3.573 \(\int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=148 \[ -\frac {3 \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 d \sqrt {a^2+b^2}}-\frac {3 a \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))}-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2} \]

[Out]

-3*a*arctanh(sin(d*x+c))/b^4/d-3/2*(2*a^2+b^2)*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/b^4/d/(a^2
+b^2)^(1/2)-1/2*sec(d*x+c)^3/b/d/(a+b*tan(d*x+c))^2+3/2*sec(d*x+c)*(2*a+b*tan(d*x+c))/b^3/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 189, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3512, 733, 813, 844, 215, 725, 206} \[ -\frac {3 \left (2 a^2+b^2\right ) \sec (c+d x) \tanh ^{-1}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right )}{2 b^4 d \sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))}-\frac {3 a \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{b^4 d \sqrt {\sec ^2(c+d x)}}-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Tan[c + d*x])^3,x]

[Out]

(-3*a*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(b^4*d*Sqrt[Sec[c + d*x]^2]) - (3*(2*a^2 + b^2)*ArcTanh[(b - a*Tan[c
+ d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Sec[c + d*x])/(2*b^4*Sqrt[a^2 + b^2]*d*Sqrt[Sec[c + d*x]^2])
- Sec[c + d*x]^3/(2*b*d*(a + b*Tan[c + d*x])^2) + (3*Sec[c + d*x]*(2*a + b*Tan[c + d*x]))/(2*b^3*d*(a + b*Tan[
c + d*x]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac {\sec (c+d x) \operatorname {Subst}\left (\int \frac {\left (1+\frac {x^2}{b^2}\right )^{3/2}}{(a+x)^3} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt {\sec ^2(c+d x)}}\\ &=-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {(3 \sec (c+d x)) \operatorname {Subst}\left (\int \frac {x \sqrt {1+\frac {x^2}{b^2}}}{(a+x)^2} \, dx,x,b \tan (c+d x)\right )}{2 b^3 d \sqrt {\sec ^2(c+d x)}}\\ &=-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))}-\frac {(3 \sec (c+d x)) \operatorname {Subst}\left (\int \frac {-2+\frac {4 a x}{b^2}}{(a+x) \sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{4 b^3 d \sqrt {\sec ^2(c+d x)}}\\ &=-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))}-\frac {(3 a \sec (c+d x)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{b^5 d \sqrt {\sec ^2(c+d x)}}+\frac {\left (3 \left (1+\frac {2 a^2}{b^2}\right ) \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{(a+x) \sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 b^3 d \sqrt {\sec ^2(c+d x)}}\\ &=-\frac {3 a \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{b^4 d \sqrt {\sec ^2(c+d x)}}-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))}-\frac {\left (3 \left (1+\frac {2 a^2}{b^2}\right ) \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a^2}{b^2}-x^2} \, dx,x,\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\sec ^2(c+d x)}}\right )}{2 b^3 d \sqrt {\sec ^2(c+d x)}}\\ &=-\frac {3 a \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{b^4 d \sqrt {\sec ^2(c+d x)}}-\frac {3 \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {b \left (1-\frac {a \tan (c+d x)}{b}\right )}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \sec (c+d x)}{2 b^4 \sqrt {a^2+b^2} d \sqrt {\sec ^2(c+d x)}}-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 2.45, size = 396, normalized size = 2.68 \[ \frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac {b^2 \left (a^2+b^2\right ) \sin (c+d x)}{a}+\frac {6 \left (2 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {2 b \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {2 b \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+2 b (a \cos (c+d x)+b \sin (c+d x))^2+\frac {b (2 a-b) (2 a+b) (a \cos (c+d x)+b \sin (c+d x))}{a}+6 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2-6 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2\right )}{2 b^4 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Tan[c + d*x])^3,x]

[Out]

