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

Optimal. Leaf size=371 \[ \frac{a^{2/3} \sin \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b^{5/3}}+\frac{(-1)^{2/3} a^{2/3} \sin \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b^{5/3}}-\frac{\sqrt [3]{-1} a^{2/3} \sin \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b^{5/3}}-\frac{(-1)^{2/3} a^{2/3} \cos \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{Si}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b^{5/3}}+\frac{a^{2/3} \cos \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}-\frac{\sqrt [3]{-1} a^{2/3} \cos \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}+\frac{\sin (c+d x)}{b d^2}-\frac{x \cos (c+d x)}{b d} \]

[Out]

-((x*Cos[c + d*x])/(b*d)) + (a^(2/3)*CosIntegral[(a^(1/3)*d)/b^(1/3) + d*x]*Sin[c - (a^(1/3)*d)/b^(1/3)])/(3*b
^(5/3)) + ((-1)^(2/3)*a^(2/3)*CosIntegral[((-1)^(1/3)*a^(1/3)*d)/b^(1/3) - d*x]*Sin[c + ((-1)^(1/3)*a^(1/3)*d)
/b^(1/3)])/(3*b^(5/3)) - ((-1)^(1/3)*a^(2/3)*CosIntegral[((-1)^(2/3)*a^(1/3)*d)/b^(1/3) + d*x]*Sin[c - ((-1)^(
2/3)*a^(1/3)*d)/b^(1/3)])/(3*b^(5/3)) + Sin[c + d*x]/(b*d^2) - ((-1)^(2/3)*a^(2/3)*Cos[c + ((-1)^(1/3)*a^(1/3)
*d)/b^(1/3)]*SinIntegral[((-1)^(1/3)*a^(1/3)*d)/b^(1/3) - d*x])/(3*b^(5/3)) + (a^(2/3)*Cos[c - (a^(1/3)*d)/b^(
1/3)]*SinIntegral[(a^(1/3)*d)/b^(1/3) + d*x])/(3*b^(5/3)) - ((-1)^(1/3)*a^(2/3)*Cos[c - ((-1)^(2/3)*a^(1/3)*d)
/b^(1/3)]*SinIntegral[((-1)^(2/3)*a^(1/3)*d)/b^(1/3) + d*x])/(3*b^(5/3))

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Rubi [A]  time = 0.91509, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3345, 3296, 2637, 3303, 3299, 3302} \[ \frac{a^{2/3} \sin \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b^{5/3}}+\frac{(-1)^{2/3} a^{2/3} \sin \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b^{5/3}}-\frac{\sqrt [3]{-1} a^{2/3} \sin \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b^{5/3}}-\frac{(-1)^{2/3} a^{2/3} \cos \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{Si}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b^{5/3}}+\frac{a^{2/3} \cos \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}-\frac{\sqrt [3]{-1} a^{2/3} \cos \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}+\frac{\sin (c+d x)}{b d^2}-\frac{x \cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*Sin[c + d*x])/(a + b*x^3),x]

[Out]

-((x*Cos[c + d*x])/(b*d)) + (a^(2/3)*CosIntegral[(a^(1/3)*d)/b^(1/3) + d*x]*Sin[c - (a^(1/3)*d)/b^(1/3)])/(3*b
^(5/3)) + ((-1)^(2/3)*a^(2/3)*CosIntegral[((-1)^(1/3)*a^(1/3)*d)/b^(1/3) - d*x]*Sin[c + ((-1)^(1/3)*a^(1/3)*d)
/b^(1/3)])/(3*b^(5/3)) - ((-1)^(1/3)*a^(2/3)*CosIntegral[((-1)^(2/3)*a^(1/3)*d)/b^(1/3) + d*x]*Sin[c - ((-1)^(
2/3)*a^(1/3)*d)/b^(1/3)])/(3*b^(5/3)) + Sin[c + d*x]/(b*d^2) - ((-1)^(2/3)*a^(2/3)*Cos[c + ((-1)^(1/3)*a^(1/3)
*d)/b^(1/3)]*SinIntegral[((-1)^(1/3)*a^(1/3)*d)/b^(1/3) - d*x])/(3*b^(5/3)) + (a^(2/3)*Cos[c - (a^(1/3)*d)/b^(
1/3)]*SinIntegral[(a^(1/3)*d)/b^(1/3) + d*x])/(3*b^(5/3)) - ((-1)^(1/3)*a^(2/3)*Cos[c - ((-1)^(2/3)*a^(1/3)*d)
/b^(1/3)]*SinIntegral[((-1)^(2/3)*a^(1/3)*d)/b^(1/3) + d*x])/(3*b^(5/3))

Rule 3345

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegrand[Sin[c +
 d*x], x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ
[p, -1]) && IntegerQ[m]

