3.13 \(\int \frac{\tan ^3(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{b^4 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^2}-\frac{1}{4 (a+b) (1-\csc (x))}-\frac{1}{4 (a-b) (\csc (x)+1)}+\frac{(2 a+3 b) \log (1-\csc (x))}{4 (a+b)^2}+\frac{(2 a-3 b) \log (\csc (x)+1)}{4 (a-b)^2}+\frac{\log (\sin (x))}{a} \]

[Out]

-1/(4*(a + b)*(1 - Csc[x])) - 1/(4*(a - b)*(1 + Csc[x])) + ((2*a + 3*b)*Log[1 - Csc[x]])/(4*(a + b)^2) + ((2*a
- 3*b)*Log[1 + Csc[x]])/(4*(a - b)^2) + (b^4*Log[a + b*Csc[x]])/(a*(a^2 - b^2)^2) + Log[Sin[x]]/a

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Rubi [A]  time = 0.166837, antiderivative size = 113, 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} \[ \frac{b^4 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^2}-\frac{1}{4 (a+b) (1-\csc (x))}-\frac{1}{4 (a-b) (\csc (x)+1)}+\frac{(2 a+3 b) \log (1-\csc (x))}{4 (a+b)^2}+\frac{(2 a-3 b) \log (\csc (x)+1)}{4 (a-b)^2}+\frac{\log (\sin (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3/(a + b*Csc[x]),x]

[Out]

-1/(4*(a + b)*(1 - Csc[x])) - 1/(4*(a - b)*(1 + Csc[x])) + ((2*a + 3*b)*Log[1 - Csc[x]])/(4*(a + b)^2) + ((2*a
- 3*b)*Log[1 + Csc[x]])/(4*(a - b)^2) + (b^4*Log[a + b*Csc[x]])/(a*(a^2 - b^2)^2) + 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 ^3(x)}{a+b \csc (x)} \, dx &=-\left (b^4 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \csc (x)\right )\right )\\ &=-\left (b^4 \operatorname{Subst}\left (\int \left (\frac{1}{4 b^3 (a+b) (b-x)^2}+\frac{2 a+3 b}{4 b^4 (a+b)^2 (b-x)}+\frac{1}{a b^4 x}-\frac{1}{a (a-b)^2 (a+b)^2 (a+x)}-\frac{1}{4 (a-b) b^3 (b+x)^2}+\frac{-2 a+3 b}{4 (a-b)^2 b^4 (b+x)}\right ) \, dx,x,b \csc (x)\right )\right )\\ &=-\frac{1}{4 (a+b) (1-\csc (x))}-\frac{1}{4 (a-b) (1+\csc (x))}+\frac{(2 a+3 b) \log (1-\csc (x))}{4 (a+b)^2}+\frac{(2 a-3 b) \log (1+\csc (x))}{4 (a-b)^2}+\frac{b^4 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^2}+\frac{\log (\sin (x))}{a}\\ \end{align*}

Mathematica [A]  time = 0.485476, size = 115, normalized size = 1.02 \[ \frac{\csc (x) (a \sin (x)+b) \left (\frac{4 b^4 \log (a \sin (x)+b)}{a (a-b)^2 (a+b)^2}-\frac{1}{(a+b) (\sin (x)-1)}+\frac{1}{(a-b) (\sin (x)+1)}+\frac{(2 a+3 b) \log (1-\sin (x))}{(a+b)^2}+\frac{(2 a-3 b) \log (\sin (x)+1)}{(a-b)^2}\right )}{4 (a+b \csc (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3/(a + b*Csc[x]),x]

[Out]

(Csc[x]*(b + a*Sin[x])*(((2*a + 3*b)*Log[1 - Sin[x]])/(a + b)^2 + ((2*a - 3*b)*Log[1 + Sin[x]])/(a - b)^2 + (4
*b^4*Log[b + a*Sin[x]])/(a*(a - b)^2*(a + b)^2) - 1/((a + b)*(-1 + Sin[x])) + 1/((a - b)*(1 + Sin[x]))))/(4*(a
+ b*Csc[x]))

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Maple [A]  time = 0.058, size = 117, normalized size = 1. \begin{align*}{\frac{{b}^{4}\ln \left ( b+a\sin \left ( x \right ) \right ) }{ \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a}}+{\frac{1}{ \left ( 4\,a-4\,b \right ) \left ( \sin \left ( x \right ) +1 \right ) }}+{\frac{a\ln \left ( \sin \left ( x \right ) +1 \right ) }{2\, \left ( a-b \right ) ^{2}}}-{\frac{3\,\ln \left ( \sin \left ( x \right ) +1 \right ) b}{4\, \left ( a-b \right ) ^{2}}}-{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( \sin \left ( x \right ) -1 \right ) }}+{\frac{a\ln \left ( \sin \left ( x \right ) -1 \right ) }{2\, \left ( a+b \right ) ^{2}}}+{\frac{3\,\ln \left ( \sin \left ( x \right ) -1 \right ) b}{4\, \left ( a+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3/(a+b*csc(x)),x)

[Out]

b^4/(a+b)^2/(a-b)^2/a*ln(b+a*sin(x))+1/(4*a-4*b)/(sin(x)+1)+1/2*a*ln(sin(x)+1)/(a-b)^2-3/4/(a-b)^2*ln(sin(x)+1
)*b-1/(4*a+4*b)/(sin(x)-1)+1/2*a/(a+b)^2*ln(sin(x)-1)+3/4/(a+b)^2*ln(sin(x)-1)*b

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Maxima [A]  time = 0.964176, size = 162, normalized size = 1.43 \begin{align*} \frac{b^{4} \log \left (a \sin \left (x\right ) + b\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (x\right ) - 1\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{b \sin \left (x\right ) - a}{2 \,{\left ({\left (a^{2} - b^{2}\right )} \sin \left (x\right )^{2} - a^{2} + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*csc(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.705348, size = 344, normalized size = 3.04 \begin{align*} \frac{4 \, b^{4} \cos \left (x\right )^{2} \log \left (a \sin \left (x\right ) + b\right ) + 2 \, a^{4} - 2 \, a^{2} b^{2} +{\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) +{\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (x\right )}{4 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*csc(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\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)**3/(a+b*csc(x)),x)

[Out]

Integral(tan(x)**3/(a + b*csc(x)), x)

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Giac [A]  time = 1.36105, size = 188, normalized size = 1.66 \begin{align*} \frac{b^{4} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a + 3 \, b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a^{3} - a b^{2} -{\left (a^{2} b - b^{3}\right )} \sin \left (x\right )}{2 \,{\left (a + b\right )}^{2}{\left (a - b\right )}^{2}{\left (\sin \left (x\right ) + 1\right )}{\left (\sin \left (x\right ) - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*csc(x)),x, algorithm="giac")

[Out]

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