3.561 \(\int \frac{\sec ^5(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=176 \[ \frac{3 a \sqrt{a^2+b^2} \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{b^4 d \sqrt{\sec ^2(c+d x)}}+\frac{3 \left (2 a^2+b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{2 b^4 d \sqrt{\sec ^2(c+d x)}}-\frac{3 \sec (c+d x) (2 a-b \tan (c+d x))}{2 b^3 d}-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.167884, antiderivative size = 176, normalized size of antiderivative = 1., 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, 815, 844, 215, 725, 206} \[ \frac{3 a \sqrt{a^2+b^2} \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{b^4 d \sqrt{\sec ^2(c+d x)}}+\frac{3 \left (2 a^2+b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{2 b^4 d \sqrt{\sec ^2(c+d x)}}-\frac{3 \sec (c+d x) (2 a-b \tan (c+d x))}{2 b^3 d}-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Tan[c + d*x])^2,x]

[Out]

(3*(2*a^2 + b^2)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(2*b^4*d*Sqrt[Sec[c + d*x]^2]) + (3*a*Sqrt[a^2 + b^2]*Arc
Tanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Sec[c + d*x])/(b^4*d*Sqrt[Sec[c + d*x]^2]) -
(3*Sec[c + d*x]*(2*a - b*Tan[c + d*x]))/(2*b^3*d) - Sec[c + d*x]^3/(b*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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 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))^2} \, dx &=\frac{\sec (c+d x) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{3/2}}{(a+x)^2} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))}+\frac{(3 \sec (c+d x)) \operatorname{Subst}\left (\int \frac{x \sqrt{1+\frac{x^2}{b^2}}}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{3 \sec (c+d x) (2 a-b \tan (c+d x))}{2 b^3 d}-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))}+\frac{(3 \sec (c+d x)) \operatorname{Subst}\left (\int \frac{-\frac{a}{b^2}+\frac{\left (2 a^2+b^2\right ) x}{b^4}}{(a+x) \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 b d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{3 \sec (c+d x) (2 a-b \tan (c+d x))}{2 b^3 d}-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))}-\frac{\left (3 a \left (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 )}{b^5 d \sqrt{\sec ^2(c+d x)}}+\frac{\left (3 \left (2 a^2+b^2\right ) \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 b^5 d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{3 \left (2 a^2+b^2\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{2 b^4 d \sqrt{\sec ^2(c+d x)}}-\frac{3 \sec (c+d x) (2 a-b \tan (c+d x))}{2 b^3 d}-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))}+\frac{\left (3 a \left (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 )}{b^5 d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{3 \left (2 a^2+b^2\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{2 b^4 d \sqrt{\sec ^2(c+d x)}}+\frac{3 a \sqrt{a^2+b^2} \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)}{b^4 d \sqrt{\sec ^2(c+d x)}}-\frac{3 \sec (c+d x) (2 a-b \tan (c+d x))}{2 b^3 d}-\frac{\sec ^3(c+d x)}{b d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.12165, size = 709, normalized size = 4.03 \[ -\frac{3 \left (2 a^2+b^2\right ) \sec ^2(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{2 b^4 d (a+b \tan (c+d x))^2}+\frac{3 \left (2 a^2+b^2\right ) \sec ^2(c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{2 b^4 d (a+b \tan (c+d x))^2}-\frac{6 a \sqrt{a^2+b^2} \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac{\sqrt{a^2+b^2} \left (a \sin \left (\frac{1}{2} (c+d x)\right )-b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \cos \left (\frac{1}{2} (c+d x)\right )+b^2 \cos \left (\frac{1}{2} (c+d x)\right )}\right )}{b^4 d (a+b \tan (c+d x))^2}-\frac{2 a \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}-\frac{2 a \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d (a+b \tan (c+d x))^2}+\frac{2 a \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}+\frac{\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{4 b^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}-\frac{\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{4 b^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}-\frac{(a-i b) (a+i b) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{b^3 d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Tan[c + d*x])^2,x]

