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

Optimal. Leaf size=281 \[ \frac{\sin \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b}+\frac{\sin \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b}+\frac{\sin \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b}-\frac{\cos \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{Si}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b}+\frac{\cos \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b}+\frac{\cos \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.452726, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3345, 3303, 3299, 3302} \[ \frac{\sin \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b}+\frac{\sin \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{CosIntegral}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b}+\frac{\sin \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{CosIntegral}\left (\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 b}-\frac{\cos \left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text{Si}\left (\frac{\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 b}+\frac{\cos \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b}+\frac{\cos \left (c-\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text{Si}\left (x d+\frac{(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sin[c + d*x])/(a + b*x^3),x]

[Out]

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

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

Mathematica [C]  time = 0.319062, size = 186, normalized size = 0.66 \[ \frac{i \left (\text{RootSum}\left [\text{$\#$1}^3 b+a\& ,-i \sin (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))+\cos (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))-\sin (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))-i \cos (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))\& \right ]-\text{RootSum}\left [\text{$\#$1}^3 b+a\& ,i \sin (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))+\cos (\text{$\#$1} d+c) \text{CosIntegral}(d (x-\text{$\#$1}))-\sin (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))+i \cos (\text{$\#$1} d+c) \text{Si}(d (x-\text{$\#$1}))\& \right ]\right )}{6 b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/6)*(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)] & ] - 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]*SinIntegra
l[d*(x - #1)] - Sin[c + d*#1]*SinIntegral[d*(x - #1)] & ]))/b

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Maple [C]  time = 0.012, size = 266, normalized size = 1. \begin{align*}{\frac{1}{{d}^{3}} \left ({\frac{{d}^{3}}{3\,b}\sum _{{\it \_R1}={\it RootOf} \left ({{\it \_Z}}^{3}b-3\,{{\it \_Z}}^{2}bc+3\,{\it \_Z}\,b{c}^{2}+a{d}^{3}-{c}^{3}b \right ) }{\frac{{{\it \_R1}}^{2} \left ( -{\it Si} \left ( -dx+{\it \_R1}-c \right ) \cos \left ({\it \_R1} \right ) +{\it Ci} \left ( dx-{\it \_R1}+c \right ) \sin \left ({\it \_R1} \right ) \right ) }{{{\it \_R1}}^{2}-2\,{\it \_R1}\,c+{c}^{2}}}}-{\frac{2\,c{d}^{3}}{3\,b}\sum _{{\it \_R1}={\it RootOf} \left ({{\it \_Z}}^{3}b-3\,{{\it \_Z}}^{2}bc+3\,{\it \_Z}\,b{c}^{2}+a{d}^{3}-{c}^{3}b \right ) }{\frac{{\it \_R1}\, \left ( -{\it Si} \left ( -dx+{\it \_R1}-c \right ) \cos \left ({\it \_R1} \right ) +{\it Ci} \left ( dx-{\it \_R1}+c \right ) \sin \left ({\it \_R1} \right ) \right ) }{{{\it \_R1}}^{2}-2\,{\it \_R1}\,c+{c}^{2}}}}+{\frac{{d}^{3}{c}^{2}}{3\,b}\sum _{{\it \_R1}={\it RootOf} \left ({{\it \_Z}}^{3}b-3\,{{\it \_Z}}^{2}bc+3\,{\it \_Z}\,b{c}^{2}+a{d}^{3}-{c}^{3}b \right ) }{\frac{-{\it Si} \left ( -dx+{\it \_R1}-c \right ) \cos \left ({\it \_R1} \right ) +{\it Ci} \left ( dx-{\it \_R1}+c \right ) \sin \left ({\it \_R1} \right ) }{{{\it \_R1}}^{2}-2\,{\it \_R1}\,c+{c}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^3*(1/3*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))-2/3*d^3*c/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*d^3*c^2/b*sum(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^2*sin(d*x+c)/(b*x^3+a),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C]  time = 2.11919, size = 747, normalized size = 2.66 \begin{align*} \frac{i \,{\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 )} - i \,{\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 )} + i \,{\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 )} - i \,{\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 )} - i \,{\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 )} + i \,{\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 )}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(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
*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*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*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*Ei(I*d*x + (-I*
a*d^3/b)^(1/3))*e^(I*c - (-I*a*d^3/b)^(1/3)) + I*Ei(-I*d*x + (I*a*d^3/b)^(1/3))*e^(-I*c - (I*a*d^3/b)^(1/3)))/
b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \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^2*sin(d*x+c)/(b*x^3+a),x, algorithm="giac")

[Out]

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