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

Optimal. Leaf size=247 \[ \frac {1}{10} x^5 \left (2 a^2+b^2\right )-\frac {3 \sqrt {\frac {\pi }{2}} a b \sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{2 d^{5/2}}-\frac {3 \sqrt {\frac {\pi }{2}} a b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{2 d^{5/2}}+\frac {3 a b x \sin \left (c+d x^2\right )}{2 d^2}-\frac {a b x^3 \cos \left (c+d x^2\right )}{d}+\frac {3 \sqrt {\pi } b^2 \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{64 d^{5/2}}-\frac {3 \sqrt {\pi } b^2 \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{64 d^{5/2}}-\frac {3 b^2 x \cos \left (2 c+2 d x^2\right )}{32 d^2}-\frac {b^2 x^3 \sin \left (2 c+2 d x^2\right )}{8 d} \]

[Out]

1/10*(2*a^2+b^2)*x^5-a*b*x^3*cos(d*x^2+c)/d-3/32*b^2*x*cos(2*d*x^2+2*c)/d^2+3/2*a*b*x*sin(d*x^2+c)/d^2-1/8*b^2
*x^3*sin(2*d*x^2+2*c)/d-3/4*a*b*cos(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/d^(5/2)-3/4*a*b*F
resnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2))*sin(c)*2^(1/2)*Pi^(1/2)/d^(5/2)+3/64*b^2*cos(2*c)*FresnelC(2*x*d^(1/2)/Pi^
(1/2))*Pi^(1/2)/d^(5/2)-3/64*b^2*FresnelS(2*x*d^(1/2)/Pi^(1/2))*sin(2*c)*Pi^(1/2)/d^(5/2)

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Rubi [A]  time = 0.24, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3403, 6, 3386, 3385, 3354, 3352, 3351, 3353} \[ \frac {1}{10} x^5 \left (2 a^2+b^2\right )-\frac {3 \sqrt {\frac {\pi }{2}} a b \sin (c) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {d} x\right )}{2 d^{5/2}}-\frac {3 \sqrt {\frac {\pi }{2}} a b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{2 d^{5/2}}+\frac {3 a b x \sin \left (c+d x^2\right )}{2 d^2}-\frac {a b x^3 \cos \left (c+d x^2\right )}{d}+\frac {3 \sqrt {\pi } b^2 \cos (2 c) \text {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{64 d^{5/2}}-\frac {3 \sqrt {\pi } b^2 \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{64 d^{5/2}}-\frac {3 b^2 x \cos \left (2 c+2 d x^2\right )}{32 d^2}-\frac {b^2 x^3 \sin \left (2 c+2 d x^2\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*Sin[c + d*x^2])^2,x]

[Out]

((2*a^2 + b^2)*x^5)/10 - (a*b*x^3*Cos[c + d*x^2])/d - (3*b^2*x*Cos[2*c + 2*d*x^2])/(32*d^2) + (3*b^2*Sqrt[Pi]*
Cos[2*c]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]])/(64*d^(5/2)) - (3*a*b*Sqrt[Pi/2]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]
*x])/(2*d^(5/2)) - (3*a*b*Sqrt[Pi/2]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/(2*d^(5/2)) - (3*b^2*Sqrt[Pi]*Fres
nelS[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/(64*d^(5/2)) + (3*a*b*x*Sin[c + d*x^2])/(2*d^2) - (b^2*x^3*Sin[2*c + 2*
d*x^2])/(8*d)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 3354

