3.4.4 \(\int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [304]

3.4.4.1 Optimal result

Integrand size = 26, antiderivative size = 351 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {f x}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {i \left (a^2-b^2\right ) f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {a (e+f x) \sin (c+d x)}{b^2 d}-\frac {f \cos (c+d x) \sin (c+d x)}{4 b d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 b d} \]

output

1/4*f*x/b/d+1/2*I*(a^2-b^2)*(f*x+e)^2/b^3/f+a*f*cos(d*x+c)/b^2/d^2-(a^2-b^ 
2)*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d-(a^2-b^2)*(f 
*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d+I*(a^2-b^2)*f*pol 
ylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^2+I*(a^2-b^2)*f*polyl 
og(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^2+a*(f*x+e)*sin(d*x+c)/ 
b^2/d-1/4*f*cos(d*x+c)*sin(d*x+c)/b/d^2-1/2*(f*x+e)*sin(d*x+c)^2/b/d
 

3.4.4.2 Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(816\) vs. \(2(351)=702\).

Time = 2.42 (sec) , antiderivative size = 816, normalized size of antiderivative = 2.32 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {8 a b f \cos (c+d x)+2 b^2 d (e+f x) \cos (2 (c+d x))-8 a^2 d e \log \left (1+\frac {b \sin (c+d x)}{a}\right )+8 b^2 d e \log \left (1+\frac {b \sin (c+d x)}{a}\right )+8 a^2 c f \log \left (1+\frac {b \sin (c+d x)}{a}\right )-8 b^2 c f \log \left (1+\frac {b \sin (c+d x)}{a}\right )-a^2 f \left (i (-2 c+\pi -2 d x)^2-32 i \arcsin \left (\frac {\sqrt {\frac {a+b}{b}}}{\sqrt {2}}\right ) \arctan \left (\frac {(a-b) \cot \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}{\sqrt {a^2-b^2}}\right )-4 \left (-2 c+\pi -2 d x+4 \arcsin \left (\frac {\sqrt {\frac {a+b}{b}}}{\sqrt {2}}\right )\right ) \log \left (1-\frac {i \left (-a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )-4 \left (-2 c+\pi -2 d x-4 \arcsin \left (\frac {\sqrt {\frac {a+b}{b}}}{\sqrt {2}}\right )\right ) \log \left (1+\frac {i \left (a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )+4 (-2 c+\pi -2 d x) \log (a+b \sin (c+d x))+8 (c+d x) \log (a+b \sin (c+d x))+8 i \left (\operatorname {PolyLog}\left (2,\frac {i \left (-a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )+\operatorname {PolyLog}\left (2,-\frac {i \left (a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )\right )\right )+b^2 f \left (i (-2 c+\pi -2 d x)^2-32 i \arcsin \left (\frac {\sqrt {\frac {a+b}{b}}}{\sqrt {2}}\right ) \arctan \left (\frac {(a-b) \cot \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}{\sqrt {a^2-b^2}}\right )-4 \left (-2 c+\pi -2 d x+4 \arcsin \left (\frac {\sqrt {\frac {a+b}{b}}}{\sqrt {2}}\right )\right ) \log \left (1-\frac {i \left (-a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )-4 \left (-2 c+\pi -2 d x-4 \arcsin \left (\frac {\sqrt {\frac {a+b}{b}}}{\sqrt {2}}\right )\right ) \log \left (1+\frac {i \left (a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )+4 (-2 c+\pi -2 d x) \log (a+b \sin (c+d x))+8 (c+d x) \log (a+b \sin (c+d x))+8 i \left (\operatorname {PolyLog}\left (2,\frac {i \left (-a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )+\operatorname {PolyLog}\left (2,-\frac {i \left (a+\sqrt {a^2-b^2}\right ) e^{-i (c+d x)}}{b}\right )\right )\right )+8 a b d (e+f x) \sin (c+d x)-b^2 f \sin (2 (c+d x))}{8 b^3 d^2} \]

