3.263 \(\int (a \sec (x)+b \tan (x))^5 \, dx\)

Optimal. Leaf size=149 \[ -\frac{1}{8} a b^4 \left (7-\frac{3 a^2}{b^2}\right ) \sin (x)-\frac{1}{16} (a+b)^3 \left (3 a^2-9 a b+8 b^2\right ) \log (1-\sin (x))+\frac{1}{16} (a-b)^3 \left (3 a^2+9 a b+8 b^2\right ) \log (\sin (x)+1)+\frac{1}{8} \sec ^2(x) (a+b \sin (x))^2 \left (a \left (3 a^2-5 b^2\right ) \sin (x)+2 b \left (a^2-2 b^2\right )\right )+\frac{1}{4} \sec ^4(x) (a \sin (x)+b) (a+b \sin (x))^4 \]

[Out]

-((a + b)^3*(3*a^2 - 9*a*b + 8*b^2)*Log[1 - Sin[x]])/16 + ((a - b)^3*(3*a^2 + 9*a*b + 8*b^2)*Log[1 + Sin[x]])/
16 - (a*(7 - (3*a^2)/b^2)*b^4*Sin[x])/8 + (Sec[x]^4*(b + a*Sin[x])*(a + b*Sin[x])^4)/4 + (Sec[x]^2*(a + b*Sin[
x])^2*(2*b*(a^2 - 2*b^2) + a*(3*a^2 - 5*b^2)*Sin[x]))/8

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Rubi [A]  time = 0.211375, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {4391, 2668, 739, 819, 774, 633, 31} \[ -\frac{1}{8} a b^4 \left (7-\frac{3 a^2}{b^2}\right ) \sin (x)-\frac{1}{16} (a+b)^3 \left (3 a^2-9 a b+8 b^2\right ) \log (1-\sin (x))+\frac{1}{16} (a-b)^3 \left (3 a^2+9 a b+8 b^2\right ) \log (\sin (x)+1)+\frac{1}{8} \sec ^2(x) (a+b \sin (x))^2 \left (a \left (3 a^2-5 b^2\right ) \sin (x)+2 b \left (a^2-2 b^2\right )\right )+\frac{1}{4} \sec ^4(x) (a \sin (x)+b) (a+b \sin (x))^4 \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x] + b*Tan[x])^5,x]

[Out]

-((a + b)^3*(3*a^2 - 9*a*b + 8*b^2)*Log[1 - Sin[x]])/16 + ((a - b)^3*(3*a^2 + 9*a*b + 8*b^2)*Log[1 + Sin[x]])/
16 - (a*(7 - (3*a^2)/b^2)*b^4*Sin[x])/8 + (Sec[x]^4*(b + a*Sin[x])*(a + b*Sin[x])^4)/4 + (Sec[x]^2*(a + b*Sin[
x])^2*(2*b*(a^2 - 2*b^2) + a*(3*a^2 - 5*b^2)*Sin[x]))/8

