3.1.27 \(\int \frac {x^3 \sin (c+d x)}{(a+b x)^2} \, dx\) [27]

Optimal. Leaf size=181 \[ \frac {2 a \cos (c+d x)}{b^3 d}-\frac {x \cos (c+d x)}{b^2 d}-\frac {a^3 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (\frac {a d}{b}+d x\right )}{b^5}+\frac {3 a^2 \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^4}+\frac {\sin (c+d x)}{b^2 d^2}+\frac {a^3 \sin (c+d x)}{b^4 (a+b x)}+\frac {3 a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^4}+\frac {a^3 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^5} \]

[Out]

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

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Rubi [A]
time = 0.28, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {6874, 2718, 3377, 2717, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {a^3 d \cos \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^5}+\frac {a^3 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^5}+\frac {a^3 \sin (c+d x)}{b^4 (a+b x)}+\frac {3 a^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^4}+\frac {3 a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^4}+\frac {2 a \cos (c+d x)}{b^3 d}+\frac {\sin (c+d x)}{b^2 d^2}-\frac {x \cos (c+d x)}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*Sin[c + d*x])/(a + b*x)^2,x]

[Out]

(2*a*Cos[c + d*x])/(b^3*d) - (x*Cos[c + d*x])/(b^2*d) - (a^3*d*Cos[c - (a*d)/b]*CosIntegral[(a*d)/b + d*x])/b^
5 + (3*a^2*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b^4 + Sin[c + d*x]/(b^2*d^2) + (a^3*Sin[c + d*x])/(b^4
*(a + b*x)) + (3*a^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/b^4 + (a^3*d*Sin[c - (a*d)/b]*SinIntegral[(a
*d)/b + d*x])/b^5

Rule 2717

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

Rule 2718

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

Rule 3377

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

Rule 3378

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 3380

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 3383

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]

