3.81 \(\int f^{a+b x} \sin ^3(d+f x^2) \, dx\)

Optimal. Leaf size=298 \[ \frac{3}{16} (-1)^{3/4} \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{1}{4} i \left (\frac{b^2 \log ^2(f)}{f}+4 d\right )} \text{Erf}\left (\frac{\sqrt [4]{-1} (b \log (f)+2 i f x)}{2 \sqrt{f}}\right )+\left (\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\frac{\pi }{6}} f^{a-\frac{1}{2}} e^{\frac{i b^2 \log ^2(f)}{12 f}+3 i d} \text{Erf}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (b \log (f)+6 i f x)}{\sqrt{6} \sqrt{f}}\right )-\frac{3}{16} (-1)^{3/4} \sqrt{\pi } f^{a-\frac{1}{2}} e^{-\frac{1}{4} i \left (\frac{b^2 \log ^2(f)}{f}+4 d\right )} \text{Erfi}\left (\frac{\sqrt [4]{-1} (-b \log (f)+2 i f x)}{2 \sqrt{f}}\right )-\left (\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\frac{\pi }{6}} f^{a-\frac{1}{2}} e^{-\frac{1}{12} i \left (\frac{b^2 \log ^2(f)}{f}+36 d\right )} \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (-b \log (f)+6 i f x)}{\sqrt{6} \sqrt{f}}\right ) \]

[Out]

(3*(-1)^(3/4)*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))*f^(-1/2 + a)*Sqrt[Pi]*Erf[((-1)^(1/4)*((2*I)*f*x + b*Log[f]))
/(2*Sqrt[f])])/16 + (1/16 - I/16)*E^((3*I)*d + ((I/12)*b^2*Log[f]^2)/f)*f^(-1/2 + a)*Sqrt[Pi/6]*Erf[((1/2 + I/
2)*((6*I)*f*x + b*Log[f]))/(Sqrt[6]*Sqrt[f])] - (3*(-1)^(3/4)*f^(-1/2 + a)*Sqrt[Pi]*Erfi[((-1)^(1/4)*((2*I)*f*
x - b*Log[f]))/(2*Sqrt[f])])/(16*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))) - ((1/16 - I/16)*f^(-1/2 + a)*Sqrt[Pi/6]*
Erfi[((1/2 + I/2)*((6*I)*f*x - b*Log[f]))/(Sqrt[6]*Sqrt[f])])/E^((I/12)*(36*d + (b^2*Log[f]^2)/f))

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Rubi [A]  time = 0.366547, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4472, 2287, 2234, 2204, 2205} \[ \frac{3}{16} (-1)^{3/4} \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{1}{4} i \left (\frac{b^2 \log ^2(f)}{f}+4 d\right )} \text{Erf}\left (\frac{\sqrt [4]{-1} (b \log (f)+2 i f x)}{2 \sqrt{f}}\right )+\left (\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\frac{\pi }{6}} f^{a-\frac{1}{2}} e^{\frac{i b^2 \log ^2(f)}{12 f}+3 i d} \text{Erf}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (b \log (f)+6 i f x)}{\sqrt{6} \sqrt{f}}\right )-\frac{3}{16} (-1)^{3/4} \sqrt{\pi } f^{a-\frac{1}{2}} e^{-\frac{1}{4} i \left (\frac{b^2 \log ^2(f)}{f}+4 d\right )} \text{Erfi}\left (\frac{\sqrt [4]{-1} (-b \log (f)+2 i f x)}{2 \sqrt{f}}\right )-\left (\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\frac{\pi }{6}} f^{a-\frac{1}{2}} e^{-\frac{1}{12} i \left (\frac{b^2 \log ^2(f)}{f}+36 d\right )} \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (-b \log (f)+6 i f x)}{\sqrt{6} \sqrt{f}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x)*Sin[d + f*x^2]^3,x]

[Out]

(3*(-1)^(3/4)*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))*f^(-1/2 + a)*Sqrt[Pi]*Erf[((-1)^(1/4)*((2*I)*f*x + b*Log[f]))
/(2*Sqrt[f])])/16 + (1/16 - I/16)*E^((3*I)*d + ((I/12)*b^2*Log[f]^2)/f)*f^(-1/2 + a)*Sqrt[Pi/6]*Erf[((1/2 + I/
2)*((6*I)*f*x + b*Log[f]))/(Sqrt[6]*Sqrt[f])] - (3*(-1)^(3/4)*f^(-1/2 + a)*Sqrt[Pi]*Erfi[((-1)^(1/4)*((2*I)*f*
x - b*Log[f]))/(2*Sqrt[f])])/(16*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))) - ((1/16 - I/16)*f^(-1/2 + a)*Sqrt[Pi/6]*
Erfi[((1/2 + I/2)*((6*I)*f*x - b*Log[f]))/(Sqrt[6]*Sqrt[f])])/E^((I/12)*(36*d + (b^2*Log[f]^2)/f))

Rule 4472

Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[v]^n, x], x] /; FreeQ[F, x] && (LinearQ
[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rubi steps

\begin{align*} \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx &=\int \left (\frac{3}{8} i e^{-i d-i f x^2} f^{a+b x}-\frac{3}{8} i e^{i d+i f x^2} f^{a+b x}-\frac{1}{8} i e^{-3 i d-3 i f x^2} f^{a+b x}+\frac{1}{8} i e^{3 i d+3 i f x^2} f^{a+b x}\right ) \, dx\\ &=-\left (\frac{1}{8} i \int e^{-3 i d-3 i f x^2} f^{a+b x} \, dx\right )+\frac{1}{8} i \int e^{3 i d+3 i f x^2} f^{a+b x} \, dx+\frac{3}{8} i \int e^{-i d-i f x^2} f^{a+b x} \, dx-\frac{3}{8} i \int e^{i d+i f x^2} f^{a+b x} \, dx\\ &=-\left (\frac{1}{8} i \int e^{-3 i d-3 i f x^2+a \log (f)+b x \log (f)} \, dx\right )+\frac{1}{8} i \int e^{3 i d+3 i f x^2+a \log (f)+b x \log (f)} \, dx+\frac{3}{8} i \int e^{-i d-i f x^2+a \log (f)+b x \log (f)} \, dx-\frac{3}{8} i \int e^{i d+i f x^2+a \log (f)+b x \log (f)} \, dx\\ &=\frac{1}{8} \left (i e^{3 i d+\frac{i b^2 \log ^2(f)}{12 f}} f^a\right ) \int e^{-\frac{i (6 i f x+b \log (f))^2}{12 f}} \, dx+\frac{1}{8} \left (3 i e^{-\frac{1}{4} i \left (4 d+\frac{b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{\frac{i (-2 i f x+b \log (f))^2}{4 f}} \, dx-\frac{1}{8} \left (3 i e^{\frac{1}{4} i \left (4 d+\frac{b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{-\frac{i (2 i f x+b \log (f))^2}{4 f}} \, dx-\frac{1}{8} \left (i e^{-\frac{1}{12} i \left (36 d+\frac{b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{\frac{i (-6 i f x+b \log (f))^2}{12 f}} \, dx\\ &=\frac{3}{16} (-1)^{3/4} e^{\frac{1}{4} i \left (4 d+\frac{b^2 \log ^2(f)}{f}\right )} f^{-\frac{1}{2}+a} \sqrt{\pi } \text{erf}\left (\frac{\sqrt [4]{-1} (2 i f x+b \log (f))}{2 \sqrt{f}}\right )+\left (\frac{1}{16}-\frac{i}{16}\right ) e^{3 i d+\frac{i b^2 \log ^2(f)}{12 f}} f^{-\frac{1}{2}+a} \sqrt{\frac{\pi }{6}} \text{erf}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (6 i f x+b \log (f))}{\sqrt{6} \sqrt{f}}\right )-\frac{3}{16} (-1)^{3/4} e^{-\frac{1}{4} i \left (4 d+\frac{b^2 \log ^2(f)}{f}\right )} f^{-\frac{1}{2}+a} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt [4]{-1} (2 i f x-b \log (f))}{2 \sqrt{f}}\right )-\left (\frac{1}{16}-\frac{i}{16}\right ) e^{-\frac{1}{12} i \left (36 d+\frac{b^2 \log ^2(f)}{f}\right )} f^{-\frac{1}{2}+a} \sqrt{\frac{\pi }{6}} \text{erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (6 i f x-b \log (f))}{\sqrt{6} \sqrt{f}}\right )\\ \end{align*}

