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

Optimal. Leaf size=315 \[ -\frac{3 \sqrt{\pi } f^a e^{-\frac{(e-b \log (f))^2}{4 c \log (f)}-d} \text{Erfi}\left (\frac{-b \log (f)-2 c x \log (f)+e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}+\frac{\sqrt{\pi } f^a e^{-\frac{(3 e-b \log (f))^2}{4 c \log (f)}-3 d} \text{Erfi}\left (\frac{-b \log (f)-2 c x \log (f)+3 e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}-\frac{3 \sqrt{\pi } f^a e^{d-\frac{(b \log (f)+e)^2}{4 c \log (f)}} \text{Erfi}\left (\frac{b \log (f)+2 c x \log (f)+e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}+\frac{\sqrt{\pi } f^a e^{3 d-\frac{(b \log (f)+3 e)^2}{4 c \log (f)}} \text{Erfi}\left (\frac{b \log (f)+2 c x \log (f)+3 e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}} \]

[Out]

(-3*E^(-d - (e - b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log
[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) + (E^(-3*d - (3*e - b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(3*e - b*Log
[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) - (3*E^(d - (e + b*Log[f])^2/(4*c*Log
[f]))*f^a*Sqrt[Pi]*Erfi[(e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) + (
E^(3*d - (3*e + b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(3*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Lo
g[f]])])/(16*Sqrt[c]*Sqrt[Log[f]])

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Rubi [A]  time = 0.4624, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5512, 2287, 2234, 2204} \[ -\frac{3 \sqrt{\pi } f^a e^{-\frac{(e-b \log (f))^2}{4 c \log (f)}-d} \text{Erfi}\left (\frac{-b \log (f)-2 c x \log (f)+e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}+\frac{\sqrt{\pi } f^a e^{-\frac{(3 e-b \log (f))^2}{4 c \log (f)}-3 d} \text{Erfi}\left (\frac{-b \log (f)-2 c x \log (f)+3 e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}-\frac{3 \sqrt{\pi } f^a e^{d-\frac{(b \log (f)+e)^2}{4 c \log (f)}} \text{Erfi}\left (\frac{b \log (f)+2 c x \log (f)+e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}+\frac{\sqrt{\pi } f^a e^{3 d-\frac{(b \log (f)+3 e)^2}{4 c \log (f)}} \text{Erfi}\left (\frac{b \log (f)+2 c x \log (f)+3 e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*E^(-d - (e - b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log
[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) + (E^(-3*d - (3*e - b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(3*e - b*Log
[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) - (3*E^(d - (e + b*Log[f])^2/(4*c*Log
[f]))*f^a*Sqrt[Pi]*Erfi[(e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) + (
E^(3*d - (3*e + b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(3*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Lo
g[f]])])/(16*Sqrt[c]*Sqrt[Log[f]])

Rule 5512

Int[(F_)^(u_)*Sinh[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sinh[v]^n, x], x] /; FreeQ[F, x] && (Linea
rQ[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]

