3.60 \(\int (d x)^{3/2} \sin (f x) \, dx\)

Optimal. Leaf size=87 \[ -\frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} S\left (\frac {\sqrt {f} \sqrt {\frac {2}{\pi }} \sqrt {d x}}{\sqrt {d}}\right )}{2 f^{5/2}}+\frac {3 d \sqrt {d x} \sin (f x)}{2 f^2}-\frac {(d x)^{3/2} \cos (f x)}{f} \]

[Out]

-(d*x)^(3/2)*cos(f*x)/f-3/4*d^(3/2)*FresnelS(f^(1/2)*2^(1/2)/Pi^(1/2)*(d*x)^(1/2)/d^(1/2))*2^(1/2)*Pi^(1/2)/f^
(5/2)+3/2*d*sin(f*x)*(d*x)^(1/2)/f^2

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Rubi [A]  time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3296, 3305, 3351} \[ -\frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} S\left (\frac {\sqrt {f} \sqrt {\frac {2}{\pi }} \sqrt {d x}}{\sqrt {d}}\right )}{2 f^{5/2}}+\frac {3 d \sqrt {d x} \sin (f x)}{2 f^2}-\frac {(d x)^{3/2} \cos (f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*Sin[f*x],x]

[Out]

-(((d*x)^(3/2)*Cos[f*x])/f) - (3*d^(3/2)*Sqrt[Pi/2]*FresnelS[(Sqrt[f]*Sqrt[2/Pi]*Sqrt[d*x])/Sqrt[d]])/(2*f^(5/
2)) + (3*d*Sqrt[d*x]*Sin[f*x])/(2*f^2)

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 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 60, normalized size = 0.69 \[ \frac {d^2 \left (\sqrt {-i f x} \Gamma \left (\frac {5}{2},-i f x\right )+\sqrt {i f x} \Gamma \left (\frac {5}{2},i f x\right )\right )}{2 f^3 \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(3/2)*Sin[f*x],x]

[Out]

(d^2*(Sqrt[(-I)*f*x]*Gamma[5/2, (-I)*f*x] + Sqrt[I*f*x]*Gamma[5/2, I*f*x]))/(2*f^3*Sqrt[d*x])

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fricas [A]  time = 0.75, size = 72, normalized size = 0.83 \[ -\frac {3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {f}{\pi d}} \operatorname {S}\left (\sqrt {2} \sqrt {d x} \sqrt {\frac {f}{\pi d}}\right ) + 2 \, {\left (2 \, d f^{2} x \cos \left (f x\right ) - 3 \, d f \sin \left (f x\right )\right )} \sqrt {d x}}{4 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*sin(f*x),x, algorithm="fricas")

[Out]

-1/4*(3*sqrt(2)*pi*d^2*sqrt(f/(pi*d))*fresnel_sin(sqrt(2)*sqrt(d*x)*sqrt(f/(pi*d))) + 2*(2*d*f^2*x*cos(f*x) -
3*d*f*sin(f*x))*sqrt(d*x))/f^3

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giac [C]  time = 0.93, size = 220, normalized size = 2.53 \[ -\frac {1}{8} \, d {\left (\frac {-\frac {3 i \, \sqrt {2} \sqrt {\pi } d^{3} \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {d f} \sqrt {d x} {\left (\frac {i \, d f}{\sqrt {d^{2} f^{2}}} + 1\right )}}{2 \, d}\right )}{\sqrt {d f} {\left (\frac {i \, d f}{\sqrt {d^{2} f^{2}}} + 1\right )} f^{2}} - \frac {2 i \, {\left (2 i \, \sqrt {d x} d^{2} f x + 3 \, \sqrt {d x} d^{2}\right )} e^{\left (-i \, f x\right )}}{f^{2}}}{d^{2}} + \frac {\frac {3 i \, \sqrt {2} \sqrt {\pi } d^{3} \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {d f} \sqrt {d x} {\left (-\frac {i \, d f}{\sqrt {d^{2} f^{2}}} + 1\right )}}{2 \, d}\right )}{\sqrt {d f} {\left (-\frac {i \, d f}{\sqrt {d^{2} f^{2}}} + 1\right )} f^{2}} - \frac {2 i \, {\left (2 i \, \sqrt {d x} d^{2} f x - 3 \, \sqrt {d x} d^{2}\right )} e^{\left (i \, f x\right )}}{f^{2}}}{d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*sin(f*x),x, algorithm="giac")

[Out]

