3.12 \(\int \frac{\tan ^5(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=178 \[ \frac{b^6 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^3}-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\csc (x))}{16 (a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\csc (x)+1)}{16 (a-b)^3}+\frac{5 a+7 b}{16 (a+b)^2 (1-\csc (x))}+\frac{5 a-7 b}{16 (a-b)^2 (\csc (x)+1)}+\frac{1}{16 (a+b) (1-\csc (x))^2}+\frac{1}{16 (a-b) (\csc (x)+1)^2}-\frac{\log (\sin (x))}{a} \]

[Out]

1/(16*(a + b)*(1 - Csc[x])^2) + (5*a + 7*b)/(16*(a + b)^2*(1 - Csc[x])) + 1/(16*(a - b)*(1 + Csc[x])^2) + (5*a
- 7*b)/(16*(a - b)^2*(1 + Csc[x])) - ((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Csc[x]])/(16*(a + b)^3) - ((8*a^2 - 2
1*a*b + 15*b^2)*Log[1 + Csc[x]])/(16*(a - b)^3) + (b^6*Log[a + b*Csc[x]])/(a*(a^2 - b^2)^3) - Log[Sin[x]]/a

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Rubi [A]  time = 0.271979, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3885, 894} \[ \frac{b^6 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^3}-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\csc (x))}{16 (a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\csc (x)+1)}{16 (a-b)^3}+\frac{5 a+7 b}{16 (a+b)^2 (1-\csc (x))}+\frac{5 a-7 b}{16 (a-b)^2 (\csc (x)+1)}+\frac{1}{16 (a+b) (1-\csc (x))^2}+\frac{1}{16 (a-b) (\csc (x)+1)^2}-\frac{\log (\sin (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^5/(a + b*Csc[x]),x]

[Out]

1/(16*(a + b)*(1 - Csc[x])^2) + (5*a + 7*b)/(16*(a + b)^2*(1 - Csc[x])) + 1/(16*(a - b)*(1 + Csc[x])^2) + (5*a
- 7*b)/(16*(a - b)^2*(1 + Csc[x])) - ((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Csc[x]])/(16*(a + b)^3) - ((8*a^2 - 2
1*a*b + 15*b^2)*Log[1 + Csc[x]])/(16*(a - b)^3) + (b^6*Log[a + b*Csc[x]])/(a*(a^2 - b^2)^3) - Log[Sin[x]]/a

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 894

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

Rubi steps

\begin{align*} \int \frac{\tan ^5(x)}{a+b \csc (x)} \, dx &=b^6 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \csc (x)\right )\\ &=b^6 \operatorname{Subst}\left (\int \left (\frac{1}{8 b^4 (a+b) (b-x)^3}+\frac{5 a+7 b}{16 b^5 (a+b)^2 (b-x)^2}+\frac{8 a^2+21 a b+15 b^2}{16 b^6 (a+b)^3 (b-x)}+\frac{1}{a b^6 x}+\frac{1}{a (a-b)^3 (a+b)^3 (a+x)}+\frac{1}{8 b^4 (-a+b) (b+x)^3}+\frac{-5 a+7 b}{16 (a-b)^2 b^5 (b+x)^2}+\frac{8 a^2-21 a b+15 b^2}{16 b^6 (-a+b)^3 (b+x)}\right ) \, dx,x,b \csc (x)\right )\\ &=\frac{1}{16 (a+b) (1-\csc (x))^2}+\frac{5 a+7 b}{16 (a+b)^2 (1-\csc (x))}+\frac{1}{16 (a-b) (1+\csc (x))^2}+\frac{5 a-7 b}{16 (a-b)^2 (1+\csc (x))}-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\csc (x))}{16 (a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (1+\csc (x))}{16 (a-b)^3}+\frac{b^6 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^3}-\frac{\log (\sin (x))}{a}\\ \end{align*}

