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

Optimal. Leaf size=273 \[ -\frac {(-a)^{3/2} \sin \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) \text {Ci}\left (x d+\frac {\sqrt {-a} d}{\sqrt {b}}\right )}{2 b^{5/2}}+\frac {(-a)^{3/2} \sin \left (\frac {\sqrt {-a} d}{\sqrt {b}}+c\right ) \text {Ci}\left (\frac {\sqrt {-a} d}{\sqrt {b}}-d x\right )}{2 b^{5/2}}-\frac {(-a)^{3/2} \cos \left (\frac {\sqrt {-a} d}{\sqrt {b}}+c\right ) \text {Si}\left (\frac {\sqrt {-a} d}{\sqrt {b}}-d x\right )}{2 b^{5/2}}-\frac {(-a)^{3/2} \cos \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) \text {Si}\left (x d+\frac {\sqrt {-a} d}{\sqrt {b}}\right )}{2 b^{5/2}}+\frac {a \cos (c+d x)}{b^2 d}+\frac {2 \cos (c+d x)}{b d^3}+\frac {2 x \sin (c+d x)}{b d^2}-\frac {x^2 \cos (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.73, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3345, 2638, 3296, 3333, 3303, 3299, 3302} \[ -\frac {(-a)^{3/2} \sin \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) \text {CosIntegral}\left (\frac {\sqrt {-a} d}{\sqrt {b}}+d x\right )}{2 b^{5/2}}+\frac {(-a)^{3/2} \sin \left (\frac {\sqrt {-a} d}{\sqrt {b}}+c\right ) \text {CosIntegral}\left (\frac {\sqrt {-a} d}{\sqrt {b}}-d x\right )}{2 b^{5/2}}-\frac {(-a)^{3/2} \cos \left (\frac {\sqrt {-a} d}{\sqrt {b}}+c\right ) \text {Si}\left (\frac {\sqrt {-a} d}{\sqrt {b}}-d x\right )}{2 b^{5/2}}-\frac {(-a)^{3/2} \cos \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) \text {Si}\left (x d+\frac {\sqrt {-a} d}{\sqrt {b}}\right )}{2 b^{5/2}}+\frac {a \cos (c+d x)}{b^2 d}+\frac {2 x \sin (c+d x)}{b d^2}+\frac {2 \cos (c+d x)}{b d^3}-\frac {x^2 \cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*Sin[c + d*x])/(a + b*x^2),x]

[Out]

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

Rule 2638

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

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 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 3333

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

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

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Mathematica [C]  time = 0.50, size = 275, normalized size = 1.01 \[ \frac {i a^{3/2} d^3 \sin \left (c-\frac {i \sqrt {a} d}{\sqrt {b}}\right ) \text {Ci}\left (d \left (x+\frac {i \sqrt {a}}{\sqrt {b}}\right )\right )-i a^{3/2} d^3 \sin \left (c+\frac {i \sqrt {a} d}{\sqrt {b}}\right ) \text {Ci}\left (d \left (x-\frac {i \sqrt {a}}{\sqrt {b}}\right )\right )+i a^{3/2} d^3 \cos \left (c-\frac {i \sqrt {a} d}{\sqrt {b}}\right ) \text {Si}\left (d \left (x+\frac {i \sqrt {a}}{\sqrt {b}}\right )\right )+i a^{3/2} d^3 \cos \left (c+\frac {i \sqrt {a} d}{\sqrt {b}}\right ) \text {Si}\left (\frac {i \sqrt {a} d}{\sqrt {b}}-d x\right )+2 a \sqrt {b} d^2 \cos (c+d x)-2 b^{3/2} d^2 x^2 \cos (c+d x)+4 b^{3/2} d x \sin (c+d x)+4 b^{3/2} \cos (c+d x)}{2 b^{5/2} d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*b^(3/2)*Cos[c + d*x] + 2*a*Sqrt[b]*d^2*Cos[c + d*x] - 2*b^(3/2)*d^2*x^2*Cos[c + d*x] + I*a^(3/2)*d^3*CosInt
egral[d*((I*Sqrt[a])/Sqrt[b] + x)]*Sin[c - (I*Sqrt[a]*d)/Sqrt[b]] - I*a^(3/2)*d^3*CosIntegral[d*(((-I)*Sqrt[a]
)/Sqrt[b] + x)]*Sin[c + (I*Sqrt[a]*d)/Sqrt[b]] + 4*b^(3/2)*d*x*Sin[c + d*x] + I*a^(3/2)*d^3*Cos[c - (I*Sqrt[a]
*d)/Sqrt[b]]*SinIntegral[d*((I*Sqrt[a])/Sqrt[b] + x)] + I*a^(3/2)*d^3*Cos[c + (I*Sqrt[a]*d)/Sqrt[b]]*SinIntegr
al[(I*Sqrt[a]*d)/Sqrt[b] - d*x])/(2*b^(5/2)*d^3)

