\(\int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [249]

Optimal result

Integrand size = 26, antiderivative size = 1584 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Output:

-1/2*a*(f*x+e)^2*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^2-3/2*a^2*(f*x+e) 
^2*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))-3/2*I*a^2*(f*x+e)^2/b/(a^2-b^ 
2)^2/d-a*f*(f*x+e)/b/(a^2-b^2)/d^2/(a+b*sin(d*x+c))+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*f^2*p 
olylog(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+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^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*e 
xp(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)*pol 
ylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-3/2*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*x+c))/(a+(a^2-b^2)^(1/2) 
))/b/(a^2-b^2)^(5/2)/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^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+3/2*I*a*(f*x+e)^2*ln(1-I*b...
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(13567\) vs. \(2(1584)=3168\).

Time = 23.90 (sec) , antiderivative size = 13567, normalized size of antiderivative = 8.57 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \] Input:

Integrate[((e + f*x)^2*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 6.00 (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))}\)

Input:

Int[((e + f*x)^2*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

(((-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...
 

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]

\[\int \frac {\left (f x +e \right )^{2} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]

Input:

int((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

int((f*x+e)^2*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. 5755 vs. \(2 (1385) = 2770\).

Time = 0.44 (sec) , antiderivative size = 5755, normalized size of antiderivative = 3.63 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((f*x+e)**2*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x+e)^2*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]

\[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((f*x + e)^2*sin(d*x + c)/(b*sin(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \] Input:

int((sin(c + d*x)*(e + f*x)^2)/(a + b*sin(c + d*x))^3,x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {too large to display} \] Input:

int((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

(80*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**2*f**2 - 48*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**4*d**2*e**2 - 120*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**4*f**2 + 96*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sq 
rt(a**2 - b**2))*sin(c + d*x)**2*a**7*b**5*d*e*f + 240*sqrt(a**2 - b**2)*a 
tan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**6*b**6* 
d**2*e**2 - 440*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**6*f**2 - 528*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**7*d*e*f - 3 
00*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
c + d*x)**2*a**4*b**8*d**2*e**2 + 432*sqrt(a**2 - b**2)*atan((tan((c + d*x 
)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**4*b**8*f**2 + 912*sqrt(a 
**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)* 
*2*a**3*b**9*d*e*f + 368*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/s 
qrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**10*f**2 - 480*sqrt(a**2 - b**2)* 
atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a*b**11*d 
*e*f - 320*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b** 
2))*sin(c + d*x)**2*b**12*f**2 + 160*sqrt(a**2 - b**2)*atan((tan((c + d*x) 
/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**11*b*f**2 - 96*sqrt(a**2 ...