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int (a \sec (x)+b \tan (x))^5 \, dx &=\int \sec ^5(x) (a+b \sin (x))^5 \, dx\\ &=b^5 \operatorname{Subst}\left (\int \frac{(a+x)^5}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (x)\right )\\ &=\frac{1}{4} \sec ^4(x) (b+a \sin (x)) (a+b \sin (x))^4-\frac{1}{4} b^3 \operatorname{Subst}\left (\int \frac{(a+x)^3 \left (-3 a^2+4 b^2+a x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (x)\right )\\ &=\frac{1}{4} \sec ^4(x) (b+a \sin (x)) (a+b \sin (x))^4+\frac{1}{8} \sec ^2(x) (a+b \sin (x))^2 \left (2 b \left (a^2-2 b^2\right )+a \left (3 a^2-5 b^2\right ) \sin (x)\right )+\frac{1}{8} b \operatorname{Subst}\left (\int \frac{(a+x) \left (3 a^4-7 a^2 b^2+8 b^4-a \left (3 a^2-7 b^2\right ) x\right )}{b^2-x^2} \, dx,x,b \sin (x)\right )\\ &=\frac{1}{8} a b^2 \left (3 a^2-7 b^2\right ) \sin (x)+\frac{1}{4} \sec ^4(x) (b+a \sin (x)) (a+b \sin (x))^4+\frac{1}{8} \sec ^2(x) (a+b \sin (x))^2 \left (2 b \left (a^2-2 b^2\right )+a \left (3 a^2-5 b^2\right ) \sin (x)\right )-\frac{1}{8} b \operatorname{Subst}\left (\int \frac{a b^2 \left (3 a^2-7 b^2\right )-a \left (3 a^4-7 a^2 b^2+8 b^4\right )-\left (3 a^4-7 a^2 b^2+8 b^4-a^2 \left (3 a^2-7 b^2\right )\right ) x}{b^2-x^2} \, dx,x,b \sin (x)\right )\\ &=\frac{1}{8} a b^2 \left (3 a^2-7 b^2\right ) \sin (x)+\frac{1}{4} \sec ^4(x) (b+a \sin (x)) (a+b \sin (x))^4+\frac{1}{8} \sec ^2(x) (a+b \sin (x))^2 \left (2 b \left (a^2-2 b^2\right )+a \left (3 a^2-5 b^2\right ) \sin (x)\right )+\frac{1}{16} \left ((a+b)^3 \left (3 a^2-9 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (x)\right )-\frac{1}{16} \left ((a-b)^3 \left (3 a^2+9 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (x)\right )\\ &=-\frac{1}{16} (a+b)^3 \left (3 a^2-9 a b+8 b^2\right ) \log (1-\sin (x))+\frac{1}{16} (a-b)^3 \left (3 a^2+9 a b+8 b^2\right ) \log (1+\sin (x))+\frac{1}{8} a b^2 \left (3 a^2-7 b^2\right ) \sin (x)+\frac{1}{4} \sec ^4(x) (b+a \sin (x)) (a+b \sin (x))^4+\frac{1}{8} \sec ^2(x) (a+b \sin (x))^2 \left (2 b \left (a^2-2 b^2\right )+a \left (3 a^2-5 b^2\right ) \sin (x)\right )\\ \end{align*}

Mathematica [B]  time = 1.22735, size = 303, normalized size = 2.03 \[ -\frac{-2 a b^6 \left (3 a^2+5 b^2\right ) \sin ^5(x)+4 b^5 \left (-12 a^2 b^2-9 a^4+b^4\right ) \sin ^4(x)-10 a b^4 \left (8 a^2 b^2+9 a^4-b^4\right ) \sin ^3(x)+8 b^3 \left (-4 a^4 b^2-2 a^2 b^4-15 a^6+b^6\right ) \sin ^2(x)-10 a b^2 \left (-6 a^4 b^2+8 a^2 b^4+9 a^6-3 b^6\right ) \sin (x)+\left (a^2-b^2\right )^2 \left ((a+b)^3 \left (3 a^2-9 a b+8 b^2\right ) \log (1-\sin (x))-(a-b)^3 \left (3 a^2+9 a b+8 b^2\right ) \log (\sin (x)+1)\right )+4 \left (b^2-a^2\right ) \sec ^4(x) (a \sin (x)-b) (a+b \sin (x))^6+2 \sec ^2(x) (a+b \sin (x))^6 \left (6 a^2 b-3 a^3 \sin (x)-5 a b^2 \sin (x)+2 b^3\right )}{16 \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x] + b*Tan[x])^5,x]

[Out]

