3.7 \(\int x^4 (a+b \sin (c+d x^2)) \, dx\)

Optimal. Leaf size=121 \[ \frac {a x^5}{5}-\frac {3 \sqrt {\frac {\pi }{2}} b \sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{4 d^{5/2}}-\frac {3 \sqrt {\frac {\pi }{2}} b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{4 d^{5/2}}+\frac {3 b x \sin \left (c+d x^2\right )}{4 d^2}-\frac {b x^3 \cos \left (c+d x^2\right )}{2 d} \]

[Out]

1/5*a*x^5-1/2*b*x^3*cos(d*x^2+c)/d+3/4*b*x*sin(d*x^2+c)/d^2-3/8*b*cos(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))*
2^(1/2)*Pi^(1/2)/d^(5/2)-3/8*b*FresnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2))*sin(c)*2^(1/2)*Pi^(1/2)/d^(5/2)

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Rubi [A]  time = 0.13, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {14, 3385, 3386, 3353, 3352, 3351} \[ \frac {a x^5}{5}-\frac {3 \sqrt {\frac {\pi }{2}} b \sin (c) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {d} x\right )}{4 d^{5/2}}-\frac {3 \sqrt {\frac {\pi }{2}} b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{4 d^{5/2}}+\frac {3 b x \sin \left (c+d x^2\right )}{4 d^2}-\frac {b x^3 \cos \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*Sin[c + d*x^2]),x]

[Out]

(a*x^5)/5 - (b*x^3*Cos[c + d*x^2])/(2*d) - (3*b*Sqrt[Pi/2]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x])/(4*d^(5/2))
- (3*b*Sqrt[Pi/2]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/(4*d^(5/2)) + (3*b*x*Sin[c + d*x^2])/(4*d^2)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 125, normalized size = 1.03 \[ \frac {a x^5}{5}-\frac {3 \sqrt {\frac {\pi }{2}} b \left (\sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+\cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )\right )}{4 d^{5/2}}-\frac {b x \cos \left (d x^2\right ) \left (2 d x^2 \cos (c)-3 \sin (c)\right )}{4 d^2}+\frac {b x \sin \left (d x^2\right ) \left (2 d x^2 \sin (c)+3 \cos (c)\right )}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^5)/5 - (b*x*Cos[d*x^2]*(2*d*x^2*Cos[c] - 3*Sin[c]))/(4*d^2) - (3*b*Sqrt[Pi/2]*(Cos[c]*FresnelS[Sqrt[d]*Sq
rt[2/Pi]*x] + FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c]))/(4*d^(5/2)) + (b*x*(3*Cos[c] + 2*d*x^2*Sin[c])*Sin[d*x^2
])/(4*d^2)

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fricas [A]  time = 0.82, size = 103, normalized size = 0.85 \[ \frac {8 \, a d^{3} x^{5} - 20 \, b d^{2} x^{3} \cos \left (d x^{2} + c\right ) - 15 \, \sqrt {2} \pi b \sqrt {\frac {d}{\pi }} \cos \relax (c) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) - 15 \, \sqrt {2} \pi b \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \relax (c) + 30 \, b d x \sin \left (d x^{2} + c\right )}{40 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*a*d^3*x^5 - 20*b*d^2*x^3*cos(d*x^2 + c) - 15*sqrt(2)*pi*b*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt
(d/pi)) - 15*sqrt(2)*pi*b*sqrt(d/pi)*fresnel_cos(sqrt(2)*x*sqrt(d/pi))*sin(c) + 30*b*d*x*sin(d*x^2 + c))/d^3

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giac [C]  time = 0.50, size = 165, normalized size = 1.36 \[ \frac {1}{5} \, a x^{5} - \frac {3 i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (i \, c\right )}}{16 \, d^{2} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} + \frac {3 i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (-i \, c\right )}}{16 \, d^{2} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} + \frac {i \, {\left (2 i \, b d x^{3} - 3 \, b x\right )} e^{\left (i \, d x^{2} + i \, c\right )}}{8 \, d^{2}} + \frac {i \, {\left (2 i \, b d x^{3} + 3 \, b x\right )} e^{\left (-i \, d x^{2} - i \, c\right )}}{8 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5 - 3/16*I*sqrt(2)*sqrt(pi)*b*erf(-1/2*sqrt(2)*x*(-I*d/abs(d) + 1)*sqrt(abs(d)))*e^(I*c)/(d^2*(-I*d/ab
s(d) + 1)*sqrt(abs(d))) + 3/16*I*sqrt(2)*sqrt(pi)*b*erf(-1/2*sqrt(2)*x*(I*d/abs(d) + 1)*sqrt(abs(d)))*e^(-I*c)
/(d^2*(I*d/abs(d) + 1)*sqrt(abs(d))) + 1/8*I*(2*I*b*d*x^3 - 3*b*x)*e^(I*d*x^2 + I*c)/d^2 + 1/8*I*(2*I*b*d*x^3
+ 3*b*x)*e^(-I*d*x^2 - I*c)/d^2

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maple [A]  time = 0.03, size = 89, normalized size = 0.74 \[ \frac {a \,x^{5}}{5}+b \left (-\frac {x^{3} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {3 x \sin \left (d \,x^{2}+c \right )}{4 d}-\frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (c ) \mathrm {S}\left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \relax (c ) \FresnelC \left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )}{8 d^{\frac {3}{2}}}}{d}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5+b*(-1/2/d*x^3*cos(d*x^2+c)+3/2/d*(1/2/d*x*sin(d*x^2+c)-1/4/d^(3/2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelS
(x*d^(1/2)*2^(1/2)/Pi^(1/2))+sin(c)*FresnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2)))))

