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

Optimal. Leaf size=340 \[ \frac {3}{16} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )+\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (b \log (f)+3 i e)^2}{12 f}+3 i d} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (3 e+i b \log (f))^2}{12 f}-3 i d} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right ) \]

[Out]

(1/96-1/96*I)*exp(3*I*d+1/12*I*(3*I*e+b*ln(f))^2/f)*f^(-1/2+a)*erf((1/12+1/12*I)*(3*I*e+6*I*f*x+b*ln(f))*6^(1/
2)/f^(1/2))*6^(1/2)*Pi^(1/2)+(-1/96+1/96*I)*exp(-3*I*d+1/12*I*(3*e+I*b*ln(f))^2/f)*f^(-1/2+a)*erfi((1/12+1/12*
I)*(3*I*e+6*I*f*x-b*ln(f))*6^(1/2)/f^(1/2))*6^(1/2)*Pi^(1/2)+3/16*(-1)^(3/4)*exp(1/4*I*(4*d+(I*e+b*ln(f))^2/f)
)*f^(-1/2+a)*erf(1/2*(-1)^(1/4)*(I*e+2*I*f*x+b*ln(f))/f^(1/2))*Pi^(1/2)-3/16*(-1)^(3/4)*exp(-I*d+1/4*I*(e+I*b*
ln(f))^2/f)*f^(-1/2+a)*erfi(1/2*(-1)^(1/4)*(I*e+2*I*f*x-b*ln(f))/f^(1/2))*Pi^(1/2)

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Rubi [A]  time = 0.60, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4472, 2287, 2234, 2204, 2205} \[ \frac {3}{16} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {Erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )+\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (b \log (f)+3 i e)^2}{12 f}+3 i d} \text {Erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {Erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (3 e+i b \log (f))^2}{12 f}-3 i d} \text {Erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 1.50, size = 323, normalized size = 0.95 \[ \frac {1}{48} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {b e+f}{2 f}} e^{-\frac {i \left (b^2 \log ^2(f)+3 e^2\right )}{4 f}} \left (9 i (\cos (d)+i \sin (d)) e^{\frac {i \left (b^2 \log ^2(f)+e^2\right )}{2 f}} \text {erfi}\left (\frac {\sqrt [4]{-1} (-i b \log (f)+e+2 f x)}{2 \sqrt {f}}\right )+e^{\frac {i e^2}{f}} \left (\sqrt {3} (\cos (3 d)-i \sin (3 d)) e^{\frac {i \left (b^2 \log ^2(f)+3 e^2\right )}{6 f}} \text {erfi}\left (\frac {(-1)^{3/4} (i b \log (f)+3 e+6 f x)}{2 \sqrt {3} \sqrt {f}}\right )-9 (\cos (d)-i \sin (d)) \text {erfi}\left (\frac {(-1)^{3/4} (i b \log (f)+e+2 f x)}{2 \sqrt {f}}\right )\right )+\sqrt {3} e^{\frac {i b^2 \log ^2(f)}{3 f}} (\sin (3 d)-i \cos (3 d)) \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-i b \log (f)+3 e+6 f x)}{\sqrt {6} \sqrt {f}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.56, size = 629, normalized size = 1.85 \[ \frac {-i \, \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + 9 i \, e^{2} - 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{12 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {6} {\left (6 \, f x + i \, b \log \relax (f) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - i \, \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} - 9 i \, e^{2} + 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{12 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {6} {\left (6 \, f x - i \, b \log \relax (f) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) + 9 i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) + 9 i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + 9 i \, e^{2} - 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{12 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {6} {\left (6 \, f x + i \, b \log \relax (f) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) + \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} - 9 i \, e^{2} + 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{12 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {6} {\left (6 \, f x - i \, b \log \relax (f) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) + 9 \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - 9 \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right )}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(-I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 9*I*e^2 - 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fres
nel_cos(1/6*sqrt(6)*(6*f*x + I*b*log(f) + 3*e)*sqrt(f/pi)/f) - I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2
- 9*I*e^2 + 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/6*sqrt(6)*(6*f*x - I*b*log(f) + 3*e)*sqrt(f/
pi)/f) + 9*I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fres
nel_cos(1/2*sqrt(2)*(2*f*x + I*b*log(f) + e)*sqrt(f/pi)/f) + 9*I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2
- I*e^2 + 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f)
- sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 9*I*e^2 - 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_si
n(1/6*sqrt(6)*(6*f*x + I*b*log(f) + 3*e)*sqrt(f/pi)/f) + sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 - 9*I*e
^2 + 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1/6*sqrt(6)*(6*f*x - I*b*log(f) + 3*e)*sqrt(f/pi)/f) +
9*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(1/
2*sqrt(2)*(2*f*x + I*b*log(f) + e)*sqrt(f/pi)/f) - 9*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 - I*e^2 + 4*
I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1/2*sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f))/f

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giac [B]  time = 0.55, size = 763, normalized size = 2.24 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*sin(f*x^2+e*x+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)) - 2*e)/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) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*
pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs(f))/f + I*d - 1/4*I*e^2/f)/((-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)) - 6*e)/f)*(-I*f/a
bs(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*l
og(abs(f))/f + 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*
b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs(f))/f + 3*I*d - 3/4*I*e^2/f)/(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)) + 6*e)/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) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/f
+ a*log(abs(f)) - 1/2*b*e*log(abs(f))/f - 3*I*d + 3/4*I*e^2/f)/(sqrt(f)*(I*f/abs(f) + 1)) - 3/16*I*sqrt(2)*sqr
t(pi)*erf(-1/8*sqrt(2)*(4*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)) + 2*e)/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) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/f + a*log(a
bs(f)) - 1/2*b*e*log(abs(f))/f - I*d + 1/4*I*e^2/f)/((I*f/abs(f) + 1)*sqrt(abs(f)))