-((a^2 - b^2)^2*((a + b)^3*(3*a^2 - 9*a*b + 8*b^2)*Log[1 - Sin[x]] - (a - b)^3*(3*a^2 + 9*a*b + 8*b^2)*Log[1 +
Sin[x]]) - 10*a*b^2*(9*a^6 - 6*a^4*b^2 + 8*a^2*b^4 - 3*b^6)*Sin[x] + 8*b^3*(-15*a^6 - 4*a^4*b^2 - 2*a^2*b^4 +
b^6)*Sin[x]^2 - 10*a*b^4*(9*a^4 + 8*a^2*b^2 - b^4)*Sin[x]^3 + 4*b^5*(-9*a^4 - 12*a^2*b^2 + b^4)*Sin[x]^4 - 2*
a*b^6*(3*a^2 + 5*b^2)*Sin[x]^5 + 4*(-a^2 + b^2)*Sec[x]^4*(-b + a*Sin[x])*(a + b*Sin[x])^6 + 2*Sec[x]^2*(a + b*
Sin[x])^6*(6*a^2*b + 2*b^3 - 3*a^3*Sin[x] - 5*a*b^2*Sin[x]))/(16*(a^2 - b^2)^2)

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Maple [A]  time = 0.075, size = 199, normalized size = 1.3 \begin{align*}{\frac{{a}^{5}\tan \left ( x \right ) \left ( \sec \left ( x \right ) \right ) ^{3}}{4}}+{\frac{3\,{a}^{5}\sec \left ( x \right ) \tan \left ( x \right ) }{8}}+{\frac{3\,{a}^{5}\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) }{8}}+{\frac{5\,{a}^{4}b}{4\, \left ( \cos \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{a}^{3}{b}^{2} \left ( \sin \left ( x \right ) \right ) ^{3}}{2\, \left ( \cos \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{a}^{3}{b}^{2} \left ( \sin \left ( x \right ) \right ) ^{3}}{4\, \left ( \cos \left ( x \right ) \right ) ^{2}}}+{\frac{5\,\sin \left ( x \right ){a}^{3}{b}^{2}}{4}}-{\frac{5\,{a}^{3}{b}^{2}\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) }{4}}+{\frac{5\,{a}^{2}{b}^{3} \left ( \sin \left ( x \right ) \right ) ^{4}}{2\, \left ( \cos \left ( x \right ) \right ) ^{4}}}+{\frac{5\,a{b}^{4} \left ( \sin \left ( x \right ) \right ) ^{5}}{4\, \left ( \cos \left ( x \right ) \right ) ^{4}}}-{\frac{5\,a{b}^{4} \left ( \sin \left ( x \right ) \right ) ^{5}}{8\, \left ( \cos \left ( x \right ) \right ) ^{2}}}-{\frac{5\,a{b}^{4} \left ( \sin \left ( x \right ) \right ) ^{3}}{8}}-{\frac{15\,\sin \left ( x \right ) a{b}^{4}}{8}}+{\frac{15\,a{b}^{4}\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) }{8}}+{\frac{{b}^{5} \left ( \tan \left ( x \right ) \right ) ^{4}}{4}}-{\frac{{b}^{5} \left ( \tan \left ( x \right ) \right ) ^{2}}{2}}-{b}^{5}\ln \left ( \cos \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)+b*tan(x))^5,x)

[Out]

1/4*a^5*tan(x)*sec(x)^3+3/8*a^5*sec(x)*tan(x)+3/8*a^5*ln(sec(x)+tan(x))+5/4*a^4*b/cos(x)^4+5/2*a^3*b^2*sin(x)^
3/cos(x)^4+5/4*a^3*b^2*sin(x)^3/cos(x)^2+5/4*sin(x)*a^3*b^2-5/4*a^3*b^2*ln(sec(x)+tan(x))+5/2*a^2*b^3*sin(x)^4
/cos(x)^4+5/4*a*b^4*sin(x)^5/cos(x)^4-5/8*a*b^4*sin(x)^5/cos(x)^2-5/8*a*b^4*sin(x)^3-15/8*sin(x)*a*b^4+15/8*a*
b^4*ln(sec(x)+tan(x))+1/4*b^5*tan(x)^4-1/2*b^5*tan(x)^2-b^5*ln(cos(x))