-1/8*d*((-3*I*sqrt(2)*sqrt(pi)*d^3*erf(-1/2*sqrt(2)*sqrt(d*f)*sqrt(d*x)*(I*d*f/sqrt(d^2*f^2) + 1)/d)/(sqrt(d*f
)*(I*d*f/sqrt(d^2*f^2) + 1)*f^2) - 2*I*(2*I*sqrt(d*x)*d^2*f*x + 3*sqrt(d*x)*d^2)*e^(-I*f*x)/f^2)/d^2 + (3*I*sq
rt(2)*sqrt(pi)*d^3*erf(-1/2*sqrt(2)*sqrt(d*f)*sqrt(d*x)*(-I*d*f/sqrt(d^2*f^2) + 1)/d)/(sqrt(d*f)*(-I*d*f/sqrt(
d^2*f^2) + 1)*f^2) - 2*I*(2*I*sqrt(d*x)*d^2*f*x - 3*sqrt(d*x)*d^2)*e^(I*f*x)/f^2)/d^2)

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maple [A]  time = 0.02, size = 87, normalized size = 1.00 \[ \frac {-\frac {d \left (d x \right )^{\frac {3}{2}} \cos \left (f x \right )}{f}+\frac {3 d \left (\frac {d \sqrt {d x}\, \sin \left (f x \right )}{2 f}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {d x}\, f}{\sqrt {\pi }\, \sqrt {\frac {f}{d}}\, d}\right )}{4 f \sqrt {\frac {f}{d}}}\right )}{f}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*sin(f*x),x)

[Out]

2/d*(-1/2*d/f*(d*x)^(3/2)*cos(f*x)+3/2*d/f*(1/2*d/f*(d*x)^(1/2)*sin(f*x)-1/4*d/f*2^(1/2)*Pi^(1/2)/(1/d*f)^(1/2
)*FresnelS(2^(1/2)/Pi^(1/2)/(1/d*f)^(1/2)*(d*x)^(1/2)/d*f)))

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maxima [C]  time = 1.69, size = 106, normalized size = 1.22 \[ -\frac {\sqrt {2} {\left (8 \, \sqrt {2} \left (d x\right )^{\frac {3}{2}} f^{2} \cos \left (f x\right ) - 12 \, \sqrt {2} \sqrt {d x} d f \sin \left (f x\right ) + \left (3 i + 3\right ) \, \sqrt {\pi } d^{2} \left (\frac {f^{2}}{d^{2}}\right )^{\frac {1}{4}} \operatorname {erf}\left (\sqrt {d x} \sqrt {\frac {i \, f}{d}}\right ) - \left (3 i - 3\right ) \, \sqrt {\pi } d^{2} \left (\frac {f^{2}}{d^{2}}\right )^{\frac {1}{4}} \operatorname {erf}\left (\sqrt {d x} \sqrt {-\frac {i \, f}{d}}\right )\right )}}{16 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*sin(f*x),x, algorithm="maxima")

[Out]

-1/16*sqrt(2)*(8*sqrt(2)*(d*x)^(3/2)*f^2*cos(f*x) - 12*sqrt(2)*sqrt(d*x)*d*f*sin(f*x) + (3*I + 3)*sqrt(pi)*d^2
*(f^2/d^2)^(1/4)*erf(sqrt(d*x)*sqrt(I*f/d)) - (3*I - 3)*sqrt(pi)*d^2*(f^2/d^2)^(1/4)*erf(sqrt(d*x)*sqrt(-I*f/d
)))/f^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (f\,x\right )\,{\left (d\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x)*(d*x)^(3/2),x)

[Out]

int(sin(f*x)*(d*x)^(3/2), x)

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sympy [A]  time = 21.83, size = 117, normalized size = 1.34 \[ - \frac {7 d^{\frac {3}{2}} x^{\frac {3}{2}} \cos {\left (f x \right )} \Gamma \left (\frac {7}{4}\right )}{4 f \Gamma \left (\frac {11}{4}\right )} + \frac {21 d^{\frac {3}{2}} \sqrt {x} \sin {\left (f x \right )} \Gamma \left (\frac {7}{4}\right )}{8 f^{2} \Gamma \left (\frac {11}{4}\right )} - \frac {21 \sqrt {2} \sqrt {\pi } d^{\frac {3}{2}} S\left (\frac {\sqrt {2} \sqrt {f} \sqrt {x}}{\sqrt {\pi }}\right ) \Gamma \left (\frac {7}{4}\right )}{16 f^{\frac {5}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(3/2)*sin(f*x),x)

[Out]

-7*d**(3/2)*x**(3/2)*cos(f*x)*gamma(7/4)/(4*f*gamma(11/4)) + 21*d**(3/2)*sqrt(x)*sin(f*x)*gamma(7/4)/(8*f**2*g
amma(11/4)) - 21*sqrt(2)*sqrt(pi)*d**(3/2)*fresnels(sqrt(2)*sqrt(f)*sqrt(x)/sqrt(pi))*gamma(7/4)/(16*f**(5/2)*
gamma(11/4))

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