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

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x^4 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx &=\int \left (a^2 x^4+\frac {b^2 x^4}{2}-\frac {1}{2} b^2 x^4 \cos \left (2 c+2 d x^2\right )+2 a b x^4 \sin \left (c+d x^2\right )\right ) \, dx\\ &=\int \left (\left (a^2+\frac {b^2}{2}\right ) x^4-\frac {1}{2} b^2 x^4 \cos \left (2 c+2 d x^2\right )+2 a b x^4 \sin \left (c+d x^2\right )\right ) \, dx\\ &=\frac {1}{10} \left (2 a^2+b^2\right ) x^5+(2 a b) \int x^4 \sin \left (c+d x^2\right ) \, dx-\frac {1}{2} b^2 \int x^4 \cos \left (2 c+2 d x^2\right ) \, dx\\ &=\frac {1}{10} \left (2 a^2+b^2\right ) x^5-\frac {a b x^3 \cos \left (c+d x^2\right )}{d}-\frac {b^2 x^3 \sin \left (2 c+2 d x^2\right )}{8 d}+\frac {(3 a b) \int x^2 \cos \left (c+d x^2\right ) \, dx}{d}+\frac {\left (3 b^2\right ) \int x^2 \sin \left (2 c+2 d x^2\right ) \, dx}{8 d}\\ &=\frac {1}{10} \left (2 a^2+b^2\right ) x^5-\frac {a b x^3 \cos \left (c+d x^2\right )}{d}-\frac {3 b^2 x \cos \left (2 c+2 d x^2\right )}{32 d^2}+\frac {3 a b x \sin \left (c+d x^2\right )}{2 d^2}-\frac {b^2 x^3 \sin \left (2 c+2 d x^2\right )}{8 d}-\frac {(3 a b) \int \sin \left (c+d x^2\right ) \, dx}{2 d^2}+\frac {\left (3 b^2\right ) \int \cos \left (2 c+2 d x^2\right ) \, dx}{32 d^2}\\ &=\frac {1}{10} \left (2 a^2+b^2\right ) x^5-\frac {a b x^3 \cos \left (c+d x^2\right )}{d}-\frac {3 b^2 x \cos \left (2 c+2 d x^2\right )}{32 d^2}+\frac {3 a b x \sin \left (c+d x^2\right )}{2 d^2}-\frac {b^2 x^3 \sin \left (2 c+2 d x^2\right )}{8 d}-\frac {(3 a b \cos (c)) \int \sin \left (d x^2\right ) \, dx}{2 d^2}+\frac {\left (3 b^2 \cos (2 c)\right ) \int \cos \left (2 d x^2\right ) \, dx}{32 d^2}-\frac {(3 a b \sin (c)) \int \cos \left (d x^2\right ) \, dx}{2 d^2}-\frac {\left (3 b^2 \sin (2 c)\right ) \int \sin \left (2 d x^2\right ) \, dx}{32 d^2}\\ &=\frac {1}{10} \left (2 a^2+b^2\right ) x^5-\frac {a b x^3 \cos \left (c+d x^2\right )}{d}-\frac {3 b^2 x \cos \left (2 c+2 d x^2\right )}{32 d^2}+\frac {3 b^2 \sqrt {\pi } \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )}{64 d^{5/2}}-\frac {3 a b \sqrt {\frac {\pi }{2}} \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{2 d^{5/2}}-\frac {3 a b \sqrt {\frac {\pi }{2}} C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)}{2 d^{5/2}}-\frac {3 b^2 \sqrt {\pi } S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)}{64 d^{5/2}}+\frac {3 a b x \sin \left (c+d x^2\right )}{2 d^2}-\frac {b^2 x^3 \sin \left (2 c+2 d x^2\right )}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 234, normalized size = 0.95 \[ \frac {64 a^2 d^{5/2} x^5-320 a b d^{3/2} x^3 \cos \left (c+d x^2\right )-240 \sqrt {2 \pi } a b \sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-240 \sqrt {2 \pi } a b \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+480 a b \sqrt {d} x \sin \left (c+d x^2\right )-40 b^2 d^{3/2} x^3 \sin \left (2 \left (c+d x^2\right )\right )+15 \sqrt {\pi } b^2 \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-15 \sqrt {\pi } b^2 \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-30 b^2 \sqrt {d} x \cos \left (2 \left (c+d x^2\right )\right )+32 b^2 d^{5/2} x^5}{320 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^2*d^(5/2)*x^5 + 32*b^2*d^(5/2)*x^5 - 320*a*b*d^(3/2)*x^3*Cos[c + d*x^2] - 30*b^2*Sqrt[d]*x*Cos[2*(c + d*
x^2)] + 15*b^2*Sqrt[Pi]*Cos[2*c]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]] - 240*a*b*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]
*Sqrt[2/Pi]*x] - 240*a*b*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c] - 15*b^2*Sqrt[Pi]*FresnelS[(2*Sqrt[d
]*x)/Sqrt[Pi]]*Sin[2*c] + 480*a*b*Sqrt[d]*x*Sin[c + d*x^2] - 40*b^2*d^(3/2)*x^3*Sin[2*(c + d*x^2)])/(320*d^(5/
2))