input

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

output

(8*a*b*f*Cos[c + d*x] + 2*b^2*d*(e + f*x)*Cos[2*(c + d*x)] - 8*a^2*d*e*Log 
[1 + (b*Sin[c + d*x])/a] + 8*b^2*d*e*Log[1 + (b*Sin[c + d*x])/a] + 8*a^2*c 
*f*Log[1 + (b*Sin[c + d*x])/a] - 8*b^2*c*f*Log[1 + (b*Sin[c + d*x])/a] - a 
^2*f*(I*(-2*c + Pi - 2*d*x)^2 - (32*I)*ArcSin[Sqrt[(a + b)/b]/Sqrt[2]]*Arc 
Tan[((a - b)*Cot[(2*c + Pi + 2*d*x)/4])/Sqrt[a^2 - b^2]] - 4*(-2*c + Pi - 
2*d*x + 4*ArcSin[Sqrt[(a + b)/b]/Sqrt[2]])*Log[1 - (I*(-a + Sqrt[a^2 - b^2 
]))/(b*E^(I*(c + d*x)))] - 4*(-2*c + Pi - 2*d*x - 4*ArcSin[Sqrt[(a + b)/b] 
/Sqrt[2]])*Log[1 + (I*(a + Sqrt[a^2 - b^2]))/(b*E^(I*(c + d*x)))] + 4*(-2* 
c + Pi - 2*d*x)*Log[a + b*Sin[c + d*x]] + 8*(c + d*x)*Log[a + b*Sin[c + d* 
x]] + (8*I)*(PolyLog[2, (I*(-a + Sqrt[a^2 - b^2]))/(b*E^(I*(c + d*x)))] + 
PolyLog[2, ((-I)*(a + Sqrt[a^2 - b^2]))/(b*E^(I*(c + d*x)))])) + b^2*f*(I* 
(-2*c + Pi - 2*d*x)^2 - (32*I)*ArcSin[Sqrt[(a + b)/b]/Sqrt[2]]*ArcTan[((a 
- b)*Cot[(2*c + Pi + 2*d*x)/4])/Sqrt[a^2 - b^2]] - 4*(-2*c + Pi - 2*d*x + 
4*ArcSin[Sqrt[(a + b)/b]/Sqrt[2]])*Log[1 - (I*(-a + Sqrt[a^2 - b^2]))/(b*E 
^(I*(c + d*x)))] - 4*(-2*c + Pi - 2*d*x - 4*ArcSin[Sqrt[(a + b)/b]/Sqrt[2] 
])*Log[1 + (I*(a + Sqrt[a^2 - b^2]))/(b*E^(I*(c + d*x)))] + 4*(-2*c + Pi - 
 2*d*x)*Log[a + b*Sin[c + d*x]] + 8*(c + d*x)*Log[a + b*Sin[c + d*x]] + (8 
*I)*(PolyLog[2, (I*(-a + Sqrt[a^2 - b^2]))/(b*E^(I*(c + d*x)))] + PolyLog[ 
2, ((-I)*(a + Sqrt[a^2 - b^2]))/(b*E^(I*(c + d*x)))])) + 8*a*b*d*(e + f*x) 
*Sin[c + d*x] - b^2*f*Sin[2*(c + d*x)])/(8*b^3*d^2)
 

3.4.4.3 Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.91, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5036, 3042, 3777, 25, 3042, 3118, 4904, 3042, 3115, 24, 5030, 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.