Rule 3296

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 2637

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

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{x^4 \sin (c+d x)}{a+b x^3} \, dx &=\int \left (\frac{x \sin (c+d x)}{b}-\frac{a x \sin (c+d x)}{b \left (a+b x^3\right )}\right ) \, dx\\ &=\frac{\int x \sin (c+d x) \, dx}{b}-\frac{a \int \frac{x \sin (c+d x)}{a+b x^3} \, dx}{b}\\ &=-\frac{x \cos (c+d x)}{b d}-\frac{a \int \left (-\frac{\sin (c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac{(-1)^{2/3} \sin (c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}+\frac{\sqrt [3]{-1} \sin (c+d x)}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}\right ) \, dx}{b}+\frac{\int \cos (c+d x) \, dx}{b d}\\ &=-\frac{x \cos (c+d x)}{b d}+\frac{\sin (c+d x)}{b d^2}+\frac{a^{2/3} \int \frac{\sin (c+d x)}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{4/3}}-\frac{\left (\sqrt [3]{-1} a^{2/3}\right ) \int \frac{\sin (c+d x)}{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}+\frac{\left ((-1)^{2/3} a^{2/3}\right ) \int \frac{\sin (c+d x)}{\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}\\ &=-\frac{x \cos (c+d x)}{b d}+\frac{\sin (c+d x)}{b d^2}+\frac{\left (a^{2/3} \cos \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac{\sin \left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{4/3}}+\frac{\left (\sqrt [3]{-1} a^{2/3} \cos \left (c+\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac{\sin \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}+\frac{\left ((-1)^{2/3} a^{2/3} \cos \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac{\sin \left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}+\frac{\left (a^{2/3} \sin \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac{\cos \left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{4/3}}-\frac{\left (\sqrt [3]{-1} a^{2/3} \sin \left (c+\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac{\cos \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}+\frac{\left ((-1)^{2/3} a^{2/3} \sin \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac{\cos \left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x} \, dx}{3 b^{4/3}}\\ &=-\frac{x \cos (c+d x)}{b d}+\frac{a^{2/3} \text{Ci}\left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right ) \sin \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}+\frac{(-1)^{2/3} a^{2/3} \text{Ci}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right ) \sin \left (c+\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}-\frac{\sqrt [3]{-1} a^{2/3} \text{Ci}\left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right ) \sin \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b^{5/3}}+\frac{\sin (c+d x)}{b d^2}-\frac{(-1)^{2/3} a^{2/3} \cos \left (c+\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b^{5/3}}+\frac{a^{2/3} \cos \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b^{5/3}}-\frac{\sqrt [3]{-1} a^{2/3} \cos \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b^{5/3}}\\ \end{align*}

Mathematica [C]  time = 0.527392, size = 231, normalized size = 0.62 \[ \frac{-i a d^2 \text{RootSum}\left [\text{$\#$1}^3 b+a\& ,\frac{-i \sin (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))+\cos (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))-\sin (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))-i \cos (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))}{\text{$\#$1}}\& \right ]+i a d^2 \text{RootSum}\left [\text{$\#$1}^3 b+a\& ,\frac{i \sin (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))+\cos (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))-\sin (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))+i \cos (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))}{\text{$\#$1}}\& \right ]+6 b (\sin (c+d x)-d x \cos (c+d x))}{6 b^2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*a*d^2*RootSum[a + b*#1^3 & , (Cos[c + d*#1]*CosIntegral[d*(x - #1)] - I*CosIntegral[d*(x - #1)]*Sin[c +
d*#1] - I*Cos[c + d*#1]*SinIntegral[d*(x - #1)] - Sin[c + d*#1]*SinIntegral[d*(x - #1)])/#1 & ] + I*a*d^2*Root
Sum[a + b*#1^3 & , (Cos[c + d*#1]*CosIntegral[d*(x - #1)] + I*CosIntegral[d*(x - #1)]*Sin[c + d*#1] + I*Cos[c
+ d*#1]*SinIntegral[d*(x - #1)] - Sin[c + d*#1]*SinIntegral[d*(x - #1)])/#1 & ] + 6*b*(-(d*x*Cos[c + d*x]) + S
in[c + d*x]))/(6*b^2*d^2)

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Maple [C]  time = 0.026, size = 559, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*sin(d*x+c)/(b*x^3+a),x)

[Out]

