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

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Rubi [A]  time = 0.160246, 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 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]

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 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 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])

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*}

Mathematica [B]  time = 2.34927, 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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Maple [B]  time = 0.127, size = 611, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.53068, size = 1175, normalized size = 7.94 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

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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Giac [B]  time = 1.71835, size = 424, normalized size = 2.86 \begin{align*} -\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} \end{align*}

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