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

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fricas [B]  time = 0.71, size = 513, normalized size = 3.47 \[ \frac {4 \, a^{2} b^{3} + 4 \, b^{5} + 6 \, {\left (2 \, a^{4} b + a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 18 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 6 \, {\left ({\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left ({\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \, {\left ({\left (a^{4} b^{4} - b^{8}\right )} d \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(4*a^2*b^3 + 4*b^5 + 6*(2*a^4*b + a^2*b^3 - b^5)*cos(d*x + c)^2 + 18*(a^3*b^2 + a*b^4)*cos(d*x + c)*sin(d*
x + c) + 3*((2*a^4 - a^2*b^2 - b^4)*cos(d*x + c)^3 + 2*(2*a^3*b + a*b^3)*cos(d*x + c)^2*sin(d*x + c) + (2*a^2*
b^2 + b^4)*cos(d*x + c))*sqrt(a^2 + b^2)*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 -
2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b
^2)*cos(d*x + c)^2 + b^2)) - 6*((a^5 - a*b^4)*cos(d*x + c)^3 + 2*(a^4*b + a^2*b^3)*cos(d*x + c)^2*sin(d*x + c)
+ (a^3*b^2 + a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((a^5 - a*b^4)*cos(d*x + c)^3 + 2*(a^4*b + a^2*b^
3)*cos(d*x + c)^2*sin(d*x + c) + (a^3*b^2 + a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1))/((a^4*b^4 - b^8)*d*co
s(d*x + c)^3 + 2*(a^3*b^5 + a*b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^2*b^6 + b^8)*d*cos(d*x + c))

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giac [B]  time = 3.69, size = 314, normalized size = 2.12 \[ -\frac {\frac {6 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {6 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {4}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b^{3}} + \frac {2 \, {\left (3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{4} + a^{2} b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2} a^{2} b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(6*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 6*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 + 3*(2*a^2 + b^2
)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2
+ b^2)))/(sqrt(a^2 + b^2)*b^4) + 4/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3) + 2*(3*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2*
a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 2*b^4*tan(1/2
*d*x + 1/2*c)^2 - 13*a^3*b*tan(1/2*d*x + 1/2*c) + 2*a*b^3*tan(1/2*d*x + 1/2*c) - 4*a^4 + a^2*b^2)/((a*tan(1/2*
d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2*a^2*b^3))/d

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maple [B]  time = 0.50, size = 611, normalized size = 4.13 \[ -\frac {3 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2} a}-\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}+\frac {9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2} a^{2}}+\frac {13 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 )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2} a}+\frac {4 a^{2}}{d \,b^{3} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}-\frac {1}{d b \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}+\frac {6 \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right ) a^{2}}{d \,b^{4} \sqrt {a^{2}+b^{2}}}+\frac {3 \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{4}}+\frac {1}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*tan(d*x+c))^3,x)

[Out]

-3/d/b^2/(a*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)*b-a)^2*a*tan(1/2*d*x+1/2*c)^3+2/d/(a*tan(1/2*d*x+1/2*c)^
2-2*tan(1/2*d*x+1/2*c)*b-a)^2/a*tan(1/2*d*x+1/2*c)^3-4/d/b^3/(a*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)*b-a)
^2*a^2*tan(1/2*d*x+1/2*c)^2+9/d/b/(a*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)*b-a)^2*tan(1/2*d*x+1/2*c)^2-2/d
*b/(a*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)*b-a)^2/a^2*tan(1/2*d*x+1/2*c)^2+13/d/b^2/(a*tan(1/2*d*x+1/2*c)
^2-2*tan(1/2*d*x+1/2*c)*b-a)^2*a*tan(1/2*d*x+1/2*c)-2/d/(a*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)*b-a)^2/a*
tan(1/2*d*x+1/2*c)+4/d/b^3/(a*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^2-1/d/b/(a*tan(1/2*d*x+1/2*c)
^2-2*tan(1/2*d*x+1/2*c)*b-a)^2+6/d/b^4/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2
))*a^2+3/d/b^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-1/d/b^3/(tan(1/2*d*x+
1/2*c)-1)+3/d*a/b^4*ln(tan(1/2*d*x+1/2*c)-1)+1/d/b^3/(tan(1/2*d*x+1/2*c)+1)-3/d*a/b^4*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.45, size = 518, normalized size = 3.50 \[ \frac {\frac {2 \, {\left (6 \, a^{4} - a^{2} b^{2} + \frac {{\left (21 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (6 \, a^{4} - 9 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (6 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (6 \, a^{4} - 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{4} b^{3} + \frac {4 \, a^{3} b^{4} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {8 \, a^{3} b^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {4 \, a^{3} b^{4} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{4} b^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {{\left (3 \, a^{4} b^{3} - 4 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (3 \, a^{4} b^{3} - 4 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {6 \, a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{4}} + \frac {6 \, a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{4}} - \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(2*(6*a^4 - a^2*b^2 + (21*a^3*b - 2*a*b^3)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(6*a^4 - 9*a^2*b^2 + b^4)*s
in(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*(6*a^3*b - a*b^3)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + (6*a^4 - 9*a^2*
b^2 + 2*b^4)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (3*a^3*b - 2*a*b^3)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a
^4*b^3 + 4*a^3*b^4*sin(d*x + c)/(cos(d*x + c) + 1) - 8*a^3*b^4*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 4*a^3*b^4
*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - a^4*b^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - (3*a^4*b^3 - 4*a^2*b^5)*s
in(d*x + c)^2/(cos(d*x + c) + 1)^2 + (3*a^4*b^3 - 4*a^2*b^5)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 6*a*log(si
n(d*x + c)/(cos(d*x + c) + 1) + 1)/b^4 + 6*a*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^4 - 3*(2*a^2 + b^2)*lo
g((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2
+ b^2)))/(sqrt(a^2 + b^2)*b^4))/d