Rule 3384

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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {x^3 \sin (c+d x)}{(a+b x)^2} \, dx &=\int \left (-\frac {2 a \sin (c+d x)}{b^3}+\frac {x \sin (c+d x)}{b^2}-\frac {a^3 \sin (c+d x)}{b^3 (a+b x)^2}+\frac {3 a^2 \sin (c+d x)}{b^3 (a+b x)}\right ) \, dx\\ &=-\frac {(2 a) \int \sin (c+d x) \, dx}{b^3}+\frac {\left (3 a^2\right ) \int \frac {\sin (c+d x)}{a+b x} \, dx}{b^3}-\frac {a^3 \int \frac {\sin (c+d x)}{(a+b x)^2} \, dx}{b^3}+\frac {\int x \sin (c+d x) \, dx}{b^2}\\ &=\frac {2 a \cos (c+d x)}{b^3 d}-\frac {x \cos (c+d x)}{b^2 d}+\frac {a^3 \sin (c+d x)}{b^4 (a+b x)}+\frac {\int \cos (c+d x) \, dx}{b^2 d}-\frac {\left (a^3 d\right ) \int \frac {\cos (c+d x)}{a+b x} \, dx}{b^4}+\frac {\left (3 a^2 \cos \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sin \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^3}+\frac {\left (3 a^2 \sin \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cos \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^3}\\ &=\frac {2 a \cos (c+d x)}{b^3 d}-\frac {x \cos (c+d x)}{b^2 d}+\frac {3 a^2 \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^4}+\frac {\sin (c+d x)}{b^2 d^2}+\frac {a^3 \sin (c+d x)}{b^4 (a+b x)}+\frac {3 a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^4}-\frac {\left (a^3 d \cos \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cos \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^4}+\frac {\left (a^3 d \sin \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sin \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^4}\\ &=\frac {2 a \cos (c+d x)}{b^3 d}-\frac {x \cos (c+d x)}{b^2 d}-\frac {a^3 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (\frac {a d}{b}+d x\right )}{b^5}+\frac {3 a^2 \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^4}+\frac {\sin (c+d x)}{b^2 d^2}+\frac {a^3 \sin (c+d x)}{b^4 (a+b x)}+\frac {3 a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^4}+\frac {a^3 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^5}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 153, normalized size = 0.85 \begin {gather*} \frac {-a^2 \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \cos \left (c-\frac {a d}{b}\right )-3 b \sin \left (c-\frac {a d}{b}\right )\right )+\frac {b \left (b d \left (2 a^2+a b x-b^2 x^2\right ) \cos (c+d x)+\left (a b^2+a^3 d^2+b^3 x\right ) \sin (c+d x)\right )}{d^2 (a+b x)}+a^2 \left (3 b \cos \left (c-\frac {a d}{b}\right )+a d \sin \left (c-\frac {a d}{b}\right )\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(850\) vs. \(2(186)=372\).
time = 0.09, size = 851, normalized size = 4.70 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^4*(-d^2*c^3*(-sin(d*x+c)/(d*a-c*b+b*(d*x+c))/b+(Si(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c
)/b)*cos((a*d-b*c)/b)/b)/b)-3*d^2*c^2*(a*d-b*c)/b*(-sin(d*x+c)/(d*a-c*b+b*(d*x+c))/b+(Si(d*x+c+(a*d-b*c)/b)*si
n((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b)/b)+3*d^2*c^2/b*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)
/b)/b-Ci(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b)-3*(a^2*d^2-2*a*b*c*d+b^2*c^2)*d^2*c/b^2*(-sin(d*x+c)/(d*a-c*b+
