3.96 \(\int \frac{\cos ^2(a+b x) \sin ^3(a+b x)}{(c+d x)^4} \, dx\)

Optimal. Leaf size=413 \[ -\frac{b^3 \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{48 d^4}-\frac{9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^4}+\frac{125 b^3 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b c}{d}+5 b x\right )}{96 d^4}+\frac{b^3 \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{48 d^4}+\frac{9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^4}-\frac{125 b^3 \sin \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{96 d^4}+\frac{b^2 \sin (a+b x)}{48 d^3 (c+d x)}+\frac{3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}-\frac{25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}-\frac{b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac{5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{24 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3} \]

[Out]

-(b*Cos[a + b*x])/(48*d^2*(c + d*x)^2) - (b*Cos[3*a + 3*b*x])/(32*d^2*(c + d*x)^2) + (5*b*Cos[5*a + 5*b*x])/(9
6*d^2*(c + d*x)^2) - (b^3*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(48*d^4) - (9*b^3*Cos[3*a - (3*b*c)/d]*
CosIntegral[(3*b*c)/d + 3*b*x])/(32*d^4) + (125*b^3*Cos[5*a - (5*b*c)/d]*CosIntegral[(5*b*c)/d + 5*b*x])/(96*d
^4) - Sin[a + b*x]/(24*d*(c + d*x)^3) + (b^2*Sin[a + b*x])/(48*d^3*(c + d*x)) - Sin[3*a + 3*b*x]/(48*d*(c + d*
x)^3) + (3*b^2*Sin[3*a + 3*b*x])/(32*d^3*(c + d*x)) + Sin[5*a + 5*b*x]/(48*d*(c + d*x)^3) - (25*b^2*Sin[5*a +
5*b*x])/(96*d^3*(c + d*x)) + (b^3*Sin[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(48*d^4) + (9*b^3*Sin[3*a - (3*
b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(32*d^4) - (125*b^3*Sin[5*a - (5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x
])/(96*d^4)

________________________________________________________________________________________

Rubi [A]  time = 0.5904, antiderivative size = 413, normalized size of antiderivative = 1., number of steps used = 20, 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} \[ -\frac{b^3 \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{48 d^4}-\frac{9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^4}+\frac{125 b^3 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b c}{d}+5 b x\right )}{96 d^4}+\frac{b^3 \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{48 d^4}+\frac{9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^4}-\frac{125 b^3 \sin \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{96 d^4}+\frac{b^2 \sin (a+b x)}{48 d^3 (c+d x)}+\frac{3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}-\frac{25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}-\frac{b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac{5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{24 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]^2*Sin[a + b*x]^3)/(c + d*x)^4,x]

[Out]

-(b*Cos[a + b*x])/(48*d^2*(c + d*x)^2) - (b*Cos[3*a + 3*b*x])/(32*d^2*(c + d*x)^2) + (5*b*Cos[5*a + 5*b*x])/(9
6*d^2*(c + d*x)^2) - (b^3*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(48*d^4) - (9*b^3*Cos[3*a - (3*b*c)/d]*
CosIntegral[(3*b*c)/d + 3*b*x])/(32*d^4) + (125*b^3*Cos[5*a - (5*b*c)/d]*CosIntegral[(5*b*c)/d + 5*b*x])/(96*d
^4) - Sin[a + b*x]/(24*d*(c + d*x)^3) + (b^2*Sin[a + b*x])/(48*d^3*(c + d*x)) - Sin[3*a + 3*b*x]/(48*d*(c + d*
x)^3) + (3*b^2*Sin[3*a + 3*b*x])/(32*d^3*(c + d*x)) + Sin[5*a + 5*b*x]/(48*d*(c + d*x)^3) - (25*b^2*Sin[5*a +
5*b*x])/(96*d^3*(c + d*x)) + (b^3*Sin[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(48*d^4) + (9*b^3*Sin[3*a - (3*
b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(32*d^4) - (125*b^3*Sin[5*a - (5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x
])/(96*d^4)

