3.16 \(\int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx\)

Optimal. Leaf size=120 \[ -\frac {\sin ^4(x) (a \cot (x)+b)}{4 \left (a^2+b^2\right )}-\frac {b^5 \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^3}-\frac {\sin ^2(x) \left (a \left (3 a^2+7 b^2\right ) \cot (x)+4 b^3\right )}{8 \left (a^2+b^2\right )^2}+\frac {a x \left (3 a^4+10 a^2 b^2+15 b^4\right )}{8 \left (a^2+b^2\right )^3} \]

[Out]

1/8*a*(3*a^4+10*a^2*b^2+15*b^4)*x/(a^2+b^2)^3-b^5*ln(b*cos(x)+a*sin(x))/(a^2+b^2)^3-1/8*(4*b^3+a*(3*a^2+7*b^2)
*cot(x))*sin(x)^2/(a^2+b^2)^2-1/4*(b+a*cot(x))*sin(x)^4/(a^2+b^2)

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Rubi [A]  time = 0.19, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3506, 741, 823, 801, 635, 203, 260} \[ \frac {a x \left (10 a^2 b^2+3 a^4+15 b^4\right )}{8 \left (a^2+b^2\right )^3}-\frac {\sin ^4(x) (a \cot (x)+b)}{4 \left (a^2+b^2\right )}-\frac {\sin ^2(x) \left (a \left (3 a^2+7 b^2\right ) \cot (x)+4 b^3\right )}{8 \left (a^2+b^2\right )^2}-\frac {b^5 \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Cot[x]),x]

[Out]

(a*(3*a^4 + 10*a^2*b^2 + 15*b^4)*x)/(8*(a^2 + b^2)^3) - (b^5*Log[b*Cos[x] + a*Sin[x]])/(a^2 + b^2)^3 - ((4*b^3
+ a*(3*a^2 + 7*b^2)*Cot[x])*Sin[x]^2)/(8*(a^2 + b^2)^2) - ((b + a*Cot[x])*Sin[x]^4)/(4*(a^2 + b^2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 635

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

Rule 741

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*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{(a+x) \left (1+\frac {x^2}{b^2}\right )^3} \, dx,x,b \cot (x)\right )}{b}\\ &=-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac {b \operatorname {Subst}\left (\int \frac {-4-\frac {3 a^2}{b^2}-\frac {3 a x}{b^2}}{(a+x) \left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \cot (x)\right )}{4 \left (a^2+b^2\right )}\\ &=-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {b^5 \operatorname {Subst}\left (\int \frac {\frac {3 a^4+7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (3 a^2+7 b^2\right ) x}{b^6}}{(a+x) \left (1+\frac {x^2}{b^2}\right )} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^2}\\ &=-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {b^5 \operatorname {Subst}\left (\int \left (\frac {8}{\left (a^2+b^2\right ) (a+x)}+\frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^4 x}{b^4 \left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^2}\\ &=-\frac {b^5 \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {b \operatorname {Subst}\left (\int \frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^4 x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3}\\ &=-\frac {b^5 \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac {b^5 \operatorname {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{\left (a^2+b^2\right )^3}-\frac {\left (a b \left (3 a^4+10 a^2 b^2+15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3}\\ &=\frac {a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x}{8 \left (a^2+b^2\right )^3}-\frac {b^5 \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {b^5 \log (\sin (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 151, normalized size = 1.26 \[ \frac {12 a^5 x-8 a^5 \sin (2 x)+a^5 \sin (4 x)+40 a^3 b^2 x-24 a^3 b^2 \sin (2 x)+2 a^3 b^2 \sin (4 x)-b \left (a^2+b^2\right )^2 \cos (4 x)+4 b \left (a^4+4 a^2 b^2+3 b^4\right ) \cos (2 x)-32 b^5 \log (a \sin (x)+b \cos (x))+60 a b^4 x-16 a b^4 \sin (2 x)+a b^4 \sin (4 x)}{32 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Cot[x]),x]

[Out]