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fricas [A]  time = 0.75, size = 216, normalized size = 0.87 \[ \frac {32 \, {\left (2 \, a^{2} + b^{2}\right )} d^{3} x^{5} - 320 \, a b d^{2} x^{3} \cos \left (d x^{2} + c\right ) - 60 \, b^{2} d x \cos \left (d x^{2} + c\right )^{2} - 240 \, \sqrt {2} \pi a b \sqrt {\frac {d}{\pi }} \cos \relax (c) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) - 240 \, \sqrt {2} \pi a b \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \relax (c) + 15 \, \pi b^{2} \sqrt {\frac {d}{\pi }} \cos \left (2 \, c\right ) \operatorname {C}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) - 15 \, \pi b^{2} \sqrt {\frac {d}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) \sin \left (2 \, c\right ) + 30 \, b^{2} d x - 80 \, {\left (b^{2} d^{2} x^{3} \cos \left (d x^{2} + c\right ) - 6 \, a b d x\right )} \sin \left (d x^{2} + c\right )}{320 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(32*(2*a^2 + b^2)*d^3*x^5 - 320*a*b*d^2*x^3*cos(d*x^2 + c) - 60*b^2*d*x*cos(d*x^2 + c)^2 - 240*sqrt(2)*p
i*a*b*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt(d/pi)) - 240*sqrt(2)*pi*a*b*sqrt(d/pi)*fresnel_cos(sqrt(2)*
x*sqrt(d/pi))*sin(c) + 15*pi*b^2*sqrt(d/pi)*cos(2*c)*fresnel_cos(2*x*sqrt(d/pi)) - 15*pi*b^2*sqrt(d/pi)*fresne
l_sin(2*x*sqrt(d/pi))*sin(2*c) + 30*b^2*d*x - 80*(b^2*d^2*x^3*cos(d*x^2 + c) - 6*a*b*d*x)*sin(d*x^2 + c))/d^3

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giac [C]  time = 0.48, size = 329, normalized size = 1.33 \[ \frac {1}{5} \, a^{2} x^{5} + \frac {1}{10} \, b^{2} x^{5} - \frac {3 i \, \sqrt {2} \sqrt {\pi } a b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (i \, c\right )}}{8 \, d^{2} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} + \frac {3 i \, \sqrt {2} \sqrt {\pi } a b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (-i \, c\right )}}{8 \, d^{2} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} - \frac {3 \, \sqrt {\pi } b^{2} \operatorname {erf}\left (-\sqrt {d} x {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}\right ) e^{\left (2 i \, c\right )}}{128 \, d^{\frac {5}{2}} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}} - \frac {3 \, \sqrt {\pi } b^{2} \operatorname {erf}\left (-\sqrt {d} x {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}\right ) e^{\left (-2 i \, c\right )}}{128 \, d^{\frac {5}{2}} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} - \frac {{\left (-4 i \, b^{2} d x^{3} + 3 \, b^{2} x\right )} e^{\left (2 i \, d x^{2} + 2 i \, c\right )}}{64 \, d^{2}} + \frac {i \, {\left (2 i \, a b d x^{3} - 3 \, a b x\right )} e^{\left (i \, d x^{2} + i \, c\right )}}{4 \, d^{2}} + \frac {i \, {\left (2 i \, a b d x^{3} + 3 \, a b x\right )} e^{\left (-i \, d x^{2} - i \, c\right )}}{4 \, d^{2}} - \frac {{\left (4 i \, b^{2} d x^{3} + 3 \, b^{2} x\right )} e^{\left (-2 i \, d x^{2} - 2 i \, c\right )}}{64 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*x^5 + 1/10*b^2*x^5 - 3/8*I*sqrt(2)*sqrt(pi)*a*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))) + 3/8*I*sqrt(2)*sqrt(pi)*a*b*erf(-1/2*sqrt(2)*x*(I*d/abs(d) + 1)*sqr
t(abs(d)))*e^(-I*c)/(d^2*(I*d/abs(d) + 1)*sqrt(abs(d))) - 3/128*sqrt(pi)*b^2*erf(-sqrt(d)*x*(-I*d/abs(d) + 1))
*e^(2*I*c)/(d^(5/2)*(-I*d/abs(d) + 1)) - 3/128*sqrt(pi)*b^2*erf(-sqrt(d)*x*(I*d/abs(d) + 1))*e^(-2*I*c)/(d^(5/
2)*(I*d/abs(d) + 1)) - 1/64*(-4*I*b^2*d*x^3 + 3*b^2*x)*e^(2*I*d*x^2 + 2*I*c)/d^2 + 1/4*I*(2*I*a*b*d*x^3 - 3*a*
b*x)*e^(I*d*x^2 + I*c)/d^2 + 1/4*I*(2*I*a*b*d*x^3 + 3*a*b*x)*e^(-I*d*x^2 - I*c)/d^2 - 1/64*(4*I*b^2*d*x^3 + 3*
b^2*x)*e^(-2*I*d*x^2 - 2*I*c)/d^2