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)^4} \, dx &=\int \left (\frac{\sin (a+b x)}{8 (c+d x)^4}+\frac{\sin (3 a+3 b x)}{16 (c+d x)^4}-\frac{\sin (5 a+5 b x)}{16 (c+d x)^4}\right ) \, dx\\ &=\frac{1}{16} \int \frac{\sin (3 a+3 b x)}{(c+d x)^4} \, dx-\frac{1}{16} \int \frac{\sin (5 a+5 b x)}{(c+d x)^4} \, dx+\frac{1}{8} \int \frac{\sin (a+b x)}{(c+d x)^4} \, dx\\ &=-\frac{\sin (a+b x)}{24 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3}+\frac{b \int \frac{\cos (a+b x)}{(c+d x)^3} \, dx}{24 d}+\frac{b \int \frac{\cos (3 a+3 b x)}{(c+d x)^3} \, dx}{16 d}-\frac{(5 b) \int \frac{\cos (5 a+5 b x)}{(c+d x)^3} \, dx}{48 d}\\ &=-\frac{b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac{5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{24 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac{b^2 \int \frac{\sin (a+b x)}{(c+d x)^2} \, dx}{48 d^2}-\frac{\left (3 b^2\right ) \int \frac{\sin (3 a+3 b x)}{(c+d x)^2} \, dx}{32 d^2}+\frac{\left (25 b^2\right ) \int \frac{\sin (5 a+5 b x)}{(c+d x)^2} \, dx}{96 d^2}\\ &=-\frac{b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac{5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{24 d (c+d x)^3}+\frac{b^2 \sin (a+b x)}{48 d^3 (c+d x)}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac{25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}-\frac{b^3 \int \frac{\cos (a+b x)}{c+d x} \, dx}{48 d^3}-\frac{\left (9 b^3\right ) \int \frac{\cos (3 a+3 b x)}{c+d x} \, dx}{32 d^3}+\frac{\left (125 b^3\right ) \int \frac{\cos (5 a+5 b x)}{c+d x} \, dx}{96 d^3}\\ &=-\frac{b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac{5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{24 d (c+d x)^3}+\frac{b^2 \sin (a+b x)}{48 d^3 (c+d x)}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac{25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}+\frac{\left (125 b^3 \cos \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}{96 d^3}-\frac{\left (9 b^3 \cos \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^3}-\frac{\left (b^3 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{48 d^3}-\frac{\left (125 b^3 \sin \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}{96 d^3}+\frac{\left (9 b^3 \sin \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^3}+\frac{\left (b^3 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{48 d^3}\\ &=-\frac{b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac{5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac{b^3 \cos \left (a-\frac{b c}{d}\right ) \text{Ci}\left (\frac{b c}{d}+b x\right )}{48 d^4}-\frac{9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Ci}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^4}+\frac{125 b^3 \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Ci}\left (\frac{5 b c}{d}+5 b x\right )}{96 d^4}-\frac{\sin (a+b x)}{24 d (c+d x)^3}+\frac{b^2 \sin (a+b x)}{48 d^3 (c+d x)}-\frac{\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac{3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}+\frac{\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac{25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}+\frac{b^3 \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{48 d^4}+\frac{9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{32 d^4}-\frac{125 b^3 \sin \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{96 d^4}\\ \end{align*}