(12*a^5*x + 40*a^3*b^2*x + 60*a*b^4*x + 4*b*(a^4 + 4*a^2*b^2 + 3*b^4)*Cos[2*x] - b*(a^2 + b^2)^2*Cos[4*x] - 32
*b^5*Log[b*Cos[x] + a*Sin[x]] - 8*a^5*Sin[2*x] - 24*a^3*b^2*Sin[2*x] - 16*a*b^4*Sin[2*x] + a^5*Sin[4*x] + 2*a^
3*b^2*Sin[4*x] + a*b^4*Sin[4*x])/(32*(a^2 + b^2)^3)

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fricas [A]  time = 0.91, size = 184, normalized size = 1.53 \[ -\frac {4 \, b^{5} \log \left (2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}\right ) + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \relax (x)^{4} - 4 \, {\left (a^{4} b + 3 \, a^{2} b^{3} + 2 \, b^{5}\right )} \cos \relax (x)^{2} - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x - {\left (2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x)^{3} - {\left (5 \, a^{5} + 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )} \cos \relax (x)\right )} \sin \relax (x)}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*cot(x)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.96, size = 273, normalized size = 2.28 \[ -\frac {a b^{5} \log \left ({\left | a \tan \relax (x) + b \right |}\right )}{a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}} + \frac {b^{5} \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac {6 \, b^{5} \tan \relax (x)^{4} + 5 \, a^{5} \tan \relax (x)^{3} + 14 \, a^{3} b^{2} \tan \relax (x)^{3} + 9 \, a b^{4} \tan \relax (x)^{3} - 4 \, a^{4} b \tan \relax (x)^{2} - 12 \, a^{2} b^{3} \tan \relax (x)^{2} + 4 \, b^{5} \tan \relax (x)^{2} + 3 \, a^{5} \tan \relax (x) + 10 \, a^{3} b^{2} \tan \relax (x) + 7 \, a b^{4} \tan \relax (x) - 2 \, a^{4} b - 8 \, a^{2} b^{3}}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \relax (x)^{2} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*cot(x)),x, algorithm="giac")

[Out]

-a*b^5*log(abs(a*tan(x) + b))/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6) + 1/2*b^5*log(tan(x)^2 + 1)/(a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6) + 1/8*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*x/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 1/8*(6*b^5
*tan(x)^4 + 5*a^5*tan(x)^3 + 14*a^3*b^2*tan(x)^3 + 9*a*b^4*tan(x)^3 - 4*a^4*b*tan(x)^2 - 12*a^2*b^3*tan(x)^2 +
4*b^5*tan(x)^2 + 3*a^5*tan(x) + 10*a^3*b^2*tan(x) + 7*a*b^4*tan(x) - 2*a^4*b - 8*a^2*b^3)/((a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6)*(tan(x)^2 + 1)^2)

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maple [B]  time = 0.24, size = 407, normalized size = 3.39 \[ -\frac {b^{5} \ln \left (a \tan \relax (x )+b \right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {7 \left (\tan ^{3}\relax (x )\right ) b^{2} a^{3}}{4 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}-\frac {9 \left (\tan ^{3}\relax (x )\right ) a \,b^{4}}{8 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}-\frac {5 \left (\tan ^{3}\relax (x )\right ) a^{5}}{8 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {\left (\tan ^{2}\relax (x )\right ) b \,a^{4}}{2 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {3 \left (\tan ^{2}\relax (x )\right ) a^{2} b^{3}}{2 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {\left (\tan ^{2}\relax (x )\right ) b^{5}}{\left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}-\frac {3 \tan \relax (x ) a^{5}}{8 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}-\frac {5 \tan \relax (x ) b^{2} a^{3}}{4 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}-\frac {7 \tan \relax (x ) a \,b^{4}}{8 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {b \,a^{4}}{4 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {a^{2} b^{3}}{\left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {3 b^{5}}{4 \left (a^{2}+b^{2}\right )^{3} \left (1+\tan ^{2}\relax (x )\right )^{2}}+\frac {b^{5} \ln \left (1+\tan ^{2}\relax (x )\right )}{2 \left (a^{2}+b^{2}\right )^{3}}+\frac {15 \arctan \left (\tan \relax (x )\right ) a \,b^{4}}{8 \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \arctan \left (\tan \relax (x )\right ) a^{5}}{8 \left (a^{2}+b^{2}\right )^{3}}+\frac {5 \arctan \left (\tan \relax (x )\right ) b^{2} a^{3}}{4 \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a+b*cot(x)),x)