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mupad [B]  time = 5.68, size = 1311, normalized size = 8.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b*tan(c + d*x))^3),x)

[Out]

((6*a^2 - b^2)/b^3 - (2*tan(c/2 + (d*x)/2)^2*(6*a^4 + b^4 - 9*a^2*b^2))/(a^2*b^3) + (tan(c/2 + (d*x)/2)*(21*a^
2 - 2*b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^4*(6*a^4 + 2*b^4 - 9*a^2*b^2))/(a^2*b^3) - (4*tan(c/2 + (d*x)/2)^3*(
6*a^2 - b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^5*(3*a^2 - 2*b^2))/(a*b^2))/(d*(tan(c/2 + (d*x)/2)^4*(3*a^2 - 4*b^
2) - tan(c/2 + (d*x)/2)^2*(3*a^2 - 4*b^2) - a^2*tan(c/2 + (d*x)/2)^6 + a^2 - 8*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*
b*tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2))) - (6*a*atanh(tan(c/2 + (d*x)/2)))/(b^4*d) + (atan((((2*a^2
+ b^2)*(a^2 + b^2)^(1/2)*((288*a^4)/b^5 + (8*tan(c/2 + (d*x)/2)*(9*a*b^7 + 108*a^3*b^5 + 72*a^5*b^3))/b^9 - (
3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*((8*tan(c/2 + (d*x)/2)*(12*a*b^10 + 24*a^3*b^8))/b^9 - 48*a^2 + (3*(2*a^2 +
b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 + 8*a^3*b^11))/b^9))/(2*(b^6 + a^2*b^4))
))/(2*(b^6 + a^2*b^4)))*3i)/(2*(b^6 + a^2*b^4)) + ((2*a^2 + b^2)*(a^2 + b^2)^(1/2)*((288*a^4)/b^5 + (8*tan(c/2
+ (d*x)/2)*(9*a*b^7 + 108*a^3*b^5 + 72*a^5*b^3))/b^9 - (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(48*a^2 - (8*tan(c/
2 + (d*x)/2)*(12*a*b^10 + 24*a^3*b^8))/b^9 + (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*
x)/2)*(12*a*b^13 + 8*a^3*b^11))/b^9))/(2*(b^6 + a^2*b^4))))/(2*(b^6 + a^2*b^4)))*3i)/(2*(b^6 + a^2*b^4)))/((16
*(54*a^4 + 27*a^2*b^2))/b^8 - (16*tan(c/2 + (d*x)/2)*(216*a^5 + 108*a^3*b^2))/b^9 - (3*(2*a^2 + b^2)*(a^2 + b^
2)^(1/2)*((288*a^4)/b^5 + (8*tan(c/2 + (d*x)/2)*(9*a*b^7 + 108*a^3*b^5 + 72*a^5*b^3))/b^9 - (3*(2*a^2 + b^2)*(
a^2 + b^2)^(1/2)*((8*tan(c/2 + (d*x)/2)*(12*a*b^10 + 24*a^3*b^8))/b^9 - 48*a^2 + (3*(2*a^2 + b^2)*(a^2 + b^2)^
(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 + 8*a^3*b^11))/b^9))/(2*(b^6 + a^2*b^4))))/(2*(b^6 + a^2*
b^4))))/(2*(b^6 + a^2*b^4)) + (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*((288*a^4)/b^5 + (8*tan(c/2 + (d*x)/2)*(9*a*b
^7 + 108*a^3*b^5 + 72*a^5*b^3))/b^9 - (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(48*a^2 - (8*tan(c/2 + (d*x)/2)*(12*a
*b^10 + 24*a^3*b^8))/b^9 + (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 +
8*a^3*b^11))/b^9))/(2*(b^6 + a^2*b^4))))/(2*(b^6 + a^2*b^4))))/(2*(b^6 + a^2*b^4))))*(2*a^2 + b^2)*(a^2 + b^2
)^(1/2)*3i)/(d*(b^6 + a^2*b^4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**5/(a + b*tan(c + d*x))**3, x)

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