b*(d*x+c))/b+(Si(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b)/b)+6/b^2*(a*d
-b*c)*d^2*c*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b)+3*d^2*c/b^2*co
s(d*x+c)-(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*d^2/b^3*(-sin(d*x+c)/(d*a-c*b+b*(d*x+c))/b+(Si(d*x+c+(a
*d-b*c)/b)*sin((a*d-b*c)/b)/b+Ci(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b)/b)+3/b^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)*d
^2*(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a*d-b*c)/b)*sin((a*d-b*c)/b)/b)-d^2*(2*a*d-2*b*c-b)/b^3
*(sin(d*x+c)-(d*x+c)*cos(d*x+c)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((b^2*cos(c)^2 + b^2*sin(c)^2)*d*x^3*cos(d*x + c) - 2*((a^2*(I*exp_integral_e(3, (I*b*d*x + I*a*d)/b) - I
*exp_integral_e(3, -(I*b*d*x + I*a*d)/b))*cos(c)^2 + a^2*(I*exp_integral_e(3, (I*b*d*x + I*a*d)/b) - I*exp_int
egral_e(3, -(I*b*d*x + I*a*d)/b))*sin(c)^2)*cos(-(b*c - a*d)/b) - (a^2*(exp_integral_e(3, (I*b*d*x + I*a*d)/b)
+ exp_integral_e(3, -(I*b*d*x + I*a*d)/b))*cos(c)^2 + a^2*(exp_integral_e(3, (I*b*d*x + I*a*d)/b) + exp_integ
ral_e(3, -(I*b*d*x + I*a*d)/b))*sin(c)^2)*sin(-(b*c - a*d)/b))*cos(d*x + c)^2 - 2*((a^2*(I*exp_integral_e(3, (
I*b*d*x + I*a*d)/b) - I*exp_integral_e(3, -(I*b*d*x + I*a*d)/b))*cos(c)^2 + a^2*(I*exp_integral_e(3, (I*b*d*x
+ I*a*d)/b) - I*exp_integral_e(3, -(I*b*d*x + I*a*d)/b))*sin(c)^2)*cos(-(b*c - a*d)/b) - (a^2*(exp_integral_e(
3, (I*b*d*x + I*a*d)/b) + exp_integral_e(3, -(I*b*d*x + I*a*d)/b))*cos(c)^2 + a^2*(exp_integral_e(3, (I*b*d*x
+ I*a*d)/b) + exp_integral_e(3, -(I*b*d*x + I*a*d)/b))*sin(c)^2)*sin(-(b*c - a*d)/b))*sin(d*x + c)^2 + ((b^2*d
*x^3*cos(c) + b^2*x^2*sin(c) + 2*a*b*x*sin(c))*cos(d*x + c)^2 + (b^2*d*x^3*cos(c) + b^2*x^2*sin(c) + 2*a*b*x*s
in(c))*sin(d*x + c)^2)*cos(d*x + 2*c) + 2*(((a^2*b^4*cos(c)^2 + a^2*b^4*sin(c)^2)*d^3*x^2 + 2*(a^3*b^3*cos(c)^
2 + a^3*b^3*sin(c)^2)*d^3*x + (a^4*b^2*cos(c)^2 + a^4*b^2*sin(c)^2)*d^3)*cos(d*x + c)^2 + ((a^2*b^4*cos(c)^2 +
a^2*b^4*sin(c)^2)*d^3*x^2 + 2*(a^3*b^3*cos(c)^2 + a^3*b^3*sin(c)^2)*d^3*x + (a^4*b^2*cos(c)^2 + a^4*b^2*sin(c
)^2)*d^3)*sin(d*x + c)^2)*integrate(x*cos(d*x + c)/(b^4*d^2*x^3 + 3*a*b^3*d^2*x^2 + 3*a^2*b^2*d^2*x + a^3*b*d^
2), x) + 2*(((a^2*b^4*cos(c)^2 + a^2*b^4*sin(c)^2)*d^3*x^2 + 2*(a^3*b^3*cos(c)^2 + a^3*b^3*sin(c)^2)*d^3*x + (
a^4*b^2*cos(c)^2 + a^4*b^2*sin(c)^2)*d^3)*cos(d*x + c)^2 + ((a^2*b^4*cos(c)^2 + a^2*b^4*sin(c)^2)*d^3*x^2 + 2*
(a^3*b^3*cos(c)^2 + a^3*b^3*sin(c)^2)*d^3*x + (a^4*b^2*cos(c)^2 + a^4*b^2*sin(c)^2)*d^3)*sin(d*x + c)^2)*integ
rate(x*cos(d*x + c)/((b^4*d^2*x^3 + 3*a*b^3*d^2*x^2 + 3*a^2*b^2*d^2*x + a^3*b*d^2)*cos(d*x + c)^2 + (b^4*d^2*x
^3 + 3*a*b^3*d^2*x^2 + 3*a^2*b^2*d^2*x + a^3*b*d^2)*sin(d*x + c)^2), x) + ((b^2*d*x^3*sin(c) - b^2*x^2*cos(c)
- 2*a*b*x*cos(c))*cos(d*x + c)^2 + (b^2*d*x^3*sin(c) - b^2*x^2*cos(c) - 2*a*b*x*cos(c))*sin(d*x + c)^2)*sin(d*
x + 2*c) - ((b^2*cos(c)^2 + b^2*sin(c)^2)*x^2 + 2*(a*b*cos(c)^2 + a*b*sin(c)^2)*x)*sin(d*x + c))/(((b^4*cos(c)
^2 + b^4*sin(c)^2)*d^2*x^2 + 2*(a*b^3*cos(c)^2 + a*b^3*sin(c)^2)*d^2*x + (a^2*b^2*cos(c)^2 + a^2*b^2*sin(c)^2)
*d^2)*cos(d*x + c)^2 + ((b^4*cos(c)^2 + b^4*sin(c)^2)*d^2*x^2 + 2*(a*b^3*cos(c)^2 + a*b^3*sin(c)^2)*d^2*x + (a
^2*b^2*cos(c)^2 + a^2*b^2*sin(c)^2)*d^2)*sin(d*x + c)^2)

