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

Optimal. Leaf size=354 \[ -\frac {3 i \sqrt {\pi } f^a e^{\frac {(e+i b \log (f))^2}{4 c \log (f)}-i d} \text {erfi}\left (\frac {-b \log (f)-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i \sqrt {\pi } f^a e^{\frac {(3 e+i b \log (f))^2}{4 c \log (f)}-3 i d} \text {erfi}\left (\frac {-b \log (f)-2 c x \log (f)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {3 i \sqrt {\pi } f^a e^{\frac {(e-i b \log (f))^2}{4 c \log (f)}+i d} \text {erfi}\left (\frac {b \log (f)+2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i \sqrt {\pi } f^a e^{3 i d-\frac {(b \log (f)+3 i e)^2}{4 c \log (f)}} \text {erfi}\left (\frac {b \log (f)+2 c x \log (f)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

3/16*I*exp(-I*d+1/4*(e+I*b*ln(f))^2/c/ln(f))*f^a*erfi(1/2*(-I*e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(
1/2)/c^(1/2)/ln(f)^(1/2)-1/16*I*exp(-3*I*d+1/4*(3*e+I*b*ln(f))^2/c/ln(f))*f^a*erfi(1/2*(-3*I*e+b*ln(f)+2*c*x*l
n(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)-3/16*I*exp(I*d+1/4*(e-I*b*ln(f))^2/c/ln(f))*f^a*erfi(1
/2*(I*e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)+1/16*I*exp(3*I*d-1/4*(3*I*e+b*l
n(f))^2/c/ln(f))*f^a*erfi(1/2*(3*I*e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 354, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4472, 2287, 2234, 2204} \[ -\frac {3 i \sqrt {\pi } f^a e^{\frac {(e+i b \log (f))^2}{4 c \log (f)}-i d} \text {Erfi}\left (\frac {-b \log (f)-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i \sqrt {\pi } f^a e^{\frac {(3 e+i b \log (f))^2}{4 c \log (f)}-3 i d} \text {Erfi}\left (\frac {-b \log (f)-2 c x \log (f)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {3 i \sqrt {\pi } f^a e^{\frac {(e-i b \log (f))^2}{4 c \log (f)}+i d} \text {Erfi}\left (\frac {b \log (f)+2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i \sqrt {\pi } f^a e^{3 i d-\frac {(b \log (f)+3 i e)^2}{4 c \log (f)}} \text {Erfi}\left (\frac {b \log (f)+2 c x \log (f)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/16)*E^((-I)*d + (e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e - b*Log[f] - 2*c*x*Log[f])/(2
*Sqrt[c]*Sqrt[Log[f]])])/(Sqrt[c]*Sqrt[Log[f]]) + ((I/16)*E^((-3*I)*d + (3*e + I*b*Log[f])^2/(4*c*Log[f]))*f^a
*Sqrt[Pi]*Erfi[((3*I)*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(Sqrt[c]*Sqrt[Log[f]]) - (((3*I)
/16)*E^(I*d + (e - I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sq
rt[Log[f]])])/(Sqrt[c]*Sqrt[Log[f]]) + ((I/16)*E^((3*I)*d - ((3*I)*e + b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*
Erfi[((3*I)*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(Sqrt[c]*Sqrt[Log[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 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+c x^2} \sin ^3(d+e x) \, dx &=\int \left (\frac {3}{8} i e^{-i d-i e x} f^{a+b x+c x^2}-\frac {3}{8} i e^{i d+i e x} f^{a+b x+c x^2}-\frac {1}{8} i e^{-3 i d-3 i e x} f^{a+b x+c x^2}+\frac {1}{8} i e^{3 i d+3 i e x} f^{a+b x+c x^2}\right ) \, dx\\ &=-\left (\frac {1}{8} i \int e^{-3 i d-3 i e x} f^{a+b x+c x^2} \, dx\right )+\frac {1}{8} i \int e^{3 i d+3 i e x} f^{a+b x+c x^2} \, dx+\frac {3}{8} i \int e^{-i d-i e x} f^{a+b x+c x^2} \, dx-\frac {3}{8} i \int e^{i d+i e x} f^{a+b x+c x^2} \, dx\\ &=-\left (\frac {1}{8} i \int \exp \left (-3 i d+a \log (f)+c x^2 \log (f)-x (3 i e-b \log (f))\right ) \, dx\right )+\frac {1}{8} i \int \exp \left (3 i d+a \log (f)+c x^2 \log (f)+x (3 i e+b \log (f))\right ) \, dx+\frac {3}{8} i \int \exp \left (-i d+a \log (f)+c x^2 \log (f)-x (i e-b \log (f))\right ) \, dx-\frac {3}{8} i \int \exp \left (i d+a \log (f)+c x^2 \log (f)+x (i e+b \log (f))\right ) \, dx\\ &=-\left (\frac {1}{8} \left (3 i e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx\right )+\frac {1}{8} \left (3 i e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(-i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx-\frac {1}{8} \left (i \exp \left (-3 i d+\frac {(3 e+i b \log (f))^2}{4 c \log (f)}\right ) f^a\right ) \int \exp \left (\frac {(-3 i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx+\frac {1}{8} \left (i e^{3 i d-\frac {(3 i e+b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(3 i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx\\ &=-\frac {3 i e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i \exp \left (-3 i d+\frac {(3 e+i b \log (f))^2}{4 c \log (f)}\right ) f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {3 i e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i e^{3 i d-\frac {(3 i e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}\\ \end {align*}

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Mathematica [A]  time = 0.98, size = 391, normalized size = 1.10 \[ \frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} e^{\frac {e (e-6 i b \log (f))}{4 c \log (f)}} \left (-\sin (3 d) e^{\frac {2 e^2}{c \log (f)}} \text {erfi}\left (\frac {\log (f) (b+2 c x)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )+i \cos (3 d) e^{\frac {2 e^2}{c \log (f)}} \text {erfi}\left (\frac {\log (f) (b+2 c x)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )-\sin (3 d) e^{\frac {e (2 e+3 i b \log (f))}{c \log (f)}} \text {erfi}\left (\frac {\log (f) (b+2 c x)-3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )-i \cos (3 d) e^{\frac {e (2 e+3 i b \log (f))}{c \log (f)}} \text {erfi}\left (\frac {\log (f) (b+2 c x)-3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )+3 i e^{\frac {i b e}{c}} (\cos (d)+i \sin (d)) \text {erfi}\left (\frac {-\log (f) (b+2 c x)-i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )+3 e^{\frac {2 i b e}{c}} (\sin (d)+i \cos (d)) \text {erfi}\left (\frac {\log (f) (b+2 c x)-i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )\right )}{16 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^((e*(e - (6*I)*b*Log[f]))/(4*c*Log[f]))*f^(a - b^2/(4*c))*Sqrt[Pi]*((-I)*E^((e*(2*e + (3*I)*b*Log[f]))/(c*L
og[f]))*Cos[3*d]*Erfi[((-3*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] + I*E^((2*e^2)/(c*Log[f]))*Cos
[3*d]*Erfi[((3*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] + (3*I)*E^((I*b*e)/c)*Erfi[((-I)*e - (b +
2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[d] + I*Sin[d]) + 3*E^(((2*I)*b*e)/c)*Erfi[((-I)*e + (b + 2*c*x)*
Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(I*Cos[d] + Sin[d]) - E^((e*(2*e + (3*I)*b*Log[f]))/(c*Log[f]))*Erfi[((-3*I)
*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*Sin[3*d] - E^((2*e^2)/(c*Log[f]))*Erfi[((3*I)*e + (b + 2*c*
x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*Sin[3*d]))/(16*Sqrt[c]*Sqrt[Log[f]])

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fricas [A]  time = 1.04, size = 346, normalized size = 0.98 \[ \frac {3 i \, \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \relax (f) + i \, e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} - e^{2} - {\left (4 i \, c d - 2 i \, b e\right )} \log \relax (f)}{4 \, c \log \relax (f)}\right )} - 3 i \, \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \relax (f) - i \, e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} - e^{2} - {\left (-4 i \, c d + 2 i \, b e\right )} \log \relax (f)}{4 \, c \log \relax (f)}\right )} - i \, \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \relax (f) + 3 i \, e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} - 9 \, e^{2} - {\left (12 i \, c d - 6 i \, b e\right )} \log \relax (f)}{4 \, c \log \relax (f)}\right )} + i \, \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \relax (f) - 3 i \, e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} - 9 \, e^{2} - {\left (-12 i \, c d + 6 i \, b e\right )} \log \relax (f)}{4 \, c \log \relax (f)}\right )}}{16 \, c \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*I*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x + b)*log(f) + I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(-1/4*((b^