Mathematica [A]  time = 3.17416, size = 457, normalized size = 1.11 \[ \frac{-2 \left (b^3 (c+d x)^3 \left (\cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )-\sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )\right )+d \cos (b x) \left (b d \cos (a) (c+d x)-\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )\right )-d \sin (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )+b d \sin (a) (c+d x)\right )\right )-27 b^3 (c+d x)^3 \left (\cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )-\sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )\right )+125 b^3 (c+d x)^3 \left (\cos \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b (c+d x)}{d}\right )-\sin \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b (c+d x)}{d}\right )\right )-d \cos (3 b x) \left (3 b d \cos (3 a) (c+d x)-\sin (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )\right )+d \cos (5 b x) \left (5 b d \cos (5 a) (c+d x)-\sin (5 a) \left (25 b^2 (c+d x)^2-2 d^2\right )\right )+d \sin (3 b x) \left (\cos (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )+3 b d \sin (3 a) (c+d x)\right )-d \sin (5 b x) \left (\cos (5 a) \left (25 b^2 (c+d x)^2-2 d^2\right )+5 b d \sin (5 a) (c+d x)\right )}{96 d^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]^2*Sin[a + b*x]^3)/(c + d*x)^4,x]

[Out]

(-(d*Cos[3*b*x]*(3*b*d*(c + d*x)*Cos[3*a] - (-2*d^2 + 9*b^2*(c + d*x)^2)*Sin[3*a])) + d*Cos[5*b*x]*(5*b*d*(c +
d*x)*Cos[5*a] - (-2*d^2 + 25*b^2*(c + d*x)^2)*Sin[5*a]) + d*((-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*a] + 3*b*d*(c
+ d*x)*Sin[3*a])*Sin[3*b*x] - d*((-2*d^2 + 25*b^2*(c + d*x)^2)*Cos[5*a] + 5*b*d*(c + d*x)*Sin[5*a])*Sin[5*b*x
] - 2*(d*Cos[b*x]*(b*d*(c + d*x)*Cos[a] - (-2*d^2 + b^2*(c + d*x)^2)*Sin[a]) - d*((-2*d^2 + b^2*(c + d*x)^2)*C
os[a] + b*d*(c + d*x)*Sin[a])*Sin[b*x] + b^3*(c + d*x)^3*(Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)] - Sin[a -
(b*c)/d]*SinIntegral[b*(c/d + x)])) - 27*b^3*(c + d*x)^3*(Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*(c + d*x))/d]
- Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d]) + 125*b^3*(c + d*x)^3*(Cos[5*a - (5*b*c)/d]*CosIntegral
[(5*b*(c + d*x))/d] - Sin[5*a - (5*b*c)/d]*SinIntegral[(5*b*(c + d*x))/d]))/(96*d^4*(c + d*x)^3)

________________________________________________________________________________________

Maple [A]  time = 0.028, size = 580, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ( -{\frac{{b}^{4}}{80} \left ( -{\frac{5\,\sin \left ( 5\,bx+5\,a \right ) }{3\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{5}{3\,d} \left ( -{\frac{5\,\cos \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{\sin \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 ) \sin \left ( 5\,{\frac{-ad+bc}{d}} \right ) }+5\,{\frac{1}{d}{\it Ci} \left ( 5\,bx+5\,a+5\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 5\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) }+{\frac{{b}^{4}}{8} \left ( -{\frac{\sin \left ( bx+a \right ) }{3\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{1}{3\,d} \left ( -{\frac{\cos \left ( bx+a \right ) }{2\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}-{\frac{1}{2\,d} \left ( -{\frac{\sin \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 ) \sin \left ({\frac{-ad+bc}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \cos \left ({\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) }+{\frac{{b}^{4}}{48} \left ( -{\frac{\sin \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{1}{d} \left ( -{\frac{3\,\cos \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{\sin \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 ) \sin \left ( 3\,{\frac{-ad+bc}{d}} \right ) }+3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 3\,{\frac{-ad+bc}{d}} \right ) } \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)^4,x)

[Out]