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maple [A]  time = 0.06, size = 189, normalized size = 0.77 \[ \frac {x^{5} a^{2}}{5}+\frac {x^{5} b^{2}}{10}-\frac {b^{2} \left (\frac {x^{3} \sin \left (2 d \,x^{2}+2 c \right )}{4 d}-\frac {3 \left (-\frac {x \cos \left (2 d \,x^{2}+2 c \right )}{4 d}+\frac {\sqrt {\pi }\, \left (\cos \left (2 c \right ) \FresnelC \left (\frac {2 x \sqrt {d}}{\sqrt {\pi }}\right )-\sin \left (2 c \right ) \mathrm {S}\left (\frac {2 x \sqrt {d}}{\sqrt {\pi }}\right )\right )}{8 d^{\frac {3}{2}}}\right )}{4 d}\right )}{2}+2 a b \left (-\frac {x^{3} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {3 x \sin \left (d \,x^{2}+c \right )}{4 d}-\frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (c ) \mathrm {S}\left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \relax (c ) \FresnelC \left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )}{8 d^{\frac {3}{2}}}}{d}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*a^2+1/10*x^5*b^2-1/2*b^2*(1/4/d*x^3*sin(2*d*x^2+2*c)-3/4/d*(-1/4/d*x*cos(2*d*x^2+2*c)+1/8/d^(3/2)*Pi^(
1/2)*(cos(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))-sin(2*c)*FresnelS(2*x*d^(1/2)/Pi^(1/2)))))+2*a*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 = 0.49, size = 207, normalized size = 0.84 \[ \frac {1}{5} \, a^{2} x^{5} - \frac {{\left (16 \, d^{3} x^{3} \cos \left (d x^{2} + c\right ) - 24 \, d^{2} x \sin \left (d x^{2} + c\right ) - \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (3 i + 3\right ) \, \cos \relax (c) + \left (3 i - 3\right ) \, \sin \relax (c)\right )} \operatorname {erf}\left (\sqrt {i \, d} x\right ) + {\left (\left (3 i - 3\right ) \, \cos \relax (c) - \left (3 i + 3\right ) \, \sin \relax (c)\right )} \operatorname {erf}\left (\sqrt {-i \, d} x\right )\right )} d^{\frac {3}{2}}\right )} a b}{16 \, d^{4}} + \frac {{\left (256 \, d^{4} x^{5} - 320 \, d^{3} x^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) - 240 \, d^{2} x \cos \left (2 \, d x^{2} + 2 \, c\right ) - 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (15 i - 15\right ) \, \cos \left (2 \, c\right ) + \left (15 i + 15\right ) \, \sin \left (2 \, c\right )\right )} \operatorname {erf}\left (\sqrt {2 i \, d} x\right ) + {\left (-\left (15 i + 15\right ) \, \cos \left (2 \, c\right ) - \left (15 i - 15\right ) \, \sin \left (2 \, c\right )\right )} \operatorname {erf}\left (\sqrt {-2 i \, d} x\right )\right )} d^{\frac {3}{2}}\right )} b^{2}}{2560 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*x^5 - 1/16*(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))*a*b/
d^4 + 1/2560*(256*d^4*x^5 - 320*d^3*x^3*sin(2*d*x^2 + 2*c) - 240*d^2*x*cos(2*d*x^2 + 2*c) - 4^(1/4)*sqrt(2)*sq
rt(pi)*(((15*I - 15)*cos(2*c) + (15*I + 15)*sin(2*c))*erf(sqrt(2*I*d)*x) + (-(15*I + 15)*cos(2*c) - (15*I - 15
)*sin(2*c))*erf(sqrt(-2*I*d)*x))*d^(3/2))*b^2/d^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*sin(c + d*x^2))^2,x)

[Out]

int(x^4*(a + b*sin(c + d*x^2))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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