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

Optimal. Leaf size=126 \[ \frac {\left (2 a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b^3 d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {x}{a}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 126, 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, 2893, 3057, 2659, 208, 3770} \[ \frac {\left (2 a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {a \tan (c+d x)}{b^2 d}-\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b^3 d}+\frac {x}{a}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2893

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]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

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

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Mathematica [B]  time = 2.32, size = 287, normalized size = 2.28 \[ \frac {\sec (c+d x) (a \cos (c+d x)+b) \left (-\frac {4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^3}+\frac {4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{b^3}+\frac {8 \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^3}-\frac {4 a \tan (c+d x)}{b^2}+\frac {4 c}{a}+\frac {4 d x}{a}+\frac {1}{b \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{b \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b}-\frac {6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{b}\right )}{4 d (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*((4*c)/a + (4*d*x)/a + (8*(a^2 - b^2)^(3/2)*ArcTanh[((-a + b)*Tan[(c + d*x)
/2])/Sqrt[a^2 - b^2]])/(a*b^3) - (4*a^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b^3 + (6*Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]])/b + (4*a^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/b^3 - (6*Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]])/b + 1/(b*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - 1/(b*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
)^2) - (4*a*Tan[c + d*x])/b^2))/(4*d*(a + b*Sec[c + d*x]))

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fricas [A]  time = 0.93, size = 444, normalized size = 3.52 \[ \left [\frac {4 \, b^{3} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{2} b \cos \left (d x + c\right ) - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, a b^{3} d \cos \left (d x + c\right )^{2}}, \frac {4 \, b^{3} d x \cos \left (d x + c\right )^{2} - 4 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} + {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{2} b \cos \left (d x + c\right ) - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, a b^{3} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.48, size = 476, normalized size = 3.78 \[ -\frac {\frac {2 \, {\left ({\left (a^{2} + a b - b^{2}\right )} \sqrt {-a^{2} + b^{2}} {\left | a \right |} {\left | -a + b \right |} {\left | b \right |} + {\left (a^{3} b + a^{2} b^{2} - a b^{3} - 2 \, b^{4}\right )} \sqrt {-a^{2} + b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b^{4} + \sqrt {b^{8} + {\left (a b^{3} + b^{4}\right )} {\left (a b^{3} - b^{4}\right )}}}{a b^{3} - b^{4}}}}\right )\right )}}{{\left (a b^{2} - b^{3}\right )} a^{2} b^{2} + {\left (a b^{4} - b^{5}\right )} {\left | a \right |} {\left | b \right |}} + \frac {2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} - a b^{4} + 2 \, b^{5} - a^{3} {\left | a \right |} {\left | b \right |} + 2 \, a b^{2} {\left | a \right |} {\left | b \right |} - b^{3} {\left | a \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b^{4} - \sqrt {b^{8} + {\left (a b^{3} + b^{4}\right )} {\left (a b^{3} - b^{4}\right )}}}{a b^{3} - b^{4}}}}\right )\right )}}{a^{2} b^{4} - b^{4} {\left | a \right |} {\left | b \right |}} - \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} \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 \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, 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 )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*((a^2 + a*b - b^2)*sqrt(-a^2 + b^2)*abs(a)*abs(-a + b)*abs(b) + (a^3*b + a^2*b^2 - a*b^3 - 2*b^4)*sqrt
(-a^2 + b^2)*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b^4 + sqrt(b^
8 + (a*b^3 + b^4)*(a*b^3 - b^4)))/(a*b^3 - b^4))))/((a*b^2 - b^3)*a^2*b^2 + (a*b^4 - b^5)*abs(a)*abs(b)) + 2*(
a^4*b - 2*a^2*b^3 - a*b^4 + 2*b^5 - a^3*abs(a)*abs(b) + 2*a*b^2*abs(a)*abs(b) - b^3*abs(a)*abs(b))*(pi*floor(1
/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b^4 - sqrt(b^8 + (a*b^3 + b^4)*(a*b^3 - b^4)))/(a*
b^3 - b^4))))/(a^2*b^4 - b^4*abs(a)*abs(b)) - (2*a^2 - 3*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + (2*a^2
- 3*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - 2*(2*a*tan(1/2*d*x + 1/2*c)^3 + b*tan(1/2*d*x + 1/2*c)^3 - 2
*a*tan(1/2*d*x + 1/2*c) + b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^2))/d