1/b*(-1/80*b^4*(-5/3*sin(5*b*x+5*a)/((b*x+a)*d-a*d+b*c)^3/d+5/3*(-5/2*cos(5*b*x+5*a)/((b*x+a)*d-a*d+b*c)^2/d-5
/2*(-5*sin(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)*sin(5*(-a*d+b*c)/d)/d+5*Ci(5*b*x
+5*a+5*(-a*d+b*c)/d)*cos(5*(-a*d+b*c)/d)/d)/d)/d)/d)+1/8*b^4*(-1/3*sin(b*x+a)/((b*x+a)*d-a*d+b*c)^3/d+1/3*(-1/
2*cos(b*x+a)/((b*x+a)*d-a*d+b*c)^2/d-1/2*(-sin(b*x+a)/((b*x+a)*d-a*d+b*c)/d+(Si(b*x+a+(-a*d+b*c)/d)*sin((-a*d+
b*c)/d)/d+Ci(b*x+a+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d)/d)/d)/d)+1/48*b^4*(-sin(3*b*x+3*a)/((b*x+a)*d-a*d+b*c)^3
/d+(-3/2*cos(3*b*x+3*a)/((b*x+a)*d-a*d+b*c)^2/d-3/2*(-3*sin(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)*sin(3*(-a*d+b*c)/d)/d+3*Ci(3*b*x+3*a+3*(-a*d+b*c)/d)*cos(3*(-a*d+b*c)/d)/d)/d)/d)/d))

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Maxima [C]  time = 4.00839, size = 706, normalized size = 1.71 \begin{align*} \frac{b^{4}{\left (-2 i \, E_{4}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + 2 i \, E_{4}\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^{4}{\left (-i \, E_{4}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + i \, E_{4}\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^{4}{\left (i \, E_{4}\left (\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - i \, E_{4}\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^{4}{\left (E_{4}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{4}\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^{4}{\left (E_{4}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{4}\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^{4}{\left (E_{4}\left (\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{4}\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^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} +{\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \,{\left (b c d^{3} - a d^{4}\right )}{\left (b x + a\right )}^{2} + 3 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\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)^4,x, algorithm="maxima")

[Out]

1/32*(b^4*(-2*I*exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 2*I*exp_integral_e(4, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^4*(-I*exp_integral_e(4, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) +
I*exp_integral_e(4, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b^4*(I*exp_integral_e(
4, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) - I*exp_integral_e(4, -(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^4*(exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(4, -(I*b*c
+ I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b^4*(exp_integral_e(4, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*
d)/d) + exp_integral_e(4, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d) + b^4*(exp_integral
_e(4, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) + exp_integral_e(4, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*
sin(-5*(b*c - a*d)/d))/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3
- a*d^4)*(b*x + a)^2 + 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a))*b)

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Fricas [B]  time = 0.81492, size = 1841, normalized size = 4.46 \begin{align*} \text{result too large to display} \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)^4,x, algorithm="fricas")

[Out]

1/192*(160*(b*d^3*x + b*c*d^2)*cos(b*x + a)^5 - 224*(b*d^3*x + b*c*d^2)*cos(b*x + a)^3 - 250*(b^3*d^3*x^3 + 3*
b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*sin(-5*(b*c - a*d)/d)*sin_integral(5*(b*d*x + b*c)/d) + 54*(b^3*d^3*x
^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*sin(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) + 4*(b^3
*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*sin(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) + 64*(
b*d^3*x + b*c*d^2)*cos(b*x + a) - 2*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral((b
*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-(b*d*x + b*c)/d))*cos
(-(b*c - a*d)/d) - 27*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(3*(b*d*x + b*c)/
d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-3*(b*d*x + b*c)/d))*cos(-3*(b*c -
a*d)/d) + 125*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(5*(b*d*x + b*c)/d) + (b
^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-5*(b*d*x + b*c)/d))*cos(-5*(b*c - a*d)/d
) - 32*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d + (25*b^2*d^3*x^2 + 50*b^2*c*d^2*x + 25*b^2*c^2*d - 2*d^3)
*cos(b*x + a)^4 - (21*b^2*d^3*x^2 + 42*b^2*c*d^2*x + 21*b^2*c^2*d - 2*d^3)*cos(b*x + a)^2)*sin(b*x + a))/(d^7*
x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^4)

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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)**4,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)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError