Integrand size = 23, antiderivative size = 240 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {2 \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{2/3} d}-\frac {2 \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} b^{2/3} d}-\frac {2 \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} b^{2/3} d} \] Output:
2/3*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/( a^(2/3)-b^(2/3))^(1/2)/b^(2/3)/d-2/3*arctanh((b^(1/3)-(-1)^(1/3)*a^(1/3)*t an(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2))/(-(-1)^(2/3)*a^(2/ 3)+b^(2/3))^(1/2)/b^(2/3)/d-2/3*arctanh((b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/ 2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2))/((-1)^(1/3)*a^(2/3)+b^(2 /3))^(1/2)/b^(2/3)/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 11.10 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.96 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {i \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-4 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+2 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]}{6 d} \] Input:
Integrate[Sin[c + d*x]^2/(a + b*Sin[c + d*x]^3),x]
Output:
((I/6)*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - I*Log[1 - 2*Cos[c + d*x] *#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (2*I)*Log[ 1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^ 2 - 2*b*#1^3 + b*#1^5) & ])/d
Time = 0.56 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3699, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {1}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac {1}{3 b^{2/3} \left (\sqrt [3]{b} \sin (c+d x)-\sqrt [3]{-1} \sqrt [3]{a}\right )}+\frac {1}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{2/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}-\frac {2 \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\) |
Input:
Int[Sin[c + d*x]^2/(a + b*Sin[c + d*x]^3),x]
Output:
(2*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/( 3*Sqrt[a^(2/3) - b^(2/3)]*b^(2/3)*d) - (2*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^ (1/3)*Tan[(c + d*x)/2])/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*Sqrt[-( (-1)^(2/3)*a^(2/3)) + b^(2/3)]*b^(2/3)*d) - (2*ArcTanh[(b^(1/3) + (-1)^(2/ 3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*Sqrt[ (-1)^(1/3)*a^(2/3) + b^(2/3)]*b^(2/3)*d)
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n) ^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.32
| method | result | size |
| derivativedivides | \(\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 d}\) | \(76\) |
| default | \(\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 d}\) | \(76\) |
| risch | \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4096+\left (729 a^{2} b^{4} d^{6}-729 b^{6} d^{6}\right ) \textit {\_Z}^{6}+3888 b^{4} d^{4} \textit {\_Z}^{4}-6912 b^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {243}{1024} b^{3} d^{5} a^{2}+\frac {243}{1024} b^{5} d^{5}\right ) \textit {\_R}^{5}+\left (-\frac {81 i a \,b^{3} d^{4}}{256}+\frac {81 i b^{5} d^{4}}{256 a}\right ) \textit {\_R}^{4}-\frac {81 b^{3} d^{3} \textit {\_R}^{3}}{64}+\left (-\frac {9 i b a \,d^{2}}{16}-\frac {9 i d^{2} b^{3}}{8 a}\right ) \textit {\_R}^{2}+\frac {9 b d \textit {\_R}}{4}+\frac {i b}{a}\right )\right )}{4}\) | \(167\) |
Input:
int(sin(d*x+c)^2/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
4/3/d*sum(_R^2/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_ R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))
Result contains complex when optimal does not.
Time = 1.02 (sec) , antiderivative size = 25253, normalized size of antiderivative = 105.22 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(sin(d*x+c)**2/(a+b*sin(d*x+c)**3),x)
Output:
Integral(sin(c + d*x)**2/(a + b*sin(c + d*x)**3), x)
\[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sin(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
integrate(sin(d*x + c)^2/(b*sin(d*x + c)^3 + a), x)
\[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sin(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
integrate(sin(d*x + c)^2/(b*sin(d*x + c)^3 + a), x)
Time = 39.47 (sec) , antiderivative size = 590, normalized size of antiderivative = 2.46 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\sum _{k=1}^6\ln \left (-\frac {8192\,a^4\,\left (-729\,a^2\,b^3-81\,a^2\,b^2\,\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )+243\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+324\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^3\,\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )+162\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^2+36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^3+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^4+972\,b^5+324\,b^4\,\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )-216\,b^3\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^2-72\,b^2\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^3+12\,b\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^4+4\,{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^5\right )}{{\mathrm {root}\left (d^6-27\,b^2\,d^4+243\,b^4\,d^2+729\,b^4\,\left (a^2-b^2\right ),d,k\right )}^5}\right )\,\mathrm {root}\left (729\,a^2\,b^4\,d^6-729\,b^6\,d^6+243\,b^4\,d^4-27\,b^2\,d^2+1,d,k\right )}{d} \] Input:
int(sin(c + d*x)^2/(a + b*sin(c + d*x)^3),x)
Output:
symsum(log(-(8192*a^4*(12*b*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b^4* (a^2 - b^2), d, k)^4 + 324*b^4*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b ^4*(a^2 - b^2), d, k) + 972*b^5 + 4*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k)^5 - 729*a^2*b^3 - 72*b^2*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k)^3 - 216*b^3*root(d^6 - 27*b^2*d ^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k)^2 - 81*a^2*b^2*root(d^6 - 27 *b^2*d^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k) + 243*a*b^4*tan(c/2 + (d*x)/2) + 3*a*tan(c/2 + (d*x)/2)*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 72 9*b^4*(a^2 - b^2), d, k)^4 + 162*a*b^2*tan(c/2 + (d*x)/2)*root(d^6 - 27*b^ 2*d^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k)^2 + 36*a*b*tan(c/2 + (d*x )/2)*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k)^3 + 324*a*tan(c/2 + (d*x)/2)*b^3*root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b^4 *(a^2 - b^2), d, k)))/root(d^6 - 27*b^2*d^4 + 243*b^4*d^2 + 729*b^4*(a^2 - b^2), d, k)^5)*root(729*a^2*b^4*d^6 - 729*b^6*d^6 + 243*b^4*d^4 - 27*b^2* d^2 + 1, d, k), k, 1, 6)/d
\[ \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sin \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{3} b +a}d x \] Input:
int(sin(d*x+c)^2/(a+b*sin(d*x+c)^3),x)
Output:
int(sin(c + d*x)**2/(sin(c + d*x)**3*b + a),x)