2 - 4*a*c)*log(f)^2 - e^2 - (4*I*c*d - 2*I*b*e)*log(f))/(c*log(f))) - 3*I*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2
*c*x + b)*log(f) - I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(-1/4*((b^2 - 4*a*c)*log(f)^2 - e^2 - (-4*I*c*d + 2*I*b*
e)*log(f))/(c*log(f))) - I*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x + b)*log(f) + 3*I*e)*sqrt(-c*log(f))/(c*lo
g(f)))*e^(-1/4*((b^2 - 4*a*c)*log(f)^2 - 9*e^2 - (12*I*c*d - 6*I*b*e)*log(f))/(c*log(f))) + I*sqrt(pi)*sqrt(-c
*log(f))*erf(1/2*((2*c*x + b)*log(f) - 3*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(-1/4*((b^2 - 4*a*c)*log(f)^2 - 9*
e^2 - (-12*I*c*d + 6*I*b*e)*log(f))/(c*log(f))))/(c*log(f))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{c x^{2} + b x + a} \sin \left (e x + d\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.93, size = 338, normalized size = 0.95 \[ -\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}+6 i \ln \relax (f ) b e -12 i d \ln \relax (f ) c -9 e^{2}}{4 \ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {3 i e +b \ln \relax (f )}{2 \sqrt {-c \ln \relax (f )}}\right )}{16 \sqrt {-c \ln \relax (f )}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}-6 i \ln \relax (f ) b e +12 i d \ln \relax (f ) c -9 e^{2}}{4 \ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {b \ln \relax (f )-3 i e}{2 \sqrt {-c \ln \relax (f )}}\right )}{16 \sqrt {-c \ln \relax (f )}}-\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {-e^{2}-2 i \ln \relax (f ) b e +4 i d \ln \relax (f ) c +\ln \relax (f )^{2} b^{2}}{4 \ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {-i e +b \ln \relax (f )}{2 \sqrt {-c \ln \relax (f )}}\right )}{16 \sqrt {-c \ln \relax (f )}}+\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {-e^{2}+2 i \ln \relax (f ) b e -4 i d \ln \relax (f ) c +\ln \relax (f )^{2} b^{2}}{4 \ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {i e +b \ln \relax (f )}{2 \sqrt {-c \ln \relax (f )}}\right )}{16 \sqrt {-c \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.41, size = 684, normalized size = 1.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*sqrt(pi)*(f^a*(I*cos(-3/2*(2*c*d - b*e)/c) + sin(-3/2*(2*c*d - b*e)/c))*erf(x*conjugate(sqrt(-c*log(f)))
- 1/2*(b*log(f) + 3*I*e)*conjugate(1/sqrt(-c*log(f))))*e^(9/4*e^2/(c*log(f))) + f^a*(-I*cos(-3/2*(2*c*d - b*e
)/c) + sin(-3/2*(2*c*d - b*e)/c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*log(f) - 3*I*e)*conjugate(1/sqrt(-
c*log(f))))*e^(9/4*e^2/(c*log(f))) + f^a*(I*cos(-3/2*(2*c*d - b*e)/c) + sin(-3/2*(2*c*d - b*e)/c))*erf(1/2*(2*
c*x*log(f) + b*log(f) + 3*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(9/4*e^2/(c*log(f))) + f^a*(-I*cos(-3/2*(2*c*d -
b*e)/c) + sin(-3/2*(2*c*d - b*e)/c))*erf(1/2*(2*c*x*log(f) + b*log(f) - 3*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(
9/4*e^2/(c*log(f))) + f^a*(-3*I*cos(-1/2*(2*c*d - b*e)/c) - 3*sin(-1/2*(2*c*d - b*e)/c))*erf(x*conjugate(sqrt(
-c*log(f))) - 1/2*(b*log(f) + I*e)*conjugate(1/sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) + f^a*(3*I*cos(-1/2*(2
*c*d - b*e)/c) - 3*sin(-1/2*(2*c*d - b*e)/c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*log(f) - I*e)*conjugat
e(1/sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) + f^a*(-3*I*cos(-1/2*(2*c*d - b*e)/c) - 3*sin(-1/2*(2*c*d - b*e)/
c))*erf(1/2*(2*c*x*log(f) + b*log(f) + I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(1/4*e^2/(c*log(f))) + f^a*(3*I*cos(
-1/2*(2*c*d - b*e)/c) - 3*sin(-1/2*(2*c*d - b*e)/c))*erf(1/2*(2*c*x*log(f) + b*log(f) - I*e)*sqrt(-c*log(f))/(
c*log(f)))*e^(1/4*e^2/(c*log(f))))*sqrt(-c*log(f))/(c*f^(1/4*b^2/c)*log(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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