Mathematica [A]  time = 0.75652, size = 172, normalized size = 0.97 \[ \frac{\csc (x) (a \sin (x)+b) \left (-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (x))}{(a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (x)+1)}{(a-b)^3}+\frac{16 b^6 \log (a \sin (x)+b)}{a (a-b)^3 (a+b)^3}+\frac{7 a+9 b}{(a+b)^2 (\sin (x)-1)}+\frac{9 b-7 a}{(a-b)^2 (\sin (x)+1)}+\frac{1}{(a+b) (\sin (x)-1)^2}+\frac{1}{(a-b) (\sin (x)+1)^2}\right )}{16 (a+b \csc (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^5/(a + b*Csc[x]),x]

[Out]

(Csc[x]*(b + a*Sin[x])*(-(((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[x]])/(a + b)^3) - ((8*a^2 - 21*a*b + 15*b^2)*
Log[1 + Sin[x]])/(a - b)^3 + (16*b^6*Log[b + a*Sin[x]])/(a*(a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[x])^2)
+ (7*a + 9*b)/((a + b)^2*(-1 + Sin[x])) + 1/((a - b)*(1 + Sin[x])^2) + (-7*a + 9*b)/((a - b)^2*(1 + Sin[x]))))
/(16*(a + b*Csc[x]))

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Maple [A]  time = 0.064, size = 216, normalized size = 1.2 \begin{align*}{\frac{{b}^{6}\ln \left ( b+a\sin \left ( x \right ) \right ) }{ \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}a}}+{\frac{1}{ \left ( 16\,a-16\,b \right ) \left ( \sin \left ( x \right ) +1 \right ) ^{2}}}-{\frac{7\,a}{16\, \left ( a-b \right ) ^{2} \left ( \sin \left ( x \right ) +1 \right ) }}+{\frac{9\,b}{16\, \left ( a-b \right ) ^{2} \left ( \sin \left ( x \right ) +1 \right ) }}-{\frac{\ln \left ( \sin \left ( x \right ) +1 \right ){a}^{2}}{2\, \left ( a-b \right ) ^{3}}}+{\frac{21\,\ln \left ( \sin \left ( x \right ) +1 \right ) ab}{16\, \left ( a-b \right ) ^{3}}}-{\frac{15\,\ln \left ( \sin \left ( x \right ) +1 \right ){b}^{2}}{16\, \left ( a-b \right ) ^{3}}}+{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( \sin \left ( x \right ) -1 \right ) ^{2}}}+{\frac{7\,a}{16\, \left ( a+b \right ) ^{2} \left ( \sin \left ( x \right ) -1 \right ) }}+{\frac{9\,b}{16\, \left ( a+b \right ) ^{2} \left ( \sin \left ( x \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ){a}^{2}}{2\, \left ( a+b \right ) ^{3}}}-{\frac{21\,\ln \left ( \sin \left ( x \right ) -1 \right ) ab}{16\, \left ( a+b \right ) ^{3}}}-{\frac{15\,\ln \left ( \sin \left ( x \right ) -1 \right ){b}^{2}}{16\, \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^6/(a+b)^3/(a-b)^3/a*ln(b+a*sin(x))+1/2/(8*a-8*b)/(sin(x)+1)^2-7/16/(a-b)^2/(sin(x)+1)*a+9/16/(a-b)^2/(sin(x)
+1)*b-1/2/(a-b)^3*ln(sin(x)+1)*a^2+21/16/(a-b)^3*ln(sin(x)+1)*a*b-15/16/(a-b)^3*ln(sin(x)+1)*b^2+1/2/(8*a+8*b)
/(sin(x)-1)^2+7/16/(a+b)^2/(sin(x)-1)*a+9/16/(a+b)^2/(sin(x)-1)*b-1/2/(a+b)^3*ln(sin(x)-1)*a^2-21/16/(a+b)^3*l
n(sin(x)-1)*a*b-15/16/(a+b)^3*ln(sin(x)-1)*b^2