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Maxima [A]  time = 1.10693, size = 275, normalized size = 1.85 \begin{align*} \frac{5}{2} \, a^{2} b^{3} \tan \left (x\right )^{4} + \frac{5}{16} \, a b^{4}{\left (\frac{2 \,{\left (5 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )\right )}}{\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1} + 3 \, \log \left (\sin \left (x\right ) + 1\right ) - 3 \, \log \left (\sin \left (x\right ) - 1\right )\right )} - \frac{1}{16} \, a^{5}{\left (\frac{2 \,{\left (3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )\right )}}{\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1} - 3 \, \log \left (\sin \left (x\right ) + 1\right ) + 3 \, \log \left (\sin \left (x\right ) - 1\right )\right )} + \frac{5}{8} \, a^{3} b^{2}{\left (\frac{2 \,{\left (\sin \left (x\right )^{3} + \sin \left (x\right )\right )}}{\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1} - \log \left (\sin \left (x\right ) + 1\right ) + \log \left (\sin \left (x\right ) - 1\right )\right )} + \frac{1}{4} \, b^{5}{\left (\frac{4 \, \sin \left (x\right )^{2} - 3}{\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1} - 2 \, \log \left (\sin \left (x\right )^{2} - 1\right )\right )} + \frac{5 \, a^{4} b}{4 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*a^2*b^3*tan(x)^4 + 5/16*a*b^4*(2*(5*sin(x)^3 - 3*sin(x))/(sin(x)^4 - 2*sin(x)^2 + 1) + 3*log(sin(x) + 1) -
3*log(sin(x) - 1)) - 1/16*a^5*(2*(3*sin(x)^3 - 5*sin(x))/(sin(x)^4 - 2*sin(x)^2 + 1) - 3*log(sin(x) + 1) + 3*
log(sin(x) - 1)) + 5/8*a^3*b^2*(2*(sin(x)^3 + sin(x))/(sin(x)^4 - 2*sin(x)^2 + 1) - log(sin(x) + 1) + log(sin(
x) - 1)) + 1/4*b^5*((4*sin(x)^2 - 3)/(sin(x)^4 - 2*sin(x)^2 + 1) - 2*log(sin(x)^2 - 1)) + 5/4*a^4*b/(sin(x)^2
- 1)^2

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Fricas [A]  time = 2.39677, size = 405, normalized size = 2.72 \begin{align*} \frac{{\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} - 8 \, b^{5}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) -{\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, b^{5}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 20 \, a^{4} b + 40 \, a^{2} b^{3} + 4 \, b^{5} - 16 \,{\left (5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2} + 2 \,{\left (2 \, a^{5} + 20 \, a^{3} b^{2} + 10 \, a b^{4} +{\left (3 \, a^{5} - 10 \, a^{3} b^{2} - 25 \, a b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{16 \, \cos \left (x\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((3*a^5 - 10*a^3*b^2 + 15*a*b^4 - 8*b^5)*cos(x)^4*log(sin(x) + 1) - (3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^
5)*cos(x)^4*log(-sin(x) + 1) + 20*a^4*b + 40*a^2*b^3 + 4*b^5 - 16*(5*a^2*b^3 + b^5)*cos(x)^2 + 2*(2*a^5 + 20*a
^3*b^2 + 10*a*b^4 + (3*a^5 - 10*a^3*b^2 - 25*a*b^4)*cos(x)^2)*sin(x))/cos(x)^4