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

\(\Big \downarrow \) 5036

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

\(\Big \downarrow \) 4904

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 5030

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

input

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

output

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

3.4.4.3.1 Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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 4904

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

rule 5030

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[ 
(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[(-I)*((e + f*x)^(m + 1)/(b*f*(m + 1 
))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] - I*b*E^( 
I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] 
 - I*b*E^(I*(c + d*x)))), x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] 
 && PosQ[a^2 - b^2]
 

rule 5036

Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.) 
*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[a/b^2   Int[(e + f*x)^m*Cos[c 
+ d*x]^(n - 2), x], x] + (-Simp[1/b   Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)* 
Sin[c + d*x], x], x] - Simp[(a^2 - b^2)/b^2   Int[(e + f*x)^m*(Cos[c + d*x] 
^(n - 2)/(a + b*Sin[c + d*x])), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && 
IGtQ[n, 1] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

3.4.4.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1749 vs. \(2 (320 ) = 640\).

Time = 1.27 (sec) , antiderivative size = 1750, normalized size of antiderivative = 4.99

method	result	size
risch	\(\text {Expression too large to display}\)	\(1750\)

input

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

output

1/2*I*a*(d*x*f-I*f+d*e)/d^2/b^2*exp(-I*(d*x+c))-2/d/b*a^2*f/(-a^2+b^2)*ln( 
(I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+1/d/b^3* 
a^4*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2 
)^(1/2)))*x-2/d/b*a^2*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/ 
2))/(I*a+(-a^2+b^2)^(1/2)))*x+1/d^2/b^3*a^4*f/(-a^2+b^2)*ln((I*a+b*exp(I*( 
d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c-2/d^2/b*a^2*f/(-a^2+b^ 
2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c+1/ 
d^2/b^3*a^4*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+( 
-a^2+b^2)^(1/2)))*c-2/d^2/b*a^2*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^ 
2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*c+2*I/d^2/b*a^2*f/(-a^2+b^2)*dilog(( 
I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-I/d^2/b^3*a 
^4*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b 
^2)^(1/2)))+2*I/d^2/b*a^2*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b 
^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))+2*I/d/b^3*a^2*f*c*x-I/d^2/b^3*a^4*f/(-a 
^2+b^2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2 
)))+2/d^2/b*c*f*ln(exp(I*(d*x+c)))-1/d^2/b*c*f*ln(I*b*exp(2*I*(d*x+c))-I*b 
-2*a*exp(I*(d*x+c)))+2/d/b^3*a^2*e*ln(exp(I*(d*x+c)))-1/d/b^3*a^2*e*ln(I*b 
*exp(2*I*(d*x+c))-I*b-2*a*exp(I*(d*x+c)))-I/d^2/b*f*c^2+1/2*I/b^3*a^2*f*x^ 
2-1/2*I*a*(d*x*f+I*f+d*e)/d^2/b^2*exp(I*(d*x+c))+1/d*b*f/(-a^2+b^2)*ln((I* 
a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+1/d*b*f/...
 

3.4.4.5 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. 1037 vs. \(2 (315) = 630\).

Time = 0.43 (sec) , antiderivative size = 1037, normalized size of antiderivative = 2.95 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

input

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

output

-1/4*(b^2*d*f*x - 4*a*b*f*cos(d*x + c) - 2*(b^2*d*f*x + b^2*d*e)*cos(d*x + 
 c)^2 - 2*I*(a^2 - b^2)*f*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*co 
s(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*I*(a 
^2 - b^2)*f*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*I*(a^2 - b^2)*f*di 
log((-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*I*(a^2 - b^2)*f*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^2 - b^2)*d*e - (a^2 - b^2)*c*f)*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) + 
2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*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) + 2*((a^2 - b^2)*d*e - (a^2 - 
b^2)*c*f)*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) + 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*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) + 2*((a^2 
- b^2)*d*f*x + (a^2 - b^2)*c*f)*log(-(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) + 2*((a 
^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(-(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) + 2* 
((a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(-(-I*a*cos(d*x + c) - a*sin(d...
 

3.4.4.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.4.4.7 Maxima [F(-2)]

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

input

integrate((f*x+e)*cos(d*x+c)^3/(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
 

3.4.4.8 Giac [F]

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

input

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

output

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

3.4.4.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}