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maple [B]  time = 0.44, size = 374, normalized size = 2.97 \[ -\frac {2 a^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 a \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}{d \,b^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d b}-\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}}{d \,b^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d b}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b^3*a^3/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+4/d/b*a/((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/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2
*c)*(a-b)/((a-b)*(a+b))^(1/2))+1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2+1/d/b^2/(tan(1/2*d*x+1/2*c)-1)*a+1/2/d/b/(tan(
1/2*d*x+1/2*c)-1)-1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*a^2+3/2/d/b*ln(tan(1/2*d*x+1/2*c)-1)-1/2/d/b/(tan(1/2*d*x+1
/2*c)+1)^2+1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*a+1/2/d/b/(tan(1/2*d*x+1/2*c)+1)+1/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*a^
2-3/2/d/b*ln(tan(1/2*d*x+1/2*c)+1)+2/d/a*arctan(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)),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 = 3.29, size = 6062, normalized size = 48.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^4*3i)/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/
2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)
/2))*sin(c/2 + (d*x)/2)^4*3i)/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c
/2 + (d*x)/2)^2)) + (cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^3)/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^
4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2))/(b*d*(cos(c/2 +
(d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (2*atan((4*a^13*sin(c/2 +
(d*x)/2) + 4*a^12*b*sin(c/2 + (d*x)/2) + 12*a^2*b^11*sin(c/2 + (d*x)/2) + 12*a^3*b^10*sin(c/2 + (d*x)/2) + 15*
a^4*b^9*sin(c/2 + (d*x)/2) + 15*a^5*b^8*sin(c/2 + (d*x)/2) - 59*a^6*b^7*sin(c/2 + (d*x)/2) - 59*a^7*b^6*sin(c/
2 + (d*x)/2) + 57*a^8*b^5*sin(c/2 + (d*x)/2) + 57*a^9*b^4*sin(c/2 + (d*x)/2) - 24*a^10*b^3*sin(c/2 + (d*x)/2)
- 24*a^11*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(12*a*b^11 + 4*a^11*b + 4*a^12 + 12*a^2*b^10 + 15*a^3*
b^9 + 15*a^4*b^8 - 59*a^5*b^7 - 59*a^6*b^6 + 57*a^7*b^5 + 57*a^8*b^4 - 24*a^9*b^3 - 24*a^10*b^2)))*cos(c/2 + (
d*x)/2)^4)/(a*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) +
(2*atan((4*a^13*sin(c/2 + (d*x)/2) + 4*a^12*b*sin(c/2 + (d*x)/2) + 12*a^2*b^11*sin(c/2 + (d*x)/2) + 12*a^3*b^
10*sin(c/2 + (d*x)/2) + 15*a^4*b^9*sin(c/2 + (d*x)/2) + 15*a^5*b^8*sin(c/2 + (d*x)/2) - 59*a^6*b^7*sin(c/2 + (
d*x)/2) - 59*a^7*b^6*sin(c/2 + (d*x)/2) + 57*a^8*b^5*sin(c/2 + (d*x)/2) + 57*a^9*b^4*sin(c/2 + (d*x)/2) - 24*a
^10*b^3*sin(c/2 + (d*x)/2) - 24*a^11*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(12*a*b^11 + 4*a^11*b + 4*a
^12 + 12*a^2*b^10 + 15*a^3*b^9 + 15*a^4*b^8 - 59*a^5*b^7 - 59*a^6*b^6 + 57*a^7*b^5 + 57*a^8*b^4 - 24*a^9*b^3 -
24*a^10*b^2)))*sin(c/2 + (d*x)/2)^4)/(a*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)
^2*sin(c/2 + (d*x)/2)^2)) - (a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^4*2i)/(b^
3*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) - (a^2*atan((
sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*sin(c/2 + (d*x)/2)^4*2i)/(b^3*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (
d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) - (atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2*6i)/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 +
(d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (2*a*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^3)/(b^2*d*(cos(c/2 + (d*x)/2)^4
+ sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) - (2*a*cos(c/2 + (d*x)/2)^3*sin(c/2 +
(d*x)/2))/(b^2*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2))
- (4*atan((4*a^13*sin(c/2 + (d*x)/2) + 4*a^12*b*sin(c/2 + (d*x)/2) + 12*a^2*b^11*sin(c/2 + (d*x)/2) + 12*a^3*b
^10*sin(c/2 + (d*x)/2) + 15*a^4*b^9*sin(c/2 + (d*x)/2) + 15*a^5*b^8*sin(c/2 + (d*x)/2) - 59*a^6*b^7*sin(c/2 +
(d*x)/2) - 59*a^7*b^6*sin(c/2 + (d*x)/2) + 57*a^8*b^5*sin(c/2 + (d*x)/2) + 57*a^9*b^4*sin(c/2 + (d*x)/2) - 24*
a^10*b^3*sin(c/2 + (d*x)/2) - 24*a^11*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(12*a*b^11 + 4*a^11*b + 4*
a^12 + 12*a^2*b^10 + 15*a^3*b^9 + 15*a^4*b^8 - 59*a^5*b^7 - 59*a^6*b^6 + 57*a^7*b^5 + 57*a^8*b^4 - 24*a^9*b^3
- 24*a^10*b^2)))*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)/(a*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4
- 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(c/
2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2*4i)/(b^3*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x
)/2)^2*sin(c/2 + (d*x)/2)^2)) + (atan(((8*a^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 8
*a^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*b^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^
2*b^4 - 3*a^4*b^2)^(5/2) + 8*b^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 26*a^2*b^7*sin
(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 6*a^3*b^6*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^
4 - 3*a^4*b^2)^(3/2) + 21*a^4*b^5*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 9*a^5*b^4*sin
(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 12*a^6*b^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b
^4 - 3*a^4*b^2)^(3/2) - 20*a^7*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 22*a^2*b^13*
sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 14*a^3*b^12*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a
^2*b^4 - 3*a^4*b^2)^(1/2) + 36*a^4*b^11*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 24*a^5*
b^10*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 47*a^6*b^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 +