Rubi steps

\begin{align*} \int f^{a+b x+c x^2} \sinh ^3(d+e x) \, dx &=\int \left (-\frac{1}{8} e^{-3 d-3 e x} f^{a+b x+c x^2}+\frac{3}{8} e^{-d-e x} f^{a+b x+c x^2}-\frac{3}{8} e^{d+e x} f^{a+b x+c x^2}+\frac{1}{8} e^{3 d+3 e x} f^{a+b x+c x^2}\right ) \, dx\\ &=-\left (\frac{1}{8} \int e^{-3 d-3 e x} f^{a+b x+c x^2} \, dx\right )+\frac{1}{8} \int e^{3 d+3 e x} f^{a+b x+c x^2} \, dx+\frac{3}{8} \int e^{-d-e x} f^{a+b x+c x^2} \, dx-\frac{3}{8} \int e^{d+e x} f^{a+b x+c x^2} \, dx\\ &=-\left (\frac{1}{8} \int \exp \left (-3 d+a \log (f)+c x^2 \log (f)-x (3 e-b \log (f))\right ) \, dx\right )+\frac{1}{8} \int \exp \left (3 d+a \log (f)+c x^2 \log (f)+x (3 e+b \log (f))\right ) \, dx+\frac{3}{8} \int \exp \left (-d+a \log (f)+c x^2 \log (f)-x (e-b \log (f))\right ) \, dx-\frac{3}{8} \int \exp \left (d+a \log (f)+c x^2 \log (f)+x (e+b \log (f))\right ) \, dx\\ &=\frac{1}{8} \left (3 e^{-d-\frac{(e-b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac{(-e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx-\frac{1}{8} \left (e^{-3 d-\frac{(3 e-b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac{(-3 e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx-\frac{1}{8} \left (3 e^{d-\frac{(e+b \log (f))^2}{4 c \log (f)}} f^a\right ) \int e^{\frac{(e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}} \, dx+\frac{1}{8} \left (e^{3 d-\frac{(3 e+b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac{(3 e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx\\ &=-\frac{3 e^{-d-\frac{(e-b \log (f))^2}{4 c \log (f)}} f^a \sqrt{\pi } \text{erfi}\left (\frac{e-b \log (f)-2 c x \log (f)}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}+\frac{e^{-3 d-\frac{(3 e-b \log (f))^2}{4 c \log (f)}} f^a \sqrt{\pi } \text{erfi}\left (\frac{3 e-b \log (f)-2 c x \log (f)}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}-\frac{3 e^{d-\frac{(e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt{\pi } \text{erfi}\left (\frac{e+b \log (f)+2 c x \log (f)}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}+\frac{e^{3 d-\frac{(3 e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt{\pi } \text{erfi}\left (\frac{3 e+b \log (f)+2 c x \log (f)}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{16 \sqrt{c} \sqrt{\log (f)}}\\ \end{align*}

Mathematica [A]  time = 0.984456, size = 263, normalized size = 0.83 \[ \frac{\sqrt{\pi } f^{a-\frac{b^2}{4 c}} e^{-\frac{3 e (2 b \log (f)+3 e)}{4 c \log (f)}} \left ((\sinh (d)+\cosh (d)) \left (3 (\cosh (2 d)-\sinh (2 d)) e^{\frac{2 e (b \log (f)+e)}{c \log (f)}} \text{Erfi}\left (\frac{\log (f) (b+2 c x)-e}{2 \sqrt{c} \sqrt{\log (f)}}\right )+(\sinh (2 d)+\cosh (2 d)) \text{Erfi}\left (\frac{\log (f) (b+2 c x)+3 e}{2 \sqrt{c} \sqrt{\log (f)}}\right )-3 e^{\frac{e (b \log (f)+2 e)}{c \log (f)}} \text{Erfi}\left (\frac{\log (f) (b+2 c x)+e}{2 \sqrt{c} \sqrt{\log (f)}}\right )\right )-e^{\frac{3 b e}{c}} (\cosh (3 d)-\sinh (3 d)) \text{Erfi}\left (\frac{\log (f) (b+2 c x)-3 e}{2 \sqrt{c} \sqrt{\log (f)}}\right )\right )}{16 \sqrt{c} \sqrt{\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a - b^2/(4*c))*Sqrt[Pi]*((Cosh[d] + Sinh[d])*(-3*E^((e*(2*e + b*Log[f]))/(c*Log[f]))*Erfi[(e + (b + 2*c*x)
*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] + 3*E^((2*e*(e + b*Log[f]))/(c*Log[f]))*Erfi[(-e + (b + 2*c*x)*Log[f])/(2*S
qrt[c]*Sqrt[Log[f]])]*(Cosh[2*d] - Sinh[2*d]) + Erfi[(3*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos
h[2*d] + Sinh[2*d])) - E^((3*b*e)/c)*Erfi[(-3*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[3*d] - S
inh[3*d])))/(16*Sqrt[c]*E^((3*e*(3*e + 2*b*Log[f]))/(4*c*Log[f]))*Sqrt[Log[f]])