Mathematica [A]  time = 0.884968, size = 268, normalized size = 0.9 \[ \frac{1}{48} (-1)^{3/4} \sqrt{\pi } f^{a-\frac{1}{2}} e^{-\frac{i b^2 \log ^2(f)}{4 f}} \left (9 i e^{\frac{i b^2 \log ^2(f)}{2 f}} (\cos (d)+i \sin (d)) \text{Erfi}\left (\frac{\sqrt [4]{-1} (2 f x-i b \log (f))}{2 \sqrt{f}}\right )+\sqrt{3} e^{\frac{i b^2 \log ^2(f)}{6 f}} \left (e^{\frac{i b^2 \log ^2(f)}{6 f}} (\sin (3 d)-i \cos (3 d)) \text{Erfi}\left (\frac{(1-i) b \log (f)+(6+6 i) f x}{2 \sqrt{6} \sqrt{f}}\right )+(\cos (3 d)-i \sin (3 d)) \text{Erfi}\left (\frac{(-1)^{3/4} (6 f x+i b \log (f))}{2 \sqrt{3} \sqrt{f}}\right )\right )-9 (\cos (d)-i \sin (d)) \text{Erfi}\left (\frac{(-1)^{3/4} (2 f x+i b \log (f))}{2 \sqrt{f}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x)*Sin[d + f*x^2]^3,x]

[Out]

((-1)^(3/4)*f^(-1/2 + a)*Sqrt[Pi]*(-9*Erfi[((-1)^(3/4)*(2*f*x + I*b*Log[f]))/(2*Sqrt[f])]*(Cos[d] - I*Sin[d])
+ (9*I)*E^(((I/2)*b^2*Log[f]^2)/f)*Erfi[((-1)^(1/4)*(2*f*x - I*b*Log[f]))/(2*Sqrt[f])]*(Cos[d] + I*Sin[d]) + S
qrt[3]*E^(((I/6)*b^2*Log[f]^2)/f)*(Erfi[((-1)^(3/4)*(6*f*x + I*b*Log[f]))/(2*Sqrt[3]*Sqrt[f])]*(Cos[3*d] - I*S
in[3*d]) + E^(((I/6)*b^2*Log[f]^2)/f)*Erfi[((6 + 6*I)*f*x + (1 - I)*b*Log[f])/(2*Sqrt[6]*Sqrt[f])]*((-I)*Cos[3
*d] + Sin[3*d]))))/(48*E^(((I/4)*b^2*Log[f]^2)/f))

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Maple [A]  time = 0.481, size = 239, normalized size = 0.8 \begin{align*}{-{\frac{i}{16}}{f}^{a}\sqrt{\pi }{{\rm e}^{{\frac{{\frac{i}{12}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+36\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{-3\,if}x+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-3\,if}}}} \right ){\frac{1}{\sqrt{-3\,if}}}}+{{\frac{i}{48}}\sqrt{3}{f}^{a}\sqrt{\pi }{{\rm e}^{{\frac{-{\frac{i}{12}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+36\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{3}\sqrt{if}x+{\frac{\ln \left ( f \right ) b\sqrt{3}}{6}{\frac{1}{\sqrt{if}}}} \right ){\frac{1}{\sqrt{if}}}}-{{\frac{3\,i}{16}}{f}^{a}\sqrt{\pi }{{\rm e}^{{\frac{-{\frac{i}{4}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+4\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{if}x+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{if}}}} \right ){\frac{1}{\sqrt{if}}}}+{{\frac{3\,i}{16}}{f}^{a}\sqrt{\pi }{{\rm e}^{{\frac{{\frac{i}{4}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+4\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{-if}x+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-if}}}} \right ){\frac{1}{\sqrt{-if}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*I*Pi^(1/2)*f^a*exp(1/12*I*(ln(f)^2*b^2+36*d*f)/f)/(-3*I*f)^(1/2)*erf(-(-3*I*f)^(1/2)*x+1/2*ln(f)*b/(-3*I
*f)^(1/2))+1/48*I*Pi^(1/2)*f^a*exp(-1/12*I*(ln(f)^2*b^2+36*d*f)/f)*3^(1/2)/(I*f)^(1/2)*erf(-3^(1/2)*(I*f)^(1/2
)*x+1/6*ln(f)*b*3^(1/2)/(I*f)^(1/2))-3/16*I*Pi^(1/2)*f^a*exp(-1/4*I*(ln(f)^2*b^2+4*d*f)/f)/(I*f)^(1/2)*erf(-(I
*f)^(1/2)*x+1/2*ln(f)*b/(I*f)^(1/2))+3/16*I*Pi^(1/2)*f^a*exp(1/4*I*(ln(f)^2*b^2+4*d*f)/f)/(-I*f)^(1/2)*erf(-(-
I*f)^(1/2)*x+1/2*ln(f)*b/(-I*f)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: IndexError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: IndexError

