\(\int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [250]

Optimal result

Integrand size = 26, antiderivative size = 2348 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Output:

9*a*f^3*polylog(4,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^2-b^2)^(1/2)))/b/(a 
^2-b^2)^2/d^4+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^4-3*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^4-3*f*(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)/d^2-3*f*(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)/d^2+9*a^3*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2 
-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^4-9*a*f^3*polylog(4,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^3*f^3*polylog(4,I*b*exp(I*(d 
*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^4+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^3-3*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^3-9*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+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)))...
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(29732\) vs. \(2(2348)=4696\).

Time = 28.73 (sec) , antiderivative size = 29732, normalized size of antiderivative = 12.66 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \] Input:

Integrate[((e + f*x)^3*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 7.83 (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))}\)

Input:

Int[((e + f*x)^3*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

(((-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 ...
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\left (f x +e \right )^{3} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]

Input:

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Fricas [B] (verification not implemented)

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 = 0.88 (sec) , antiderivative size = 10614, normalized size of antiderivative = 4.52 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((f*x+e)**3*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

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
 

Giac [F]

\[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((f*x + e)^3*sin(d*x + c)/(b*sin(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \] Input:

int((sin(c + d*x)*(e + f*x)^3)/(a + b*sin(c + d*x))^3,x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {too large to display} \] Input:

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

(480*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*si 
n(c + d*x)**2*a**12*b**2*d*e*f**2 - 624*sqrt(a**2 - b**2)*atan((tan((c + d 
*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**11*b**3*f**3 - 96*sqrt 
(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x 
)**2*a**10*b**4*d**3*e**3 - 1920*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)* 
a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**10*b**4*d*e*f**2 + 288*sqrt(a 
**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)* 
*2*a**9*b**5*d**2*e**2*f - 72*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + 
 b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**9*b**5*f**3 + 720*sqrt(a**2 - b* 
*2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**8* 
b**6*d**3*e**3 - 840*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt( 
a**2 - b**2))*sin(c + d*x)**2*a**8*b**6*d*e*f**2 - 2304*sqrt(a**2 - b**2)* 
atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**7*b**7 
*d**2*e**2*f + 336*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a* 
*2 - b**2))*sin(c + d*x)**2*a**7*b**7*f**3 - 1800*sqrt(a**2 - b**2)*atan(( 
tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**6*b**8*d**3* 
e**3 + 9192*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b* 
*2))*sin(c + d*x)**2*a**6*b**8*d*e*f**2 + 6696*sqrt(a**2 - b**2)*atan((tan 
((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**5*b**9*d**2*e** 
2*f + 24840*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 -...