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Maxima [A]  time = 1.18709, size = 339, normalized size = 1.9 \begin{align*} \frac{b^{6} \log \left (a \sin \left (x\right ) + b\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac{{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac{{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) - 1\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{{\left (5 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (x\right )^{3} + 6 \, a^{3} - 10 \, a b^{2} - 4 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (x\right )^{2} -{\left (3 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (x\right )}{8 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^6*log(a*sin(x) + b)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - 1/16*(8*a^2 - 21*a*b + 15*b^2)*log(sin(x) + 1)/(
a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/16*(8*a^2 + 21*a*b + 15*b^2)*log(sin(x) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3
) - 1/8*((5*a^2*b - 9*b^3)*sin(x)^3 + 6*a^3 - 10*a*b^2 - 4*(2*a^3 - 3*a*b^2)*sin(x)^2 - (3*a^2*b - 7*b^3)*sin(
x))/((a^4 - 2*a^2*b^2 + b^4)*sin(x)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(x)^2)

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Fricas [A]  time = 1.01934, size = 602, normalized size = 3.38 \begin{align*} \frac{16 \, b^{6} \cos \left (x\right )^{4} \log \left (a \sin \left (x\right ) + b\right ) + 4 \, a^{6} - 8 \, a^{4} b^{2} + 4 \, a^{2} b^{4} -{\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) -{\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) - 8 \,{\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2} - 2 \,{\left (2 \, a^{5} b - 4 \, a^{3} b^{3} + 2 \, a b^{5} -{\left (5 \, a^{5} b - 14 \, a^{3} b^{3} + 9 \, a b^{5}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{16 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (x\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^6*cos(x)^4*log(a*sin(x) + b) + 4*a^6 - 8*a^4*b^2 + 4*a^2*b^4 - (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a
^3*b^3 + 24*a^2*b^4 + 15*a*b^5)*cos(x)^4*log(sin(x) + 1) - (8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2
*b^4 - 15*a*b^5)*cos(x)^4*log(-sin(x) + 1) - 8*(2*a^6 - 5*a^4*b^2 + 3*a^2*b^4)*cos(x)^2 - 2*(2*a^5*b - 4*a^3*b
^3 + 2*a*b^5 - (5*a^5*b - 14*a^3*b^3 + 9*a*b^5)*cos(x)^2)*sin(x))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(x
)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**5/(a+b*csc(x)),x)

[Out]

Integral(tan(x)**5/(a + b*csc(x)), x)

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Giac [A]  time = 1.33538, size = 340, normalized size = 1.91 \begin{align*} \frac{b^{6} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac{{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac{{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{6 \, a^{5} - 16 \, a^{3} b^{2} + 10 \, a b^{4} +{\left (5 \, a^{4} b - 14 \, a^{2} b^{3} + 9 \, b^{5}\right )} \sin \left (x\right )^{3} - 4 \,{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \left (x\right )^{2} -{\left (3 \, a^{4} b - 10 \, a^{2} b^{3} + 7 \, b^{5}\right )} \sin \left (x\right )}{8 \,{\left (a + b\right )}^{3}{\left (a - b\right )}^{3}{\left (\sin \left (x\right ) + 1\right )}^{2}{\left (\sin \left (x\right ) - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^6*log(abs(a*sin(x) + b))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - 1/16*(8*a^2 - 21*a*b + 15*b^2)*log(sin(x) +
1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/16*(8*a^2 + 21*a*b + 15*b^2)*log(-sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2
+ b^3) - 1/8*(6*a^5 - 16*a^3*b^2 + 10*a*b^4 + (5*a^4*b - 14*a^2*b^3 + 9*b^5)*sin(x)^3 - 4*(2*a^5 - 5*a^3*b^2
+ 3*a*b^4)*sin(x)^2 - (3*a^4*b - 10*a^2*b^3 + 7*b^5)*sin(x))/((a + b)^3*(a - b)^3*(sin(x) + 1)^2*(sin(x) - 1)^
2)