3.28 \(\int \frac {\csc ^3(x)}{(a \cos (x)+b \sin (x))^3} \, dx\)

Optimal. Leaf size=117 \[ \frac {3 b \cot (x)}{a^4}-\frac {\cot ^2(x)}{2 a^3}+\frac {2 \left (a^2+3 b^2\right ) \log (\tan (x))}{a^5}-\frac {2 \left (a^2+3 b^2\right ) \log (a+b \tan (x))}{a^5}-\frac {\left (a^2-3 b^2\right ) \left (a^2+b^2\right )}{a^4 b^2 (a+b \tan (x))}+\frac {\left (a^2+b^2\right )^2}{2 a^3 b^2 (a+b \tan (x))^2} \]

[Out]

3*b*cot(x)/a^4-1/2*cot(x)^2/a^3+2*(a^2+3*b^2)*ln(tan(x))/a^5-2*(a^2+3*b^2)*ln(a+b*tan(x))/a^5+1/2*(a^2+b^2)^2/
a^3/b^2/(a+b*tan(x))^2-(a^2-3*b^2)*(a^2+b^2)/a^4/b^2/(a+b*tan(x))

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Rubi [A]  time = 0.14, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3087, 894} \[ \frac {\left (a^2+b^2\right )^2}{2 a^3 b^2 (a+b \tan (x))^2}-\frac {\left (a^2-3 b^2\right ) \left (a^2+b^2\right )}{a^4 b^2 (a+b \tan (x))}+\frac {2 \left (a^2+3 b^2\right ) \log (\tan (x))}{a^5}-\frac {2 \left (a^2+3 b^2\right ) \log (a+b \tan (x))}{a^5}+\frac {3 b \cot (x)}{a^4}-\frac {\cot ^2(x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

(3*b*Cot[x])/a^4 - Cot[x]^2/(2*a^3) + (2*(a^2 + 3*b^2)*Log[Tan[x]])/a^5 - (2*(a^2 + 3*b^2)*Log[a + b*Tan[x]])/
a^5 + (a^2 + b^2)^2/(2*a^3*b^2*(a + b*Tan[x])^2) - ((a^2 - 3*b^2)*(a^2 + b^2))/(a^4*b^2*(a + b*Tan[x]))

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]))

Rule 3087

Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[1/d, Subst[Int[(x^m*(a + b*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Tan[c + d*x]], x] /; FreeQ[{a,
b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.83, size = 208, normalized size = 1.78 \[ \frac {a^4 \csc ^2(x)+6 a^3 b \cot ^3(x)+2 b^2 \left (2 \left (a^2+3 b^2\right ) \log (\sin (x))-2 \left (a^2+3 b^2\right ) \log (a \cos (x)+b \sin (x))-3 \left (a^2+b^2\right )\right )-2 a b \cot (x) \left (-4 \left (a^2+3 b^2\right ) \log (\sin (x))+4 a^2 \log (a \cos (x)+b \sin (x))+a^2 \csc ^2(x)+3 a^2+12 b^2 \log (a \cos (x)+b \sin (x))\right )+\cot ^2(x) \left (4 a^2 \left (\left (a^2+3 b^2\right ) \log (\sin (x))-\left (a^2+3 b^2\right ) \log (a \cos (x)+b \sin (x))+3 b^2\right )-a^4 \csc ^2(x)\right )}{2 a^5 (a \cot (x)+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

(6*a^3*b*Cot[x]^3 + a^4*Csc[x]^2 - 2*a*b*Cot[x]*(3*a^2 + a^2*Csc[x]^2 - 4*(a^2 + 3*b^2)*Log[Sin[x]] + 4*a^2*Lo
g[a*Cos[x] + b*Sin[x]] + 12*b^2*Log[a*Cos[x] + b*Sin[x]]) + 2*b^2*(-3*(a^2 + b^2) + 2*(a^2 + 3*b^2)*Log[Sin[x]
] - 2*(a^2 + 3*b^2)*Log[a*Cos[x] + b*Sin[x]]) + Cot[x]^2*(-(a^4*Csc[x]^2) + 4*a^2*(3*b^2 + (a^2 + 3*b^2)*Log[S
in[x]] - (a^2 + 3*b^2)*Log[a*Cos[x] + b*Sin[x]])))/(2*a^5*(b + a*Cot[x])^2)