[Out]

-b^5/(a^2+b^2)^3*ln(a*tan(x)+b)-7/4/(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)^3*b^2*a^3-9/8/(a^2+b^2)^3/(1+tan(x)^2)^2
*tan(x)^3*a*b^4-5/8/(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)^3*a^5+1/2/(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)^2*b*a^4+3/2/
(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)^2*a^2*b^3+1/(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)^2*b^5-3/8/(a^2+b^2)^3/(1+tan(x
)^2)^2*tan(x)*a^5-5/4/(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)*b^2*a^3-7/8/(a^2+b^2)^3/(1+tan(x)^2)^2*tan(x)*a*b^4+1/
4/(a^2+b^2)^3/(1+tan(x)^2)^2*b*a^4+1/(a^2+b^2)^3/(1+tan(x)^2)^2*a^2*b^3+3/4/(a^2+b^2)^3/(1+tan(x)^2)^2*b^5+1/2
/(a^2+b^2)^3*b^5*ln(1+tan(x)^2)+15/8/(a^2+b^2)^3*arctan(tan(x))*a*b^4+3/8/(a^2+b^2)^3*arctan(tan(x))*a^5+5/4/(
a^2+b^2)^3*arctan(tan(x))*b^2*a^3

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maxima [B]  time = 0.68, size = 244, normalized size = 2.03 \[ -\frac {b^{5} \log \left (a \tan \relax (x) + b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {b^{5} \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac {{\left (5 \, a^{3} + 9 \, a b^{2}\right )} \tan \relax (x)^{3} - 2 \, a^{2} b - 6 \, b^{3} - 4 \, {\left (a^{2} b + 2 \, b^{3}\right )} \tan \relax (x)^{2} + {\left (3 \, a^{3} + 7 \, a b^{2}\right )} \tan \relax (x)}{8 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \relax (x)^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*cot(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.50, size = 263, normalized size = 2.19 \[ \frac {\frac {a^2\,b+3\,b^3}{4\,{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {tan}\relax (x)}^3\,\left (5\,a^3+9\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\relax (x)}^2\,\left (a^2\,b+2\,b^3\right )}{2\,{\left (a^2+b^2\right )}^2}-\frac {a\,\mathrm {tan}\relax (x)\,\left (3\,a^2+7\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\relax (x)}^4+2\,{\mathrm {tan}\relax (x)}^2+1}-\frac {b^5\,\ln \left (b+a\,\mathrm {tan}\relax (x)\right )}{{\left (a^2+b^2\right )}^3}+\frac {\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,\left (-3\,a^2+a\,b\,9{}\mathrm {i}+8\,b^2\right )}{16\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}+\frac {\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )\,\left (-a^2\,3{}\mathrm {i}+9\,a\,b+b^2\,8{}\mathrm {i}\right )}{16\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + b*cot(x)),x)

[Out]

((a^2*b + 3*b^3)/(4*(a^2 + b^2)^2) - (tan(x)^3*(9*a*b^2 + 5*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) + (tan(x)^2*(a^2
*b + 2*b^3))/(2*(a^2 + b^2)^2) - (a*tan(x)*(3*a^2 + 7*b^2))/(8*(a^4 + b^4 + 2*a^2*b^2)))/(2*tan(x)^2 + tan(x)^
4 + 1) - (b^5*log(b + a*tan(x)))/(a^2 + b^2)^3 + (log(tan(x) - 1i)*(a*b*9i - 3*a^2 + 8*b^2))/(16*(a*b^2*3i - 3
*a^2*b - a^3*1i + b^3)) + (log(tan(x) + 1i)*(9*a*b - a^2*3i + b^2*8i))/(16*(3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)
)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\relax (x )}}{a + b \cot {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**4/(a+b*cot(x)),x)

[Out]

Integral(sin(x)**4/(a + b*cot(x)), x)

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