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maple [A]  time = 1.10, size = 311, normalized size = 0.91 \[ -\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (-9 e^{2}+6 i \ln \relax (f ) b e +\ln \relax (f )^{2} b^{2}+36 d f \right )}{12 f}} \erf \left (-\sqrt {-3 i f}\, x +\frac {3 i e +b \ln \relax (f )}{2 \sqrt {-3 i f}}\right )}{16 \sqrt {-3 i f}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (-9 e^{2}-6 i \ln \relax (f ) b e +\ln \relax (f )^{2} b^{2}+36 d f \right )}{12 f}} \sqrt {3}\, \erf \left (-\sqrt {3}\, \sqrt {i f}\, x +\frac {\left (b \ln \relax (f )-3 i e \right ) \sqrt {3}}{6 \sqrt {i f}}\right )}{48 \sqrt {i f}}-\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (-e^{2}-2 i \ln \relax (f ) b e +\ln \relax (f )^{2} b^{2}+4 d f \right )}{4 f}} \erf \left (-\sqrt {i f}\, x +\frac {-i e +b \ln \relax (f )}{2 \sqrt {i f}}\right )}{16 \sqrt {i f}}+\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (-e^{2}+2 i \ln \relax (f ) b e +\ln \relax (f )^{2} b^{2}+4 d f \right )}{4 f}} \erf \left (-\sqrt {-i f}\, x +\frac {i e +b \ln \relax (f )}{2 \sqrt {-i f}}\right )}{16 \sqrt {-i f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*I*Pi^(1/2)*f^a*exp(1/12*I*(-9*e^2+6*I*ln(f)*b*e+ln(f)^2*b^2+36*d*f)/f)/(-3*I*f)^(1/2)*erf(-(-3*I*f)^(1/2
)*x+1/2*(3*I*e+b*ln(f))/(-3*I*f)^(1/2))+1/48*I*Pi^(1/2)*f^a*exp(-1/12*I*(-9*e^2-6*I*ln(f)*b*e+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*(b*ln(f)-3*I*e)*3^(1/2)/(I*f)^(1/2))-3/16*I*Pi^(1/2)
*f^a*exp(-1/4*I*(-e^2-2*I*ln(f)*b*e+ln(f)^2*b^2+4*d*f)/f)/(I*f)^(1/2)*erf(-(I*f)^(1/2)*x+1/2*(-I*e+b*ln(f))/(I
*f)^(1/2))+3/16*I*Pi^(1/2)*f^a*exp(1/4*I*(-e^2+2*I*ln(f)*b*e+ln(f)^2*b^2+4*d*f)/f)/(-I*f)^(1/2)*erf(-(-I*f)^(1
/2)*x+1/2*(I*e+b*ln(f))/(-I*f)^(1/2))

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maxima [A]  time = 0.49, size = 377, normalized size = 1.11 \[ \frac {3 \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right ) - \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (6 i \, f x - b \log \relax (f) + 3 i \, e\right )} \sqrt {3 i \, f}}{6 \, f}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (6 i \, f x + b \log \relax (f) + 3 i \, e\right )} \sqrt {-3 i \, f}}{6 \, f}\right )\right )} f^{\frac {3}{2}} + \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (27 i + 27\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) + \left (27 i - 27\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x - b \log \relax (f) + i \, e\right )} \sqrt {i \, f}}{2 \, f}\right ) + {\left (\left (27 i - 27\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) - \left (27 i + 27\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x + b \log \relax (f) + i \, e\right )} \sqrt {-i \, f}}{2 \, f}\right )\right )} f^{\frac {3}{2}}}{288 \, f^{2} f^{\frac {b e}{2 \, f}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(3*9^(1/4)*sqrt(2)*sqrt(pi)*(((I + 1)*f^a*cos(1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f) - (I - 1)*f^a*sin(
1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f))*erf(1/6*I*(6*I*f*x - b*log(f) + 3*I*e)*sqrt(3*I*f)/f) + (-(I - 1)*f^a
*cos(1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f) + (I + 1)*f^a*sin(1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f))*erf(1/
6*I*(6*I*f*x + b*log(f) + 3*I*e)*sqrt(-3*I*f)/f))*f^(3/2) + sqrt(2)*sqrt(pi)*((-(27*I + 27)*f^a*cos(1/4*(b^2*l
og(f)^2 - e^2 + 4*d*f)/f) + (27*I - 27)*f^a*sin(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f))*erf(1/2*I*(2*I*f*x - b*lo
g(f) + I*e)*sqrt(I*f)/f) + ((27*I - 27)*f^a*cos(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f) - (27*I + 27)*f^a*sin(1/4*
(b^2*log(f)^2 - e^2 + 4*d*f)/f))*erf(1/2*I*(2*I*f*x + b*log(f) + I*e)*sqrt(-I*f)/f))*f^(3/2))/(f^2*f^(1/2*b*e/
f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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(f**(b*x+a)*sin(f*x**2+e*x+d)**3,x)

[Out]

Timed out

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