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fricas [C]  time = 0.79, size = 240, normalized size = 0.88 \[ \frac {\sqrt {\frac {a d^{2}}{b}} a d^{2} {\rm Ei}\left (i \, d x - \sqrt {\frac {a d^{2}}{b}}\right ) e^{\left (i \, c + \sqrt {\frac {a d^{2}}{b}}\right )} - \sqrt {\frac {a d^{2}}{b}} a d^{2} {\rm Ei}\left (i \, d x + \sqrt {\frac {a d^{2}}{b}}\right ) e^{\left (i \, c - \sqrt {\frac {a d^{2}}{b}}\right )} + \sqrt {\frac {a d^{2}}{b}} a d^{2} {\rm Ei}\left (-i \, d x - \sqrt {\frac {a d^{2}}{b}}\right ) e^{\left (-i \, c + \sqrt {\frac {a d^{2}}{b}}\right )} - \sqrt {\frac {a d^{2}}{b}} a d^{2} {\rm Ei}\left (-i \, d x + \sqrt {\frac {a d^{2}}{b}}\right ) e^{\left (-i \, c - \sqrt {\frac {a d^{2}}{b}}\right )} + 8 \, b d x \sin \left (d x + c\right ) - 4 \, {\left (b d^{2} x^{2} - a d^{2} - 2 \, b\right )} \cos \left (d x + c\right )}{4 \, b^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*sin(d*x+c)/(b*x^2+a),x, algorithm="fricas")

[Out]

1/4*(sqrt(a*d^2/b)*a*d^2*Ei(I*d*x - sqrt(a*d^2/b))*e^(I*c + sqrt(a*d^2/b)) - sqrt(a*d^2/b)*a*d^2*Ei(I*d*x + sq
rt(a*d^2/b))*e^(I*c - sqrt(a*d^2/b)) + sqrt(a*d^2/b)*a*d^2*Ei(-I*d*x - sqrt(a*d^2/b))*e^(-I*c + sqrt(a*d^2/b))
- sqrt(a*d^2/b)*a*d^2*Ei(-I*d*x + sqrt(a*d^2/b))*e^(-I*c - sqrt(a*d^2/b)) + 8*b*d*x*sin(d*x + c) - 4*(b*d^2*x
^2 - a*d^2 - 2*b)*cos(d*x + c))/(b^2*d^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \sin \left (d x + c\right )}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*sin(d*x+c)/(b*x^2+a),x, algorithm="giac")

[Out]

integrate(x^4*sin(d*x + c)/(b*x^2 + a), x)

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maple [B]  time = 0.08, size = 1656, normalized size = 6.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*sin(d*x+c)/(b*x^2+a),x)

[Out]

