Integrand size = 26, antiderivative size = 2348 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \]
3*I*a*f^2*(f*x+e)*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+9*I*a^3*f^2*(f*x+e)*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+9*I*a*f^2*(f*x+e)*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)^3*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)^3* ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+6*I*f^2*( f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3+6 *I*f^2*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^ 2)/d^3-9*I*a^3*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2) ))/b/(a^2-b^2)^(5/2)/d^3-3/2*a*f*(f*x+e)^2/b/(a^2-b^2)/d^2/(a+b*sin(d*x+c) )+9/2*a^3*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/ (a^2-b^2)^(5/2)/d^2-9/2*a*f*(f*x+e)^2*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*I*a*f^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+ c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3-9*I*a^2*f^2*(f*x+e)*polylog (2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^3-9*I*a^2*f^2*( f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^3 -9*I*a*f^2*(f*x+e)*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*a*f^3*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2))) /b/(a^2-b^2)^(3/2)/d^4+9*a^2*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2) ^(1/2)))/b/(a^2-b^2)^2/d^4+9*a^2*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a+(a...
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(29732\) vs. \(2(2348)=4696\).
Time = 22.19 (sec) , antiderivative size = 29732, normalized size of antiderivative = 12.66 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \]
Time = 7.35 (sec) , antiderivative size = 2348, 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.
\(\displaystyle \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(e+f x)^3}{b (a+b \sin (c+d x))^2}-\frac {a (e+f x)^3}{b (a+b \sin (c+d x))^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 i (e+f x)^3 \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)^3 \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 {9 f (e+f x)^2 \operatorname {PolyLog}\left (2,\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^2}-\frac {9 f (e+f x)^2 \operatorname {PolyLog}\left (2,\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^2}+\frac {9 i f^2 (e+f x) \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 {9 i f^2 (e+f x) \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 {9 f^3 \operatorname {PolyLog}\left (4,\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^4}+\frac {9 f^3 \operatorname {PolyLog}\left (4,\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^4}-\frac {3 i (e+f x)^3 a^2}{2 b \left (a^2-b^2\right )^2 d}+\frac {9 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{2 b \left (a^2-b^2\right )^2 d^2}+\frac {9 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {9 i f^2 (e+f x) \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 {9 i f^2 (e+f x) \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 {9 f^3 \operatorname {PolyLog}\left (3,\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^4}+\frac {9 f^3 \operatorname {PolyLog}\left (3,\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^4}-\frac {3 (e+f x)^3 \cos (c+d x) a^2}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {3 i (e+f x)^3 \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 f^2 (e+f x) \log \left (1-\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 (e+f x)^3 \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 f^2 (e+f x) \log \left (1-\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 f^3 \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^4}-\frac {9 f (e+f x)^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 )^{3/2} d^2}+\frac {3 f^3 \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^4}+\frac {9 f (e+f x)^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 )^{3/2} d^2}-\frac {9 i f^2 (e+f x) \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 {9 i f^2 (e+f x) \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 {9 f^3 \operatorname {PolyLog}\left (4,\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^4}-\frac {9 f^3 \operatorname {PolyLog}\left (4,\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^4}-\frac {3 f (e+f x)^2 a}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {(e+f x)^3 \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)^3}{b \left (a^2-b^2\right ) d}-\frac {3 f (e+f x)^2 \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 {3 f (e+f x)^2 \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 {6 i f^2 (e+f x) \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 {6 i f^2 (e+f x) \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 {6 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {6 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {(e+f x)^3 \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)^3)/(b*(a^2 - b^2)^2*d) + (I*(e + f*x)^3)/(b*(a^2 - b^2)*d) - ((3*I)*a*f^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqr t[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) + (9*a^2*f*(e + f*x)^2*Log[1 - ( I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^2*d^2) - (3* f*(e + f*x)^2*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)^3*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)^3* 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*f^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) + (9*a^2*f*(e + f*x)^2*Log[1 - (I*b*E^( I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^2*d^2) - (3*f*(e + f*x)^2*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)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqr t[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d) + (((3*I)/2)*a*(e + f*x)^3*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*f^3*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) - ((9*I)*a^2*f^2*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x) ))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^3) + ((6*I)*f^2*(e + f*x)*Po lyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^3) + (9*a^3*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 ...
3.3.50.3.1 Defintions of rubi rules used
\[\int \frac {\left (f x +e \right )^{3} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 10614 vs. \(2 (2046) = 4092\).
Time = 0.94 (sec) , antiderivative size = 10614, normalized size of antiderivative = 4.52 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \]