3.95 \(\int \frac{\cos ^2(a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=338 \[ \frac{25 b^2 \sin \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b c}{d}+5 b x\right )}{32 d^3}-\frac{9 b^2 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^3}-\frac{b^2 \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{16 d^3}-\frac{b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{16 d^3}-\frac{9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^3}+\frac{25 b^2 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{32 d^3}-\frac{b \cos (a+b x)}{16 d^2 (c+d x)}-\frac{3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac{5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac{\sin (a+b x)}{16 d (c+d x)^2}-\frac{\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac{\sin (5 a+5 b x)}{32 d (c+d x)^2} \]

[Out]

-(b*Cos[a + b*x])/(16*d^2*(c + d*x)) - (3*b*Cos[3*a + 3*b*x])/(32*d^2*(c + d*x)) + (5*b*Cos[5*a + 5*b*x])/(32*
d^2*(c + d*x)) + (25*b^2*CosIntegral[(5*b*c)/d + 5*b*x]*Sin[5*a - (5*b*c)/d])/(32*d^3) - (9*b^2*CosIntegral[(3
*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/(32*d^3) - (b^2*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/(16*d^3) -
Sin[a + b*x]/(16*d*(c + d*x)^2) - Sin[3*a + 3*b*x]/(32*d*(c + d*x)^2) + Sin[5*a + 5*b*x]/(32*d*(c + d*x)^2) -
(b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(16*d^3) - (9*b^2*Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/
d + 3*b*x])/(32*d^3) + (25*b^2*Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x])/(32*d^3)

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Rubi [A]  time = 0.50475, antiderivative size = 338, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac{25 b^2 \sin \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b c}{d}+5 b x\right )}{32 d^3}-\frac{9 b^2 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^3}-\frac{b^2 \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{16 d^3}-\frac{b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{16 d^3}-\frac{9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^3}+\frac{25 b^2 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{32 d^3}-\frac{b \cos (a+b x)}{16 d^2 (c+d x)}-\frac{3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac{5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac{\sin (a+b x)}{16 d (c+d x)^2}-\frac{\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac{\sin (5 a+5 b x)}{32 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[a + b*x])/(16*d^2*(c + d*x)) - (3*b*Cos[3*a + 3*b*x])/(32*d^2*(c + d*x)) + (5*b*Cos[5*a + 5*b*x])/(32*
d^2*(c + d*x)) + (25*b^2*CosIntegral[(5*b*c)/d + 5*b*x]*Sin[5*a - (5*b*c)/d])/(32*d^3) - (9*b^2*CosIntegral[(3
*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/(32*d^3) - (b^2*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/(16*d^3) -
Sin[a + b*x]/(16*d*(c + d*x)^2) - Sin[3*a + 3*b*x]/(32*d*(c + d*x)^2) + Sin[5*a + 5*b*x]/(32*d*(c + d*x)^2) -
(b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(16*d^3) - (9*b^2*Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/
d + 3*b*x])/(32*d^3) + (25*b^2*Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x])/(32*d^3)

