Integrand size = 26, antiderivative size = 1584 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \]
2*I*f^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3+ 2*I*f^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3+ 3*a^2*f*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2 /d^2+3*a^3*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/( a^2-b^2)^(5/2)/d^2-3*a*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2) ^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-3*a^3*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c) )/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2+3*a*f*(f*x+e)*polylog(2,I*b*e xp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2+3*a^2*f*(f*x+e)*l n(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^2-3/2*I*a*(f*x +e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d-3/2 *I*a^3*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^ (5/2)/d-3*I*a^2*f^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a ^2-b^2)^2/d^3-3*I*a^2*f^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)) )/b/(a^2-b^2)^2/d^3-3*I*a*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1 /2)))/b/(a^2-b^2)^(3/2)/d^3-3*I*a^3*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a ^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^3-a*f*(f*x+e)/b/(a^2-b^2)/d^2/(a+b*sin (d*x+c))+3*I*a*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^ 2-b^2)^(3/2)/d^3+3/2*I*a^3*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^ (1/2)))/b/(a^2-b^2)^(5/2)/d+3/2*I*a*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+( a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+3*I*a^3*f^2*polylog(3,I*b*exp(I*(d...
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(13567\) vs. \(2(1584)=3168\).
Time = 19.73 (sec) , antiderivative size = 13567, normalized size of antiderivative = 8.57 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \]
Time = 5.65 (sec) , antiderivative size = 1584, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {7293, 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.
\(\displaystyle \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(e+f x)^2}{b (a+b \sin (c+d x))^2}-\frac {a (e+f x)^2}{b (a+b \sin (c+d x))^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^2}-\frac {3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^2}+\frac {3 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^3}-\frac {3 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^3}-\frac {3 i (e+f x)^2 a^2}{2 b \left (a^2-b^2\right )^2 d}+\frac {3 f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^2}+\frac {3 f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^2}-\frac {3 i f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^3}-\frac {3 i f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^3}-\frac {3 (e+f x)^2 \cos (c+d x) a^2}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {2 f^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {3 i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {3 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {f (e+f x) a}{b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {(e+f x)^2 \cos (c+d x) a}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {i (e+f x)^2}{b \left (a^2-b^2\right ) d}-\frac {2 f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {2 f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {2 i f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {2 i f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {(e+f x)^2 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}\) |
(((-3*I)/2)*a^2*(e + f*x)^2)/(b*(a^2 - b^2)^2*d) + (I*(e + f*x)^2)/(b*(a^2 - b^2)*d) + (2*a*f^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b *(a^2 - b^2)^(3/2)*d^3) + (3*a^2*f*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x))) /(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^2) - (2*f*(e + f*x)*Log[1 - (I *b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^2) + (((3*I)/ 2)*a^3*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/( b*(a^2 - b^2)^(5/2)*d) - (((3*I)/2)*a*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d *x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + (3*a^2*f*(e + f*x) *Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^ 2) - (2*f*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/ (b*(a^2 - b^2)*d^2) - (((3*I)/2)*a^3*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d* x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d) + (((3*I)/2)*a*(e + f *x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2) ^(3/2)*d) - ((3*I)*a^2*f^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^3) + ((2*I)*f^2*PolyLog[2, (I*b*E^(I*(c + d*x )))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^3) + (3*a^3*f*(e + f*x)*PolyL og[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d ^2) - (3*a*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^ 2])])/(b*(a^2 - b^2)^(3/2)*d^2) - ((3*I)*a^2*f^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^3) + ((2*I)*f^2*Poly...
3.3.49.3.1 Defintions of rubi rules used
\[\int \frac {\left (f x +e \right )^{2} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5755 vs. \(2 (1385) = 2770\).
Time = 0.67 (sec) , antiderivative size = 5755, normalized size of antiderivative = 3.63 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
\[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \]