\(\int \frac {(e+f x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [226]

Optimal result

Integrand size = 26, antiderivative size = 309 \[ \int \frac {(e+f x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a (e+f x)^2}{2 b^2 f}-\frac {(e+f x) \cos (c+d x)}{b d}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {a^2 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^2}+\frac {a^2 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^2}+\frac {f \sin (c+d x)}{b d^2} \] Output:

-1/2*a*(f*x+e)^2/b^2/f-(f*x+e)*cos(d*x+c)/b/d-I*a^2*(f*x+e)*ln(1-I*b*exp(I 
*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^2/(a^2-b^2)^(1/2)/d+I*a^2*(f*x+e)*ln(1-I* 
b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^2/(a^2-b^2)^(1/2)/d-a^2*f*polylog( 
2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^2/(a^2-b^2)^(1/2)/d^2+a^2*f*po 
lylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^2/(a^2-b^2)^(1/2)/d^2+f* 
sin(d*x+c)/b/d^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. \(1973\) vs. \(2(309)=618\).

Time = 16.53 (sec) , antiderivative size = 1973, normalized size of antiderivative = 6.39 \[ \int \frac {(e+f x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(a*(c + d*x)*(2*d*e - 2*c*f + f*(c + d*x)))/(b^2*d^2) - ((d*e - c*f + 
 f*(c + d*x))*Cos[c + d*x])/(b*d^2) + (f*Sin[c + d*x])/(b*d^2) + (((2*(d*e 
 - c*f)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] 
- (I*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + 
d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b^2] + (I*f*Log[1 - I* 
Tan[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - 
 b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] - (I*f*Log[1 - I*Tan[(c + d*x)/ 
2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^ 
2 + b^2])])/Sqrt[-a^2 + b^2] + (I*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + S 
qrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])])/Sqrt[ 
-a^2 + b^2] - (I*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqr 
t[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*f*PolyLog[2, (a*(1 + I*Tan[(c + d* 
x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*f*PolyLog[2 
, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b^ 
2] - (I*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b 
^2]))])/Sqrt[-a^2 + b^2])*((a^2*e)/(b^2*(a + b*Sin[c + d*x])) - (a^2*c*f)/ 
(b^2*d*(a + b*Sin[c + d*x])) + (a^2*f*(c + d*x))/(b^2*d*(a + b*Sin[c + d*x 
]))))/(d*((f*Log[1 - (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + 
b^2]))]*Sec[(c + d*x)/2]^2)/(2*Sqrt[-a^2 + b^2]*(1 - I*Tan[(c + d*x)/2])) 
+ (f*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[...
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5026, 3042, 3777, 3042, 3117, 5026, 17, 3042, 3804, 2694, 27, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 5026

\(\displaystyle \frac {\int (e+f x) \sin (c+d x)dx}{b}-\frac {a \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x) \sin (c+d x)dx}{b}-\frac {a \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {\frac {f \int \cos (c+d x)dx}{d}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 5026

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {\int (e+f x)dx}{b}-\frac {a \int \frac {e+f x}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {a \int \frac {e+f x}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {a \int \frac {e+f x}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3804

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {2 a \int \frac {e^{i (c+d x)} (e+f x)}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 2694

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {2 a \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {2 a \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {f \int \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {f \int \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {2 a \left (\frac {i b \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^2}{2 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\)

Input:

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

Output:

-((a*((e + f*x)^2/(2*b*f) - (2*a*(((-1/2*I)*b*(((e + f*x)*Log[1 - (I*b*E^( 
I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - (I*f*PolyLog[2, (I*b*E^(I*(c 
 + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d^2)))/Sqrt[a^2 - b^2] + ((I/2)*b*((( 
e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - (I* 
f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d^2)))/Sqrt[ 
a^2 - b^2]))/b))/b) + (-(((e + f*x)*Cos[c + d*x])/d) + (f*Sin[c + d*x])/d^ 
2)/b
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[2   Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x 
)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 5026

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. 
)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/b   Int[(e + f*x)^m*Sin[c + 
 d*x]^(n - 1), x], x] - Simp[a/b   Int[(e + f*x)^m*(Sin[c + d*x]^(n - 1)/(a 
 + b*Sin[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] & 
& IGtQ[n, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (277 ) = 554\).

Time = 2.16 (sec) , antiderivative size = 625, normalized size of antiderivative = 2.02

method	result	size
risch	\(-\frac {a f \,x^{2}}{2 b^{2}}-\frac {a e x}{b^{2}}-\frac {\left (d x f +d e +i f \right ) {\mathrm e}^{i \left (d x +c \right )}}{2 d^{2} b}-\frac {\left (d x f +d e -i f \right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 d^{2} b}+\frac {2 i a^{2} e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \,b^{2} \sqrt {-a^{2}+b^{2}}}+\frac {a^{2} f \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b -\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{d \,b^{2} \sqrt {-a^{2}+b^{2}}}-\frac {a^{2} f \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d \,b^{2} \sqrt {-a^{2}+b^{2}}}+\frac {a^{2} f \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b -\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b^{2} \sqrt {-a^{2}+b^{2}}}-\frac {a^{2} f \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i a^{2} f \operatorname {dilog}\left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b -\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i a^{2} f \operatorname {dilog}\left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 i a^{2} f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b^{2} \sqrt {-a^{2}+b^{2}}}\)	\(625\)

Input:

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

Output:

-1/2*a/b^2*f*x^2-a/b^2*e*x-1/2*(d*x*f+I*f+d*e)/d^2/b*exp(I*(d*x+c))-1/2*(d 
*x*f-I*f+d*e)/d^2/b*exp(-I*(d*x+c))+2*I/d/b^2*a^2*e/(-a^2+b^2)^(1/2)*arcta 
n(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+1/d/b^2*a^2*f/(-a^2+b^2 
)^(1/2)*ln((I*a+exp(I*(d*x+c))*b-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2))) 
*x-1/d/b^2*a^2*f/(-a^2+b^2)^(1/2)*ln((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2 
))/(I*a+(-a^2+b^2)^(1/2)))*x+1/d^2/b^2*a^2*f/(-a^2+b^2)^(1/2)*ln((I*a+exp( 
I*(d*x+c))*b-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c-1/d^2/b^2*a^2*f/( 
-a^2+b^2)^(1/2)*ln((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2) 
^(1/2)))*c-I/d^2/b^2*a^2*f/(-a^2+b^2)^(1/2)*dilog((I*a+exp(I*(d*x+c))*b-(- 
a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))+I/d^2/b^2*a^2*f/(-a^2+b^2)^(1/2)*d 
ilog((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-2*I/d 
^2/b^2*a^2*f*c/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^ 
2+b^2)^(1/2))
 

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. 1154 vs. \(2 (269) = 538\).

Time = 0.22 (sec) , antiderivative size = 1154, normalized size of antiderivative = 3.73 \[ \int \frac {(e+f x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((a^3 - a*b^2)*d^2*f*x^2 - I*a^2*b*f*sqrt(-(a^2 - b^2)/b^2)*dilog((I* 
a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt 
(-(a^2 - b^2)/b^2) - b)/b + 1) + I*a^2*b*f*sqrt(-(a^2 - b^2)/b^2)*dilog((I 
*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqr 
t(-(a^2 - b^2)/b^2) - b)/b + 1) + I*a^2*b*f*sqrt(-(a^2 - b^2)/b^2)*dilog(( 
-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*s 
qrt(-(a^2 - b^2)/b^2) - b)/b + 1) - I*a^2*b*f*sqrt(-(a^2 - b^2)/b^2)*dilog 
((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c)) 
*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 2*(a^3 - a*b^2)*d^2*e*x - 2*(a^2*b - 
 b^3)*f*sin(d*x + c) - (a^2*b*d*e - a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log( 
2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a 
) - (a^2*b*d*e - a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) - 
2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (a^2*b*d*e - a^ 
2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) 
 + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (a^2*b*d*e - a^2*b*c*f)*sqrt(-(a^ 
2 - b^2)/b^2)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 
- b^2)/b^2) - 2*I*a) + (a^2*b*d*f*x + a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*lo 
g(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c) 
)*sqrt(-(a^2 - b^2)/b^2) - b)/b) - (a^2*b*d*f*x + a^2*b*c*f)*sqrt(-(a^2 - 
b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

(2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2 
*d*e - cos(c + d*x)*a**2*b*d*e - cos(c + d*x)*a**2*b*d*f*x + cos(c + d*x)* 
b**3*d*e + cos(c + d*x)*b**3*d*f*x - int((sin(c + d*x)*x)/(sin(c + d*x)*b 
+ a),x)*a**3*b*d**2*f + int((sin(c + d*x)*x)/(sin(c + d*x)*b + a),x)*a*b** 
3*d**2*f + sin(c + d*x)*a**2*b*f - sin(c + d*x)*b**3*f - a**3*d**2*e*x + a 
*b**2*d**2*e*x)/(b**2*d**2*(a**2 - b**2))