[Out]

-(((a - I*b)*(a + I*b)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(b^3*d*(a + b*Tan[c + d*x])^2)) - (2*
a*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(a + b*Tan[c + d*x])^2) - (6*a*Sqrt[a^2 + b^2]*Ar
cTanh[(Sqrt[a^2 + b^2]*(-(b*Cos[(c + d*x)/2]) + a*Sin[(c + d*x)/2]))/(a^2*Cos[(c + d*x)/2] + b^2*Cos[(c + d*x)
/2])]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^4*d*(a + b*Tan[c + d*x])^2) - (3*(2*a^2 + b^2)*Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(2*b^4*d*(a + b*Tan
[c + d*x])^2) + (3*(2*a^2 + b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*S
in[c + d*x])^2)/(2*b^4*d*(a + b*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(4*b^2
*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a + b*Tan[c + d*x])^2) - (2*a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*C
os[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2) - (Sec[c
+ d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(4*b^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a + b*Tan[c +
d*x])^2) + (2*a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2)

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Maple [B]  time = 0.111, size = 440, normalized size = 2.5 \begin{align*} -{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-2\,{\frac{a}{d{b}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ){a}^{2}}{d{b}^{4}}}+{\frac{3}{2\,{b}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+2\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{{b}^{2}d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) a}}+2\,{\frac{{a}^{2}}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}+2\,{\frac{1}{bd \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}-6\,{\frac{\sqrt{{a}^{2}+{b}^{2}}a}{d{b}^{4}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+2\,{\frac{a}{d{b}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}+{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ){a}^{2}}{d{b}^{4}}}-{\frac{3}{2\,{b}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \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))^2,x)

[Out]

-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2-2/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)+3/d/b^4*
ln(tan(1/2*d*x+1/2*c)+1)*a^2+3/2/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)+2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+
1/2*c)*b-a)*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)/a*tan(1/2*d*x+1/2*c)+2/
d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)*a^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)-6/d/b^4*(a^2+b^2)^(1/2)*a*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))+1/2/d/b^2/(tan(1/2*d
*x+1/2*c)-1)^2+2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)-3/d/b^4*ln(tan(1/2*d*x+1/2*c)
-1)*a^2-3/2/d/b^2*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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.62475, size = 860, normalized size = 4.89 \begin{align*} -\frac{6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, b^{3} + 6 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (a^{2} \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )^{2} \sin \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 ) - 3 \,{\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \,{\left (a b^{4} d \cos \left (d x + c\right )^{3} + b^{5} d \cos \left (d x + c\right )^{2} \sin \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))^2,x, algorithm="fricas")

[Out]

-1/4*(6*a*b^2*cos(d*x + c)*sin(d*x + c) - 2*b^3 + 6*(2*a^2*b + b^3)*cos(d*x + c)^2 - 6*(a^2*cos(d*x + c)^3 + a
*b*cos(d*x + c)^2*sin(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)) - 3*((2*a^3 + a*b^2)*cos(d*x + c)^3 + (2*a^2*b + b^3)*cos(d*x + c)^2*sin(d*x
+ c))*log(sin(d*x + c) + 1) + 3*((2*a^3 + a*b^2)*cos(d*x + c)^3 + (2*a^2*b + b^3)*cos(d*x + c)^2*sin(d*x + c))
*log(-sin(d*x + c) + 1))/(a*b^4*d*cos(d*x + c)^3 + b^5*d*cos(d*x + c)^2*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.6443, size = 378, normalized size = 2.15 \begin{align*} \frac{\frac{3 \,{\left (2 \, a^{2} + b^{2}\right )} \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 ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac{6 \,{\left (a^{3} + a 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{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}} + \frac{4 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} + a 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 )} a 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))^2,x, algorithm="giac")

[Out]

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