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Fricas [A]
time = 0.35, size = 316, normalized size = 1.75 \begin {gather*} -\frac {2 \, {\left (b^{4} d x^{2} - a b^{3} d x - 2 \, a^{2} b^{2} d\right )} \cos \left (d x + c\right ) + {\left ({\left (a^{3} b d^{3} x + a^{4} d^{3}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (a^{3} b d^{3} x + a^{4} d^{3}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) - 6 \, {\left (a^{2} b^{2} d^{2} x + a^{3} b d^{2}\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left (a^{3} b d^{2} + b^{4} x + a b^{3}\right )} \sin \left (d x + c\right ) + {\left (3 \, {\left (a^{2} b^{2} d^{2} x + a^{3} b d^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + 3 \, {\left (a^{2} b^{2} d^{2} x + a^{3} b d^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) + 2 \, {\left (a^{3} b d^{3} x + a^{4} d^{3}\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, {\left (b^{6} d^{2} x + a b^{5} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(b^4*d*x^2 - a*b^3*d*x - 2*a^2*b^2*d)*cos(d*x + c) + ((a^3*b*d^3*x + a^4*d^3)*cos_integral((b*d*x + a*
d)/b) + (a^3*b*d^3*x + a^4*d^3)*cos_integral(-(b*d*x + a*d)/b) - 6*(a^2*b^2*d^2*x + a^3*b*d^2)*sin_integral((b
*d*x + a*d)/b))*cos(-(b*c - a*d)/b) - 2*(a^3*b*d^2 + b^4*x + a*b^3)*sin(d*x + c) + (3*(a^2*b^2*d^2*x + a^3*b*d
^2)*cos_integral((b*d*x + a*d)/b) + 3*(a^2*b^2*d^2*x + a^3*b*d^2)*cos_integral(-(b*d*x + a*d)/b) + 2*(a^3*b*d^
3*x + a^4*d^3)*sin_integral((b*d*x + a*d)/b))*sin(-(b*c - a*d)/b))/(b^6*d^2*x + a*b^5*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sin {\left (c + d x \right )}}{\left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1474 vs. \(2 (186) = 372\).
time = 3.79, size = 1474, normalized size = 8.14 \begin {gather*} -\frac {{\left ({\left (b x + a\right )} a^{3} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{3} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - a^{3} b c d^{3} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a^{4} d^{4} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + {\left (b x + a\right )} a^{3} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{3} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - a^{3} b c d^{3} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a^{4} d^{4} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + 3 \, {\left (b x + a\right )} a^{2} b {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{2} \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) \sin \left (-\frac {b c - a d}{b}\right ) - 3 \, a^{2} b^{2} c d^{2} \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) \sin \left (-\frac {b c - a d}{b}\right ) + 3 \, a^{3} b d^{3} \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) \sin \left (-\frac {b c - a d}{b}\right ) - 3 \, {\left (b x + a\right )} a^{2} b {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + 3 \, a^{2} b^{2} c d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - 3 \, a^{3} b d^{3} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a^{3} b d^{3} \sin \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) + {\left (b x + a\right )}^{2} b^{2} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}^{2} \cos \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) - 2 \, {\left (b x + a\right )} b^{3} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} c \cos \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) + b^{4} c^{2} \cos \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) - {\left (b x + a\right )} a b^{2} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d \cos \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) + a b^{3} c d \cos \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) - 2 \, a^{2} b^{2} d^{2} \cos \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) + {\left (b x + a\right )} b^{3} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} \sin \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) - b^{4} c \sin \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right ) + a b^{3} d \sin \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right )\right )} b^{2}}{{\left ({\left (b x + a\right )} b^{7} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d - b^{8} c d + a b^{7} d^{2}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b*x + a)*a^3*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^3*cos(-(b*c - a*d)/b)*cos_integral(((b*x + a)*(b*c/(b*x
+ a) - a*d/(b*x + a) + d) - b*c + a*d)/b) - a^3*b*c*d^3*cos(-(b*c - a*d)/b)*cos_integral(((b*x + a)*(b*c/(b*x
+ a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + a^4*d^4*cos(-(b*c - a*d)/b)*cos_integral(((b*x + a)*(b*c/(b*x + a)
- a*d/(b*x + a) + d) - b*c + a*d)/b) + (b*x + a)*a^3*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^3*sin(-(b*c - a*d)
/b)*sin_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) - a^3*b*c*d^3*sin(-(b*c - a*d)
/b)*sin_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + a^4*d^4*sin(-(b*c - a*d)/b)*
sin_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + 3*(b*x + a)*a^2*b*(b*c/(b*x + a)
- a*d/(b*x + a) + d)*d^2*cos_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*sin(-(b*
c - a*d)/b) - 3*a^2*b^2*c*d^2*cos_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*sin(
-(b*c - a*d)/b) + 3*a^3*b*d^3*cos_integral(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*sin(
-(b*c - a*d)/b) - 3*(b*x + a)*a^2*b*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^2*cos(-(b*c - a*d)/b)*sin_integral((
(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + 3*a^2*b^2*c*d^2*cos(-(b*c - a*d)/b)*sin_integr
al(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) - 3*a^3*b*d^3*cos(-(b*c - a*d)/b)*sin_integr
al(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b) + a^3*b*d^3*sin(-(b*x + a)*(b*c/(b*x + a) -
a*d/(b*x + a) + d)/b) + (b*x + a)^2*b^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)^2*cos(-(b*x + a)*(b*c/(b*x + a) -
a*d/(b*x + a) + d)/b) - 2*(b*x + a)*b^3*(b*c/(b*x + a) - a*d/(b*x + a) + d)*c*cos(-(b*x + a)*(b*c/(b*x + a) -
a*d/(b*x + a) + d)/b) + b^4*c^2*cos(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) - (b*x + a)*a*b^2*(b*c/(
b*x + a) - a*d/(b*x + a) + d)*d*cos(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) + a*b^3*c*d*cos(-(b*x +
a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) - 2*a^2*b^2*d^2*cos(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b
) + (b*x + a)*b^3*(b*c/(b*x + a) - a*d/(b*x + a) + d)*sin(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) -
b^4*c*sin(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) + a*b^3*d*sin(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x
+ a) + d)/b))*b^2/(((b*x + a)*b^7*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d - b^8*c*d + a*b^7*d^2)*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\sin \left (c+d\,x\right )}{{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*sin(c + d*x))/(a + b*x)^2,x)

[Out]

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

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