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fricas [B]  time = 1.94, size = 385, normalized size = 3.29 \[ -\frac {24 \, a^{2} b^{2} \cos \relax (x)^{4} - a^{4} + 6 \, a^{2} b^{2} + 2 \, {\left (a^{4} - 15 \, a^{2} b^{2}\right )} \cos \relax (x)^{2} - 2 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \relax (x)^{4} - a^{2} b^{2} - 3 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 6 \, b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left ({\left (a^{3} b + 3 \, a b^{3}\right )} \cos \relax (x)^{3} - {\left (a^{3} b + 3 \, a b^{3}\right )} \cos \relax (x)\right )} \sin \relax (x)\right )} \log \left (2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}\right ) + 2 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \relax (x)^{4} - a^{2} b^{2} - 3 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 6 \, b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left ({\left (a^{3} b + 3 \, a b^{3}\right )} \cos \relax (x)^{3} - {\left (a^{3} b + 3 \, a b^{3}\right )} \cos \relax (x)\right )} \sin \relax (x)\right )} \log \left (-\frac {1}{4} \, \cos \relax (x)^{2} + \frac {1}{4}\right ) - 4 \, {\left (3 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x)^{3} - {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \relax (x)\right )} \sin \relax (x)}{2 \, {\left (a^{5} b^{2} - {\left (a^{7} - a^{5} b^{2}\right )} \cos \relax (x)^{4} + {\left (a^{7} - 2 \, a^{5} b^{2}\right )} \cos \relax (x)^{2} - 2 \, {\left (a^{6} b \cos \relax (x)^{3} - a^{6} b \cos \relax (x)\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.90, size = 146, normalized size = 1.25 \[ \frac {2 \, {\left (a^{2} + 3 \, b^{2}\right )} \log \left ({\left | \tan \relax (x) \right |}\right )}{a^{5}} - \frac {2 \, {\left (a^{2} b + 3 \, b^{3}\right )} \log \left ({\left | b \tan \relax (x) + a \right |}\right )}{a^{5} b} - \frac {2 \, a^{4} b \tan \relax (x)^{3} - 4 \, a^{2} b^{3} \tan \relax (x)^{3} - 12 \, b^{5} \tan \relax (x)^{3} + a^{5} \tan \relax (x)^{2} - 6 \, a^{3} b^{2} \tan \relax (x)^{2} - 18 \, a b^{4} \tan \relax (x)^{2} - 4 \, a^{2} b^{3} \tan \relax (x) + a^{3} b^{2}}{2 \, {\left (b \tan \relax (x)^{2} + a \tan \relax (x)\right )}^{2} a^{4} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a^2 + 3*b^2)*log(abs(tan(x)))/a^5 - 2*(a^2*b + 3*b^3)*log(abs(b*tan(x) + a))/(a^5*b) - 1/2*(2*a^4*b*tan(x)^
3 - 4*a^2*b^3*tan(x)^3 - 12*b^5*tan(x)^3 + a^5*tan(x)^2 - 6*a^3*b^2*tan(x)^2 - 18*a*b^4*tan(x)^2 - 4*a^2*b^3*t
an(x) + a^3*b^2)/((b*tan(x)^2 + a*tan(x))^2*a^4*b^2)

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maple [A]  time = 0.76, size = 151, normalized size = 1.29 \[ \frac {a}{2 b^{2} \left (a +b \tan \relax (x )\right )^{2}}+\frac {1}{a \left (a +b \tan \relax (x )\right )^{2}}+\frac {b^{2}}{2 a^{3} \left (a +b \tan \relax (x )\right )^{2}}-\frac {1}{b^{2} \left (a +b \tan \relax (x )\right )}+\frac {2}{a^{2} \left (a +b \tan \relax (x )\right )}+\frac {3 b^{2}}{a^{4} \left (a +b \tan \relax (x )\right )}-\frac {2 \ln \left (a +b \tan \relax (x )\right )}{a^{3}}-\frac {6 \ln \left (a +b \tan \relax (x )\right ) b^{2}}{a^{5}}-\frac {1}{2 a^{3} \tan \relax (x )^{2}}+\frac {2 \ln \left (\tan \relax (x )\right )}{a^{3}}+\frac {6 \ln \left (\tan \relax (x )\right ) b^{2}}{a^{5}}+\frac {3 b}{a^{4} \tan \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a/b^2/(a+b*tan(x))^2+1/a/(a+b*tan(x))^2+1/2/a^3*b^2/(a+b*tan(x))^2-1/b^2/(a+b*tan(x))+2/a^2/(a+b*tan(x))+3
/a^4*b^2/(a+b*tan(x))-2*ln(a+b*tan(x))/a^3-6/a^5*ln(a+b*tan(x))*b^2-1/2/a^3/tan(x)^2+2*ln(tan(x))/a^3+6/a^5*ln
(tan(x))*b^2+3/a^4*b/tan(x)