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Maple [A]  time = 0.163, size = 316, normalized size = 1. \begin{align*} -{\frac{\sqrt{\pi }{f}^{a}}{16}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+6\,\ln \left ( f \right ) be-12\,d\ln \left ( f \right ) c+9\,{e}^{2}}{4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{3\,e+b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}}+{\frac{\sqrt{\pi }{f}^{a}}{16}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-6\,\ln \left ( f \right ) be+12\,d\ln \left ( f \right ) c+9\,{e}^{2}}{4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{b\ln \left ( f \right ) -3\,e}{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}}-{\frac{3\,\sqrt{\pi }{f}^{a}}{16}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-2\,\ln \left ( f \right ) be+4\,d\ln \left ( f \right ) c+{e}^{2}}{4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{b\ln \left ( f \right ) -e}{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}}+{\frac{3\,\sqrt{\pi }{f}^{a}}{16}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+2\,\ln \left ( f \right ) be-4\,d\ln \left ( f \right ) c+{e}^{2}}{4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{e+b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09426, size = 355, normalized size = 1.13 \begin{align*} \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right )} x - \frac{b \log \left (f\right ) + 3 \, e}{2 \, \sqrt{-c \log \left (f\right )}}\right ) e^{\left (3 \, d - \frac{{\left (b \log \left (f\right ) + 3 \, e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} - \frac{3 \, \sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right )} x - \frac{b \log \left (f\right ) + e}{2 \, \sqrt{-c \log \left (f\right )}}\right ) e^{\left (d - \frac{{\left (b \log \left (f\right ) + e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} + \frac{3 \, \sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right )} x - \frac{b \log \left (f\right ) - e}{2 \, \sqrt{-c \log \left (f\right )}}\right ) e^{\left (-d - \frac{{\left (b \log \left (f\right ) - e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} - \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right )} x - \frac{b \log \left (f\right ) - 3 \, e}{2 \, \sqrt{-c \log \left (f\right )}}\right ) e^{\left (-3 \, d - \frac{{\left (b \log \left (f\right ) - 3 \, e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.04397, size = 1434, normalized size = 4.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 9*e^2 - 6*(2*c*d - b*e)*log(f))/(c*log(f)
)) + sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 9*e^2 - 6*(2*c*d - b*e)*log(f))/(c*log(f))))*erf(1/2*((2*c*x
+ b)*log(f) + 3*e)*sqrt(-c*log(f))/(c*log(f))) - 3*sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^
2 + e^2 - 2*(2*c*d - b*e)*log(f))/(c*log(f))) + sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 - 2*(2*c*d -
b*e)*log(f))/(c*log(f))))*erf(1/2*((2*c*x + b)*log(f) + e)*sqrt(-c*log(f))/(c*log(f))) + 3*sqrt(-c*log(f))*(sq
rt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 + 2*(2*c*d - b*e)*log(f))/(c*log(f))) + sqrt(pi)*sinh(-1/4*((b^
2 - 4*a*c)*log(f)^2 + e^2 + 2*(2*c*d - b*e)*log(f))/(c*log(f))))*erf(1/2*((2*c*x + b)*log(f) - e)*sqrt(-c*log(
f))/(c*log(f))) - sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 9*e^2 + 6*(2*c*d - b*e)*log(f)
)/(c*log(f))) + sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 9*e^2 + 6*(2*c*d - b*e)*log(f))/(c*log(f))))*erf(
1/2*((2*c*x + b)*log(f) - 3*e)*sqrt(-c*log(f))/(c*log(f))))/(c*log(f))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.25175, size = 463, normalized size = 1.47 \begin{align*} \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right )}{\left (2 \, x + \frac{b \log \left (f\right ) - 3 \, e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 12 \, c d \log \left (f\right ) - 6 \, b e \log \left (f\right ) + 9 \, e^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} - \frac{3 \, \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right )}{\left (2 \, x + \frac{b \log \left (f\right ) - e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) - 2 \, b e \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} + \frac{3 \, \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right )}{\left (2 \, x + \frac{b \log \left (f\right ) + e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) + 2 \, b e \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} - \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right )}{\left (2 \, x + \frac{b \log \left (f\right ) + 3 \, e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 12 \, c d \log \left (f\right ) + 6 \, b e \log \left (f\right ) + 9 \, e^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, \sqrt{-c \log \left (f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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