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

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

[Out]

1/3*cos(c+(-1)^(1/3)*a^(1/3)*d/b^(1/3))*Si(-(-1)^(1/3)*a^(1/3)*d/b^(1/3)+d*x)/b+1/3*cos(c-a^(1/3)*d/b^(1/3))*S
i(a^(1/3)*d/b^(1/3)+d*x)/b+1/3*cos(c-(-1)^(2/3)*a^(1/3)*d/b^(1/3))*Si((-1)^(2/3)*a^(1/3)*d/b^(1/3)+d*x)/b+1/3*
Ci(a^(1/3)*d/b^(1/3)+d*x)*sin(c-a^(1/3)*d/b^(1/3))/b+1/3*Ci((-1)^(1/3)*a^(1/3)*d/b^(1/3)-d*x)*sin(c+(-1)^(1/3)
*a^(1/3)*d/b^(1/3))/b+1/3*Ci((-1)^(2/3)*a^(1/3)*d/b^(1/3)+d*x)*sin(c-(-1)^(2/3)*a^(1/3)*d/b^(1/3))/b

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Rubi [A]  time = 0.45, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.31, size = 186, normalized size = 0.66 \[ \frac {i \left (\text {RootSum}\left [\text {$\#$1}^3 b+a\& ,-i \sin (\text {$\#$1} d+c) \text {Ci}(d (x-\text {$\#$1}))+\cos (\text {$\#$1} d+c) \text {Ci}(d (x-\text {$\#$1}))-\sin (\text {$\#$1} d+c) \text {Si}(d (x-\text {$\#$1}))-i \cos (\text {$\#$1} d+c) \text {Si}(d (x-\text {$\#$1}))\& \right ]-\text {RootSum}\left [\text {$\#$1}^3 b+a\& ,i \sin (\text {$\#$1} d+c) \text {Ci}(d (x-\text {$\#$1}))+\cos (\text {$\#$1} d+c) \text {Ci}(d (x-\text {$\#$1}))-\sin (\text {$\#$1} d+c) \text {Si}(d (x-\text {$\#$1}))+i \cos (\text {$\#$1} d+c) \text {Si}(d (x-\text {$\#$1}))\& \right ]\right )}{6 b} \]

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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fricas [C]  time = 0.81, size = 292, normalized size = 1.04 \[ \frac {i \, {\rm Ei}\left (-i \, d x + \frac {1}{2} \, \left (\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} - 1\right )}\right ) e^{\left (\frac {1}{2} \, \left (\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} + 1\right )} - i \, c\right )} - i \, {\rm Ei}\left (i \, d x + \frac {1}{2} \, \left (-\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} - 1\right )}\right ) e^{\left (\frac {1}{2} \, \left (-\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} + 1\right )} + i \, c\right )} + i \, {\rm Ei}\left (-i \, d x + \frac {1}{2} \, \left (\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} - 1\right )}\right ) e^{\left (\frac {1}{2} \, \left (\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} + 1\right )} - i \, c\right )} - i \, {\rm Ei}\left (i \, d x + \frac {1}{2} \, \left (-\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} - 1\right )}\right ) e^{\left (\frac {1}{2} \, \left (-\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} + 1\right )} + i \, c\right )} - i \, {\rm Ei}\left (i \, d x + \left (-\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}}\right ) e^{\left (i \, c - \left (-\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}}\right )} + i \, {\rm Ei}\left (-i \, d x + \left (\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}}\right ) e^{\left (-i \, c - \left (\frac {i \, a d^{3}}{b}\right )^{\frac {1}{3}}\right )}}{6 \, b} \]

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

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)

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maple [C]  time = 0.04, size = 266, normalized size = 0.95 \[ \frac {\frac {d^{3} \left (\munderset {\textit {\_R1} =\RootOf \left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\textit {\_R1}^{2} \left (-\Si \left (-d x +\textit {\_R1} -c \right ) \cos \left (\textit {\_R1} \right )+\Ci \left (d x -\textit {\_R1} +c \right ) \sin \left (\textit {\_R1} \right )\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right )}{3 b}-\frac {2 d^{3} c \left (\munderset {\textit {\_R1} =\RootOf \left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\textit {\_R1} \left (-\Si \left (-d x +\textit {\_R1} -c \right ) \cos \left (\textit {\_R1} \right )+\Ci \left (d x -\textit {\_R1} +c \right ) \sin \left (\textit {\_R1} \right )\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right )}{3 b}+\frac {d^{3} c^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {-\Si \left (-d x +\textit {\_R1} -c \right ) \cos \left (\textit {\_R1} \right )+\Ci \left (d x -\textit {\_R1} +c \right ) \sin \left (\textit {\_R1} \right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right )}{3 b}}{d^{3}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*sin(c + d*x))/(a + b*x^3),x)

[Out]

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

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

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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