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maxima [B]  time = 0.36, size = 308, normalized size = 2.63 \[ -\frac {a^{4} - \frac {8 \, a^{3} b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {2 \, {\left (a^{4} + 22 \, a^{2} b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {4 \, {\left (21 \, a^{3} b + 4 \, a b^{3}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {{\left (15 \, a^{4} - 144 \, a^{2} b^{2} - 112 \, b^{4}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {4 \, {\left (19 \, a^{3} b + 16 \, a b^{3}\right )} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}}}{8 \, {\left (\frac {a^{7} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {4 \, a^{6} b \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {4 \, a^{6} b \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {a^{7} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} - \frac {2 \, {\left (a^{7} - 2 \, a^{5} b^{2}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}\right )}} - \frac {\frac {12 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}}{8 \, a^{4}} - \frac {2 \, {\left (a^{2} + 3 \, b^{2}\right )} \log \left (-a - \frac {2 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{5}} + \frac {2 \, {\left (a^{2} + 3 \, b^{2}\right )} \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(a^4 - 8*a^3*b*sin(x)/(cos(x) + 1) - 2*(a^4 + 22*a^2*b^2)*sin(x)^2/(cos(x) + 1)^2 + 4*(21*a^3*b + 4*a*b^3
)*sin(x)^3/(cos(x) + 1)^3 - (15*a^4 - 144*a^2*b^2 - 112*b^4)*sin(x)^4/(cos(x) + 1)^4 - 4*(19*a^3*b + 16*a*b^3)
*sin(x)^5/(cos(x) + 1)^5)/(a^7*sin(x)^2/(cos(x) + 1)^2 + 4*a^6*b*sin(x)^3/(cos(x) + 1)^3 - 4*a^6*b*sin(x)^5/(c
os(x) + 1)^5 + a^7*sin(x)^6/(cos(x) + 1)^6 - 2*(a^7 - 2*a^5*b^2)*sin(x)^4/(cos(x) + 1)^4) - 1/8*(12*b*sin(x)/(
cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/a^4 - 2*(a^2 + 3*b^2)*log(-a - 2*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(
cos(x) + 1)^2)/a^5 + 2*(a^2 + 3*b^2)*log(sin(x)/(cos(x) + 1))/a^5

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mupad [B]  time = 0.86, size = 253, normalized size = 2.16 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (2\,a^2+6\,b^2\right )}{a^5}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (42\,a^2\,b+8\,b^3\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (38\,a^2\,b+32\,b^3\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^3+22\,a\,b^2\right )+\frac {a^3}{2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (-15\,a^4+144\,a^2\,b^2+112\,b^4\right )}{2\,a}-4\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{4\,a^6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (8\,a^6-16\,a^4\,b^2\right )+4\,a^6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+16\,a^5\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3-16\,a^5\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}-\frac {\ln \left (-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a\right )\,\left (2\,a^2+6\,b^2\right )}{a^5}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,a^3}-\frac {3\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^3*(a*cos(x) + b*sin(x))^3),x)

[Out]

(log(tan(x/2))*(2*a^2 + 6*b^2))/a^5 - (tan(x/2)^3*(42*a^2*b + 8*b^3) - tan(x/2)^5*(38*a^2*b + 32*b^3) - tan(x/
2)^2*(22*a*b^2 + a^3) + a^3/2 + (tan(x/2)^4*(112*b^4 - 15*a^4 + 144*a^2*b^2))/(2*a) - 4*a^2*b*tan(x/2))/(4*a^6
*tan(x/2)^2 - tan(x/2)^4*(8*a^6 - 16*a^4*b^2) + 4*a^6*tan(x/2)^6 + 16*a^5*b*tan(x/2)^3 - 16*a^5*b*tan(x/2)^5)
- (log(a + 2*b*tan(x/2) - a*tan(x/2)^2)*(2*a^2 + 6*b^2))/a^5 - tan(x/2)^2/(8*a^3) - (3*b*tan(x/2))/(2*a^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\relax (x )}}{\left (a \cos {\relax (x )} + b \sin {\relax (x )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**3/(a*cos(x)+b*sin(x))**3,x)

[Out]

Integral(csc(x)**3/(a*cos(x) + b*sin(x))**3, x)

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