1/d^5*((b*d^2*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+2*c*b*d^2*(sin(d*x+c)-(d*x+c)*cos(d*x+
c))+a*d^4*cos(d*x+c)-3*b*c^2*d^2*cos(d*x+c))/b^2-1/2*d^2*(4*(d*(-a*b)^(1/2)+c*b)*a*c*d^2-4*(d*(-a*b)^(1/2)+c*b
)*b*c^3-a^2*d^4+2*a*b*c^2*d^2+3*b^2*c^4)/((d*(-a*b)^(1/2)+c*b)/b-c)/b^3*(Si(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*cos(
(d*(-a*b)^(1/2)+c*b)/b)+Ci(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*sin((d*(-a*b)^(1/2)+c*b)/b))-1/2*d^2*(-4*(d*(-a*b)^(1
/2)-c*b)*a*c*d^2+4*(d*(-a*b)^(1/2)-c*b)*b*c^3-a^2*d^4+2*a*b*c^2*d^2+3*b^2*c^4)/(-(d*(-a*b)^(1/2)-c*b)/b-c)/b^3
*(Si(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*cos((d*(-a*b)^(1/2)-c*b)/b)-Ci(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*sin((d*(-a*b)^
(1/2)-c*b)/b))+(-4*c*d^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+8*c^2*d^2*cos(d*x+c))/b+2*c*d^2*((d*(-a*b)^(1/2)+c*b)
/b*a*d^2-3*(d*(-a*b)^(1/2)+c*b)*c^2+2*a*c*d^2+2*b*c^3)/((d*(-a*b)^(1/2)+c*b)/b-c)/b^2*(Si(d*x+c-(d*(-a*b)^(1/2
)+c*b)/b)*cos((d*(-a*b)^(1/2)+c*b)/b)+Ci(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*sin((d*(-a*b)^(1/2)+c*b)/b))+2*c*d^2*(-
(d*(-a*b)^(1/2)-c*b)/b*a*d^2+3*(d*(-a*b)^(1/2)-c*b)*c^2+2*a*c*d^2+2*b*c^3)/(-(d*(-a*b)^(1/2)-c*b)/b-c)/b^2*(Si
(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*cos((d*(-a*b)^(1/2)-c*b)/b)-Ci(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*sin((d*(-a*b)^(1/2
)-c*b)/b))-6*c^2*d^2/b*cos(d*x+c)+3*c^2*d^2*(2*(d*(-a*b)^(1/2)+c*b)*c-a*d^2-b*c^2)/((d*(-a*b)^(1/2)+c*b)/b-c)/
b^2*(Si(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*cos((d*(-a*b)^(1/2)+c*b)/b)+Ci(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*sin((d*(-a*
b)^(1/2)+c*b)/b))+3*c^2*d^2*(-2*(d*(-a*b)^(1/2)-c*b)*c-a*d^2-b*c^2)/(-(d*(-a*b)^(1/2)-c*b)/b-c)/b^2*(Si(d*x+c+
(d*(-a*b)^(1/2)-c*b)/b)*cos((d*(-a*b)^(1/2)-c*b)/b)-Ci(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*sin((d*(-a*b)^(1/2)-c*b)/
b))-2*c^3*d^2*(d*(-a*b)^(1/2)+c*b)/b^2/((d*(-a*b)^(1/2)+c*b)/b-c)*(Si(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*cos((d*(-a
*b)^(1/2)+c*b)/b)+Ci(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*sin((d*(-a*b)^(1/2)+c*b)/b))+2*c^3*d^2*(d*(-a*b)^(1/2)-c*b)
/b^2/(-(d*(-a*b)^(1/2)-c*b)/b-c)*(Si(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*cos((d*(-a*b)^(1/2)-c*b)/b)-Ci(d*x+c+(d*(-a
*b)^(1/2)-c*b)/b)*sin((d*(-a*b)^(1/2)-c*b)/b))+c^4*d^2*(1/2/((d*(-a*b)^(1/2)+c*b)/b-c)/b*(Si(d*x+c-(d*(-a*b)^(
1/2)+c*b)/b)*cos((d*(-a*b)^(1/2)+c*b)/b)+Ci(d*x+c-(d*(-a*b)^(1/2)+c*b)/b)*sin((d*(-a*b)^(1/2)+c*b)/b))+1/2/(-(
d*(-a*b)^(1/2)-c*b)/b-c)/b*(Si(d*x+c+(d*(-a*b)^(1/2)-c*b)/b)*cos((d*(-a*b)^(1/2)-c*b)/b)-Ci(d*x+c+(d*(-a*b)^(1
/2)-c*b)/b)*sin((d*(-a*b)^(1/2)-c*b)/b))))

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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^4*sin(d*x+c)/(b*x^2+a),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\sin \left (c+d\,x\right )}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*sin(c + d*x))/(a + b*x^2),x)

[Out]

int((x^4*sin(c + d*x))/(a + b*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \sin {\left (c + d x \right )}}{a + b x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*sin(d*x+c)/(b*x**2+a),x)

[Out]

Integral(x**4*sin(c + d*x)/(a + b*x**2), x)

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