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Sympy [B]  time = 8.31164, size = 308, normalized size = 2.07 \begin{align*} - \frac{3 a^{5} \log{\left (\sin{\left (x \right )} - 1 \right )}}{16} + \frac{3 a^{5} \log{\left (\sin{\left (x \right )} + 1 \right )}}{16} - \frac{3 a^{5} \sin ^{3}{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac{5 a^{5} \sin{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac{5 a^{4} b \sec ^{4}{\left (x \right )}}{4} + \frac{5 a^{3} b^{2} \log{\left (\sin{\left (x \right )} - 1 \right )}}{8} - \frac{5 a^{3} b^{2} \log{\left (\sin{\left (x \right )} + 1 \right )}}{8} + \frac{10 a^{3} b^{2} \sin ^{3}{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac{10 a^{3} b^{2} \sin{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac{5 a^{2} b^{3} \tan ^{4}{\left (x \right )}}{2} - \frac{15 a b^{4} \log{\left (\sin{\left (x \right )} - 1 \right )}}{16} + \frac{15 a b^{4} \log{\left (\sin{\left (x \right )} + 1 \right )}}{16} + \frac{25 a b^{4} \sin ^{3}{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} - \frac{15 a b^{4} \sin{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac{b^{5} \log{\left (\sec ^{2}{\left (x \right )} \right )}}{2} + \frac{b^{5} \sec ^{4}{\left (x \right )}}{4} - b^{5} \sec ^{2}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)+b*tan(x))**5,x)

[Out]

-3*a**5*log(sin(x) - 1)/16 + 3*a**5*log(sin(x) + 1)/16 - 3*a**5*sin(x)**3/(8*sin(x)**4 - 16*sin(x)**2 + 8) + 5
*a**5*sin(x)/(8*sin(x)**4 - 16*sin(x)**2 + 8) + 5*a**4*b*sec(x)**4/4 + 5*a**3*b**2*log(sin(x) - 1)/8 - 5*a**3*
b**2*log(sin(x) + 1)/8 + 10*a**3*b**2*sin(x)**3/(8*sin(x)**4 - 16*sin(x)**2 + 8) + 10*a**3*b**2*sin(x)/(8*sin(
x)**4 - 16*sin(x)**2 + 8) + 5*a**2*b**3*tan(x)**4/2 - 15*a*b**4*log(sin(x) - 1)/16 + 15*a*b**4*log(sin(x) + 1)
/16 + 25*a*b**4*sin(x)**3/(8*sin(x)**4 - 16*sin(x)**2 + 8) - 15*a*b**4*sin(x)/(8*sin(x)**4 - 16*sin(x)**2 + 8)
+ b**5*log(sec(x)**2)/2 + b**5*sec(x)**4/4 - b**5*sec(x)**2

________________________________________________________________________________________

Giac [A]  time = 1.11954, size = 240, normalized size = 1.61 \begin{align*} \frac{1}{16} \,{\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} - 8 \, b^{5}\right )} \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{16} \,{\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, b^{5}\right )} \log \left (-\sin \left (x\right ) + 1\right ) + \frac{6 \, b^{5} \sin \left (x\right )^{4} - 3 \, a^{5} \sin \left (x\right )^{3} + 10 \, a^{3} b^{2} \sin \left (x\right )^{3} + 25 \, a b^{4} \sin \left (x\right )^{3} + 40 \, a^{2} b^{3} \sin \left (x\right )^{2} - 4 \, b^{5} \sin \left (x\right )^{2} + 5 \, a^{5} \sin \left (x\right ) + 10 \, a^{3} b^{2} \sin \left (x\right ) - 15 \, a b^{4} \sin \left (x\right ) + 10 \, a^{4} b - 20 \, a^{2} b^{3}}{8 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 - 8*b^5)*log(sin(x) + 1) - 1/16*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*lo
g(-sin(x) + 1) + 1/8*(6*b^5*sin(x)^4 - 3*a^5*sin(x)^3 + 10*a^3*b^2*sin(x)^3 + 25*a*b^4*sin(x)^3 + 40*a^2*b^3*s
in(x)^2 - 4*b^5*sin(x)^2 + 5*a^5*sin(x) + 10*a^3*b^2*sin(x) - 15*a*b^4*sin(x) + 10*a^4*b - 20*a^2*b^3)/(sin(x)
^2 - 1)^2