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Fricas [B]  time = 0.545741, size = 1519, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(-I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 12*a*f*log(f) - 36*I*d*f)/f)*fresnel_cos(1/6*sqrt(6)
*(6*f*x + I*b*log(f))*sqrt(f/pi)/f) - I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 + 12*a*f*log(f) + 36*I*d
*f)/f)*fresnel_cos(-1/6*sqrt(6)*(6*f*x - I*b*log(f))*sqrt(f/pi)/f) + 9*I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*
log(f)^2 + 4*a*f*log(f) - 4*I*d*f)/f)*fresnel_cos(1/2*sqrt(2)*(2*f*x + I*b*log(f))*sqrt(f/pi)/f) + 9*I*sqrt(2)
*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 + 4*a*f*log(f) + 4*I*d*f)/f)*fresnel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log(f
))*sqrt(f/pi)/f) - sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 12*a*f*log(f) - 36*I*d*f)/f)*fresnel_sin(1
/6*sqrt(6)*(6*f*x + I*b*log(f))*sqrt(f/pi)/f) + sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 + 12*a*f*log(f)
+ 36*I*d*f)/f)*fresnel_sin(-1/6*sqrt(6)*(6*f*x - I*b*log(f))*sqrt(f/pi)/f) + 9*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-
I*b^2*log(f)^2 + 4*a*f*log(f) - 4*I*d*f)/f)*fresnel_sin(1/2*sqrt(2)*(2*f*x + I*b*log(f))*sqrt(f/pi)/f) - 9*sqr
t(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 + 4*a*f*log(f) + 4*I*d*f)/f)*fresnel_sin(-1/2*sqrt(2)*(2*f*x - I*b*l
og(f))*sqrt(f/pi)/f))/f

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + b x} \sin ^{3}{\left (d + f x^{2} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x+a)*sin(f*x**2+d)**3,x)

[Out]

Integral(f**(a + b*x)*sin(d + f*x**2)**3, x)

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Giac [B]  time = 1.46531, size = 803, normalized size = 2.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*I*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x - (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)))/f)*(-I*f/abs(f) + 1)*
sqrt(abs(f)))*e^(1/8*I*pi^2*b^2*sgn(f)/f + 1/4*pi*b^2*log(abs(f))*sgn(f)/f - 1/8*I*pi^2*b^2/f - 1/4*pi*b^2*log
(abs(f))/f + 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*log(abs(f)) + I*d)/((-I*f/abs(f) +
1)*sqrt(abs(f))) - 1/48*I*sqrt(6)*sqrt(pi)*erf(-1/24*sqrt(6)*sqrt(f)*(12*x - (pi*b*sgn(f) - pi*b + 2*I*b*log(
abs(f)))/f)*(-I*f/abs(f) + 1))*e^(1/24*I*pi^2*b^2*sgn(f)/f + 1/12*pi*b^2*log(abs(f))*sgn(f)/f - 1/24*I*pi^2*b^
2/f - 1/12*pi*b^2*log(abs(f))/f + 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*log(abs(f))
+ 3*I*d)/(sqrt(f)*(-I*f/abs(f) + 1)) + 1/48*I*sqrt(6)*sqrt(pi)*erf(-1/24*sqrt(6)*sqrt(f)*(12*x + (pi*b*sgn(f)
- pi*b + 2*I*b*log(abs(f)))/f)*(I*f/abs(f) + 1))*e^(-1/24*I*pi^2*b^2*sgn(f)/f - 1/12*pi*b^2*log(abs(f))*sgn(f)
/f + 1/24*I*pi^2*b^2/f + 1/12*pi*b^2*log(abs(f))/f - 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi
*a + a*log(abs(f)) - 3*I*d)/(sqrt(f)*(I*f/abs(f) + 1)) - 3/16*I*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x + (pi*b
*sgn(f) - pi*b + 2*I*b*log(abs(f)))/f)*(I*f/abs(f) + 1)*sqrt(abs(f)))*e^(-1/8*I*pi^2*b^2*sgn(f)/f - 1/4*pi*b^2
*log(abs(f))*sgn(f)/f + 1/8*I*pi^2*b^2/f + 1/4*pi*b^2*log(abs(f))/f - 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*s
gn(f) + 1/2*I*pi*a + a*log(abs(f)) - I*d)/((I*f/abs(f) + 1)*sqrt(abs(f)))