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maxima [C]  time = 0.64, size = 92, normalized size = 0.76 \[ \frac {1}{5} \, a x^{5} - \frac {{\left (16 \, d^{3} x^{3} \cos \left (d x^{2} + c\right ) - 24 \, d^{2} x \sin \left (d x^{2} + c\right ) - \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (3 i + 3\right ) \, \cos \relax (c) + \left (3 i - 3\right ) \, \sin \relax (c)\right )} \operatorname {erf}\left (\sqrt {i \, d} x\right ) + {\left (\left (3 i - 3\right ) \, \cos \relax (c) - \left (3 i + 3\right ) \, \sin \relax (c)\right )} \operatorname {erf}\left (\sqrt {-i \, d} x\right )\right )} d^{\frac {3}{2}}\right )} b}{32 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5 - 1/32*(16*d^3*x^3*cos(d*x^2 + c) - 24*d^2*x*sin(d*x^2 + c) - sqrt(2)*sqrt(pi)*((-(3*I + 3)*cos(c) +
(3*I - 3)*sin(c))*erf(sqrt(I*d)*x) + ((3*I - 3)*cos(c) - (3*I + 3)*sin(c))*erf(sqrt(-I*d)*x))*d^(3/2))*b/d^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\left (a+b\,\sin \left (d\,x^2+c\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*sin(c + d*x^2)),x)

[Out]

int(x^4*(a + b*sin(c + d*x^2)), x)

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sympy [B]  time = 4.83, size = 488, normalized size = 4.03 \[ \frac {a x^{5}}{5} - \frac {5 \sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \sin {\relax (c )} C\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{32 \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \sin {\relax (c )} C\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right )}{2} - \frac {21 \sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \cos {\relax (c )} S\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {3}{4}\right )}{32 \Gamma \left (\frac {11}{4}\right )} + \frac {\sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \cos {\relax (c )} S\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right )}{2} - \frac {15 \sqrt {2} \sqrt {\pi } b \sqrt {\frac {1}{d}} \sin {\relax (c )} C\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{128 d^{2} \Gamma \left (\frac {9}{4}\right )} - \frac {63 \sqrt {2} \sqrt {\pi } b \sqrt {\frac {1}{d}} \cos {\relax (c )} S\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {3}{4}\right )}{128 d^{2} \Gamma \left (\frac {11}{4}\right )} + \frac {5 b x^{3} \sqrt {\frac {1}{d}} \sin {\relax (c )} \sin {\left (d x^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{32 \sqrt {d} \Gamma \left (\frac {9}{4}\right )} - \frac {21 b x^{3} \sqrt {\frac {1}{d}} \cos {\relax (c )} \cos {\left (d x^{2} \right )} \Gamma \left (\frac {3}{4}\right )}{32 \sqrt {d} \Gamma \left (\frac {11}{4}\right )} + \frac {15 b x \sqrt {\frac {1}{d}} \sin {\relax (c )} \cos {\left (d x^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{64 d^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {63 b x \sqrt {\frac {1}{d}} \sin {\left (d x^{2} \right )} \cos {\relax (c )} \Gamma \left (\frac {3}{4}\right )}{64 d^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**5/5 - 5*sqrt(2)*sqrt(pi)*b*x**4*sqrt(1/d)*sin(c)*fresnelc(sqrt(2)*sqrt(d)*x/sqrt(pi))*gamma(1/4)/(32*gamm
a(9/4)) + sqrt(2)*sqrt(pi)*b*x**4*sqrt(1/d)*sin(c)*fresnelc(sqrt(2)*sqrt(d)*x/sqrt(pi))/2 - 21*sqrt(2)*sqrt(pi
)*b*x**4*sqrt(1/d)*cos(c)*fresnels(sqrt(2)*sqrt(d)*x/sqrt(pi))*gamma(3/4)/(32*gamma(11/4)) + sqrt(2)*sqrt(pi)*
b*x**4*sqrt(1/d)*cos(c)*fresnels(sqrt(2)*sqrt(d)*x/sqrt(pi))/2 - 15*sqrt(2)*sqrt(pi)*b*sqrt(1/d)*sin(c)*fresne
lc(sqrt(2)*sqrt(d)*x/sqrt(pi))*gamma(1/4)/(128*d**2*gamma(9/4)) - 63*sqrt(2)*sqrt(pi)*b*sqrt(1/d)*cos(c)*fresn
els(sqrt(2)*sqrt(d)*x/sqrt(pi))*gamma(3/4)/(128*d**2*gamma(11/4)) + 5*b*x**3*sqrt(1/d)*sin(c)*sin(d*x**2)*gamm
a(1/4)/(32*sqrt(d)*gamma(9/4)) - 21*b*x**3*sqrt(1/d)*cos(c)*cos(d*x**2)*gamma(3/4)/(32*sqrt(d)*gamma(11/4)) +
15*b*x*sqrt(1/d)*sin(c)*cos(d*x**2)*gamma(1/4)/(64*d**(3/2)*gamma(9/4)) + 63*b*x*sqrt(1/d)*sin(d*x**2)*cos(c)*
gamma(3/4)/(64*d**(3/2)*gamma(11/4))

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