3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 79*a^7*b^8*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 49*a
^8*b^7*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 65*a^9*b^6*sin(c/2 + (d*x)/2)*(a^6 - b^6
+ 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 24*a^10*b^5*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 2
4*a^11*b^4*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 4*a^12*b^3*sin(c/2 + (d*x)/2)*(a^6 -
b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 4*a^13*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2)
- 8*a*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*a^2*b*sin(c/2 + (d*x)/2)*(a^6 - b^6
+ 3*a^2*b^4 - 3*a^4*b^2)^(5/2) - 8*a*b^8*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 8*a^8
*b*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2))*1i)/(a*b*cos(c/2 + (d*x)/2)*(12*a*b^15 - 24*a
^2*b^14 - 33*a^3*b^13 + 90*a^4*b^12 + 22*a^5*b^11 - 134*a^6*b^10 + 17*a^7*b^9 + 100*a^8*b^8 - 31*a^9*b^7 - 38*
a^10*b^6 + 16*a^11*b^5 + 6*a^12*b^4 - 3*a^13*b^3)))*cos(c/2 + (d*x)/2)^4*((a + b)^3*(a - b)^3)^(1/2)*2i)/(a*b^
3*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (atan(((8*a
^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 8*a^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*
b^4 - 3*a^4*b^2)^(5/2) + 8*b^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*b^9*sin(c/2 +
(d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 26*a^2*b^7*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*
a^4*b^2)^(3/2) - 6*a^3*b^6*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 21*a^4*b^5*sin(c/2 +
(d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 9*a^5*b^4*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*
a^4*b^2)^(3/2) + 12*a^6*b^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 20*a^7*b^2*sin(c/2
+ (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 22*a^2*b^13*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)^(1/2) + 14*a^3*b^12*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 36*a^4*b^11*sin
(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 24*a^5*b^10*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*
b^4 - 3*a^4*b^2)^(1/2) - 47*a^6*b^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 79*a^7*b^8*
sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 49*a^8*b^7*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^
2*b^4 - 3*a^4*b^2)^(1/2) + 65*a^9*b^6*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 24*a^10*b
^5*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 24*a^11*b^4*sin(c/2 + (d*x)/2)*(a^6 - b^6 +
3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 4*a^12*b^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 4*a^1
3*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 8*a*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3
*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*a^2*b*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) - 8*a*b^8*s
in(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 8*a^8*b*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^
4 - 3*a^4*b^2)^(3/2))*1i)/(a*b*cos(c/2 + (d*x)/2)*(12*a*b^15 - 24*a^2*b^14 - 33*a^3*b^13 + 90*a^4*b^12 + 22*a^
5*b^11 - 134*a^6*b^10 + 17*a^7*b^9 + 100*a^8*b^8 - 31*a^9*b^7 - 38*a^10*b^6 + 16*a^11*b^5 + 6*a^12*b^4 - 3*a^1
3*b^3)))*sin(c/2 + (d*x)/2)^4*((a + b)^3*(a - b)^3)^(1/2)*2i)/(a*b^3*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)
/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) - (atan(((8*a^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^
4 - 3*a^4*b^2)^(3/2) - 8*a^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*b^3*sin(c/2 + (d
*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*b^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2
)^(3/2) - 26*a^2*b^7*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 6*a^3*b^6*sin(c/2 + (d*x)/
2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 21*a^4*b^5*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^
2)^(3/2) + 9*a^5*b^4*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 12*a^6*b^3*sin(c/2 + (d*x)
/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 20*a^7*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b
^2)^(3/2) - 22*a^2*b^13*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 14*a^3*b^12*sin(c/2 + (
d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 36*a^4*b^11*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*
a^4*b^2)^(1/2) + 24*a^5*b^10*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 47*a^6*b^9*sin(c/2
+ (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 79*a^7*b^8*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)^(1/2) + 49*a^8*b^7*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 65*a^9*b^6*sin(c
/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 24*a^10*b^5*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^
4 - 3*a^4*b^2)^(1/2) - 24*a^11*b^4*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 4*a^12*b^3*s
in(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 4*a^13*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2
*b^4 - 3*a^4*b^2)^(1/2) - 8*a*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) + 8*a^2*b*sin(c
/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(5/2) - 8*a*b^8*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)^(3/2) - 8*a^8*b*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2))*1i)/(a*b*cos(c/2 + (d
*x)/2)*(12*a*b^15 - 24*a^2*b^14 - 33*a^3*b^13 + 90*a^4*b^12 + 22*a^5*b^11 - 134*a^6*b^10 + 17*a^7*b^9 + 100*a^
8*b^8 - 31*a^9*b^7 - 38*a^10*b^6 + 16*a^11*b^5 + 6*a^12*b^4 - 3*a^13*b^3)))*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*
x)/2)^2*((a + b)^3*(a - b)^3)^(1/2)*4i)/(a*b^3*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d
*x)/2)^2*sin(c/2 + (d*x)/2)^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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