Rule 4406

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

Rule 3297

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

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac{\sin (a+b x)}{8 (c+d x)^3}+\frac{\sin (3 a+3 b x)}{16 (c+d x)^3}-\frac{\sin (5 a+5 b x)}{16 (c+d x)^3}\right ) \, dx\\ &=\frac{1}{16} \int \frac{\sin (3 a+3 b x)}{(c+d x)^3} \, dx-\frac{1}{16} \int \frac{\sin (5 a+5 b x)}{(c+d x)^3} \, dx+\frac{1}{8} \int \frac{\sin (a+b x)}{(c+d x)^3} \, dx\\ &=-\frac{\sin (a+b x)}{16 d (c+d x)^2}-\frac{\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac{\sin (5 a+5 b x)}{32 d (c+d x)^2}+\frac{b \int \frac{\cos (a+b x)}{(c+d x)^2} \, dx}{16 d}+\frac{(3 b) \int \frac{\cos (3 a+3 b x)}{(c+d x)^2} \, dx}{32 d}-\frac{(5 b) \int \frac{\cos (5 a+5 b x)}{(c+d x)^2} \, dx}{32 d}\\ &=-\frac{b \cos (a+b x)}{16 d^2 (c+d x)}-\frac{3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac{5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac{\sin (a+b x)}{16 d (c+d x)^2}-\frac{\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac{\sin (5 a+5 b x)}{32 d (c+d x)^2}-\frac{b^2 \int \frac{\sin (a+b x)}{c+d x} \, dx}{16 d^2}-\frac{\left (9 b^2\right ) \int \frac{\sin (3 a+3 b x)}{c+d x} \, dx}{32 d^2}+\frac{\left (25 b^2\right ) \int \frac{\sin (5 a+5 b x)}{c+d x} \, dx}{32 d^2}\\ &=-\frac{b \cos (a+b x)}{16 d^2 (c+d x)}-\frac{3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac{5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac{\sin (a+b x)}{16 d (c+d x)^2}-\frac{\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac{\sin (5 a+5 b x)}{32 d (c+d x)^2}+\frac{\left (25 b^2 \cos \left (5 a-\frac{5 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{5 b c}{d}+5 b x\right )}{c+d x} \, dx}{32 d^2}-\frac{\left (9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{32 d^2}-\frac{\left (b^2 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{16 d^2}+\frac{\left (25 b^2 \sin \left (5 a-\frac{5 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{5 b c}{d}+5 b x\right )}{c+d x} \, dx}{32 d^2}-\frac{\left (9 b^2 \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{32 d^2}-\frac{\left (b^2 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{16 d^2}\\ &=-\frac{b \cos (a+b x)}{16 d^2 (c+d x)}-\frac{3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac{5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}+\frac{25 b^2 \text{Ci}\left (\frac{5 b c}{d}+5 b x\right ) \sin \left (5 a-\frac{5 b c}{d}\right )}{32 d^3}-\frac{9 b^2 \text{Ci}\left (\frac{3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac{3 b c}{d}\right )}{32 d^3}-\frac{b^2 \text{Ci}\left (\frac{b c}{d}+b x\right ) \sin \left (a-\frac{b c}{d}\right )}{16 d^3}-\frac{\sin (a+b x)}{16 d (c+d x)^2}-\frac{\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac{\sin (5 a+5 b x)}{32 d (c+d x)^2}-\frac{b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{16 d^3}-\frac{9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^3}+\frac{25 b^2 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{32 d^3}\\ \end{align*}

Mathematica [A]  time = 4.21656, size = 279, normalized size = 0.83 \[ \frac{-2 \left (b^2 \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )+b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )+\frac{d (b (c+d x) \cos (a+b x)+d \sin (a+b x))}{(c+d x)^2}\right )+25 b^2 \sin \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b (c+d x)}{d}\right )-9 b^2 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )-9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )+25 b^2 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b (c+d x)}{d}\right )-\frac{d (3 b (c+d x) \cos (3 (a+b x))+d \sin (3 (a+b x)))}{(c+d x)^2}+\frac{d (5 b (c+d x) \cos (5 (a+b x))+d \sin (5 (a+b x)))}{(c+d x)^2}}{32 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*b^2*CosIntegral[(5*b*(c + d*x))/d]*Sin[5*a - (5*b*c)/d] - 9*b^2*CosIntegral[(3*b*(c + d*x))/d]*Sin[3*a - (
3*b*c)/d] - (d*(3*b*(c + d*x)*Cos[3*(a + b*x)] + d*Sin[3*(a + b*x)]))/(c + d*x)^2 + (d*(5*b*(c + d*x)*Cos[5*(a
+ b*x)] + d*Sin[5*(a + b*x)]))/(c + d*x)^2 - 2*(b^2*CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + (d*(b*(c + d*
x)*Cos[a + b*x] + d*Sin[a + b*x]))/(c + d*x)^2 + b^2*Cos[a - (b*c)/d]*SinIntegral[b*(c/d + x)]) - 9*b^2*Cos[3*
a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d] + 25*b^2*Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*(c + d*x))/d])/(3
2*d^3)

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Maple [A]  time = 0.027, size = 475, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ( -{\frac{{b}^{3}}{80} \left ( -{\frac{5\,\sin \left ( 5\,bx+5\,a \right ) }{2\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+{\frac{5}{2\,d} \left ( -5\,{\frac{\cos \left ( 5\,bx+5\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-5\,{\frac{1}{d} \left ( 5\,{\frac{1}{d}{\it Si} \left ( 5\,bx+5\,a+5\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 5\,{\frac{-ad+bc}{d}} \right ) }-5\,{\frac{1}{d}{\it Ci} \left ( 5\,bx+5\,a+5\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 5\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) }+{\frac{{b}^{3}}{8} \left ( -{\frac{\sin \left ( bx+a \right ) }{2\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+{\frac{1}{2\,d} \left ( -{\frac{\cos \left ( bx+a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \cos \left ({\frac{-ad+bc}{d}} \right ) }-{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \sin \left ({\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) }+{\frac{{b}^{3}}{48} \left ( -{\frac{3\,\sin \left ( 3\,bx+3\,a \right ) }{2\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+{\frac{3}{2\,d} \left ( -3\,{\frac{\cos \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-3\,{\frac{1}{d} \left ( 3\,{\frac{1}{d}{\it Si} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 3\,{\frac{-ad+bc}{d}} \right ) }-3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 3\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/80*b^3*(-5/2*sin(5*b*x+5*a)/((b*x+a)*d-a*d+b*c)^2/d+5/2*(-5*cos(5*b*x+5*a)/((b*x+a)*d-a*d+b*c)/d-5*(5*
Si(5*b*x+5*a+5*(-a*d+b*c)/d)*cos(5*(-a*d+b*c)/d)/d-5*Ci(5*b*x+5*a+5*(-a*d+b*c)/d)*sin(5*(-a*d+b*c)/d)/d)/d)/d)
+1/8*b^3*(-1/2*sin(b*x+a)/((b*x+a)*d-a*d+b*c)^2/d+1/2*(-cos(b*x+a)/((b*x+a)*d-a*d+b*c)/d-(Si(b*x+a+(-a*d+b*c)/
d)*cos((-a*d+b*c)/d)/d-Ci(b*x+a+(-a*d+b*c)/d)*sin((-a*d+b*c)/d)/d)/d)/d)+1/48*b^3*(-3/2*sin(3*b*x+3*a)/((b*x+a
)*d-a*d+b*c)^2/d+3/2*(-3*cos(3*b*x+3*a)/((b*x+a)*d-a*d+b*c)/d-3*(3*Si(3*b*x+3*a+3*(-a*d+b*c)/d)*cos(3*(-a*d+b*
c)/d)/d-3*Ci(3*b*x+3*a+3*(-a*d+b*c)/d)*sin(3*(-a*d+b*c)/d)/d)/d)/d))

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Maxima [C]  time = 3.00987, size = 639, normalized size = 1.89 \begin{align*} \frac{b^{3}{\left (-2 i \, E_{3}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + 2 i \, E_{3}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac{b c - a d}{d}\right ) + b^{3}{\left (-i \, E_{3}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + i \, E_{3}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + b^{3}{\left (i \, E_{3}\left (\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \cos \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right ) - 2 \, b^{3}{\left (E_{3}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) - b^{3}{\left (E_{3}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{3}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + b^{3}{\left (E_{3}\left (\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{3}\left (-\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \sin \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right )}{32 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} +{\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \,{\left (b c d^{2} - a d^{3}\right )}{\left (b x + a\right )}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^3/(d*x+c)^3,x, algorithm="maxima")

[Out]

1/32*(b^3*(-2*I*exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 2*I*exp_integral_e(3, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^3*(-I*exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) +
I*exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b^3*(I*exp_integral_e(
3, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) - I*exp_integral_e(3, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*c
os(-5*(b*c - a*d)/d) - 2*b^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(3, -(I*b*c
+ I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b^3*(exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*
d)/d) + exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d) + b^3*(exp_integral
_e(3, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) + exp_integral_e(3, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*
sin(-5*(b*c - a*d)/d))/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*
b)

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Fricas [A]  time = 0.759665, size = 1358, normalized size = 4.02 \begin{align*} \frac{160 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{5} - 224 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} + 50 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{5 \,{\left (b d x + b c\right )}}{d}\right ) - 18 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) - 4 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{Si}\left (\frac{b d x + b c}{d}\right ) + 64 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) + 32 \,{\left (d^{2} \cos \left (b x + a\right )^{4} - d^{2} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right ) - 2 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{b d x + b c}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (-\frac{b d x + b c}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) - 9 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (-\frac{3 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + 25 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{5 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (-\frac{5 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right )}{64 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(160*(b*d^2*x + b*c*d)*cos(b*x + a)^5 - 224*(b*d^2*x + b*c*d)*cos(b*x + a)^3 + 50*(b^2*d^2*x^2 + 2*b^2*c*
d*x + b^2*c^2)*cos(-5*(b*c - a*d)/d)*sin_integral(5*(b*d*x + b*c)/d) - 18*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2
)*cos(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) - 4*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-(b*c -
a*d)/d)*sin_integral((b*d*x + b*c)/d) + 64*(b*d^2*x + b*c*d)*cos(b*x + a) + 32*(d^2*cos(b*x + a)^4 - d^2*cos(b
*x + a)^2)*sin(b*x + a) - 2*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral((b*d*x + b*c)/d) + (b^2*d^2*x^
2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-(b*d*x + b*c)/d))*sin(-(b*c - a*d)/d) - 9*((b^2*d^2*x^2 + 2*b^2*c*d*x
+ b^2*c^2)*cos_integral(3*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-3*(b*d*x + b
*c)/d))*sin(-3*(b*c - a*d)/d) + 25*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(5*(b*d*x + b*c)/d) + (b
^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-5*(b*d*x + b*c)/d))*sin(-5*(b*c - a*d)/d))/(d^5*x^2 + 2*c*d^
4*x + c^2*d^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError