)))/b^3 - (16384*(2*a*b^9 - 16*a^9*b + 16*a^10 - 2*b^10 - 16*a^2*b^8 + 24*a^3*b^7 - 18*a^4*b^6 + 26*a^5*b^5 +
12*a^6*b^4 - 36*a^7*b^3 + 8*a^8*b^2))/(a^3*b^8)))*4i)/(b^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sec(c + d*x))**2, x)

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