1/d^5*((d^3*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-3*c*d^3*cos(d*x+c))/b+1/3/b^2*d^3*sum((6*_R1^2*b*c^2-_R1*a*d^3-8*_
R1*b*c^3-3*a*c*d^3+3*b*c^4)/(_R1^2-2*_R1*c+c^2)*(-Si(-d*x+_R1-c)*cos(_R1)+Ci(d*x-_R1+c)*sin(_R1)),_R1=RootOf(_
Z^3*b-3*_Z^2*b*c+3*_Z*b*c^2+a*d^3-b*c^3))+4*c*d^3/b*cos(d*x+c)-4/3/b^2*c*d^3*sum((3*_R1^2*b*c-3*_R1*b*c^2-a*d^
3+b*c^3)/(_R1^2-2*_R1*c+c^2)*(-Si(-d*x+_R1-c)*cos(_R1)+Ci(d*x-_R1+c)*sin(_R1)),_R1=RootOf(_Z^3*b-3*_Z^2*b*c+3*
_Z*b*c^2+a*d^3-b*c^3))+2*c^2*d^3/b*sum(_R1^2/(_R1^2-2*_R1*c+c^2)*(-Si(-d*x+_R1-c)*cos(_R1)+Ci(d*x-_R1+c)*sin(_
R1)),_R1=RootOf(_Z^3*b-3*_Z^2*b*c+3*_Z*b*c^2+a*d^3-b*c^3))-4/3*c^3*d^3/b*sum(_R1/(_R1^2-2*_R1*c+c^2)*(-Si(-d*x
+_R1-c)*cos(_R1)+Ci(d*x-_R1+c)*sin(_R1)),_R1=RootOf(_Z^3*b-3*_Z^2*b*c+3*_Z*b*c^2+a*d^3-b*c^3))+1/3*c^4*d^3/b*s
um(1/(_R1^2-2*_R1*c+c^2)*(-Si(-d*x+_R1-c)*cos(_R1)+Ci(d*x-_R1+c)*sin(_R1)),_R1=RootOf(_Z^3*b-3*_Z^2*b*c+3*_Z*b
*c^2+a*d^3-b*c^3)))

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Maxima [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(x^4*sin(d*x+c)/(b*x^3+a),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C]  time = 2.41593, size = 1030, normalized size = 2.78 \begin{align*} \frac{\left (\frac{i \, a d^{3}}{b}\right )^{\frac{2}{3}}{\left (\sqrt{3} + i\right )}{\rm Ei}\left (-i \, d x + \frac{1}{2} \, \left (\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} + 1\right )} - i \, c\right )} - \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{2}{3}}{\left (\sqrt{3} + i\right )}{\rm Ei}\left (i \, d x + \frac{1}{2} \, \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} + 1\right )} + i \, c\right )} - \left (\frac{i \, a d^{3}}{b}\right )^{\frac{2}{3}}{\left (\sqrt{3} - i\right )}{\rm Ei}\left (-i \, d x + \frac{1}{2} \, \left (\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} + 1\right )} - i \, c\right )} + \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{2}{3}}{\left (\sqrt{3} - i\right )}{\rm Ei}\left (i \, d x + \frac{1}{2} \, \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} + 1\right )} + i \, c\right )} - 12 \, d x \cos \left (d x + c\right ) + 2 i \, \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{2}{3}}{\rm Ei}\left (i \, d x + \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}\right ) e^{\left (i \, c - \left (-\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}\right )} - 2 i \, \left (\frac{i \, a d^{3}}{b}\right )^{\frac{2}{3}}{\rm Ei}\left (-i \, d x + \left (\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}\right ) e^{\left (-i \, c - \left (\frac{i \, a d^{3}}{b}\right )^{\frac{1}{3}}\right )} + 12 \, \sin \left (d x + c\right )}{12 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*sin(d*x+c)/(b*x^3+a),x, algorithm="fricas")

[Out]

1/12*((I*a*d^3/b)^(2/3)*(sqrt(3) + I)*Ei(-I*d*x + 1/2*(I*a*d^3/b)^(1/3)*(-I*sqrt(3) - 1))*e^(1/2*(I*a*d^3/b)^(
1/3)*(I*sqrt(3) + 1) - I*c) - (-I*a*d^3/b)^(2/3)*(sqrt(3) + I)*Ei(I*d*x + 1/2*(-I*a*d^3/b)^(1/3)*(-I*sqrt(3) -
 1))*e^(1/2*(-I*a*d^3/b)^(1/3)*(I*sqrt(3) + 1) + I*c) - (I*a*d^3/b)^(2/3)*(sqrt(3) - I)*Ei(-I*d*x + 1/2*(I*a*d
^3/b)^(1/3)*(I*sqrt(3) - 1))*e^(1/2*(I*a*d^3/b)^(1/3)*(-I*sqrt(3) + 1) - I*c) + (-I*a*d^3/b)^(2/3)*(sqrt(3) -
I)*Ei(I*d*x + 1/2*(-I*a*d^3/b)^(1/3)*(I*sqrt(3) - 1))*e^(1/2*(-I*a*d^3/b)^(1/3)*(-I*sqrt(3) + 1) + I*c) - 12*d
*x*cos(d*x + c) + 2*I*(-I*a*d^3/b)^(2/3)*Ei(I*d*x + (-I*a*d^3/b)^(1/3))*e^(I*c - (-I*a*d^3/b)^(1/3)) - 2*I*(I*
a*d^3/b)^(2/3)*Ei(-I*d*x + (I*a*d^3/b)^(1/3))*e^(-I*c - (I*a*d^3/b)^(1/3)) + 12*sin(d*x + c))/(b*d^2)

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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(x**4*sin(d*x+c)/(b*x**3+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \sin \left (d x + c\right )}{b x^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*sin(d*x+c)/(b*x^3+a),x, algorithm="giac")

[Out]

integrate(x^4*sin(d*x + c)/(b*x^3 + a), x)