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

Optimal. Leaf size=225 \[ -\frac{\sqrt{\pi } f^a e^{\frac{b^2 \log ^2(f)}{8 f-4 c \log (f)}-2 d} \text{Erf}\left (\frac{b \log (f)-2 x (2 f-c \log (f))}{2 \sqrt{2 f-c \log (f)}}\right )}{8 \sqrt{2 f-c \log (f)}}+\frac{\sqrt{\pi } f^a e^{2 d-\frac{b^2 \log ^2(f)}{4 c \log (f)+8 f}} \text{Erfi}\left (\frac{b \log (f)+2 x (c \log (f)+2 f)}{2 \sqrt{c \log (f)+2 f}}\right )}{8 \sqrt{c \log (f)+2 f}}-\frac{\sqrt{\pi } f^{a-\frac{b^2}{4 c}} \text{Erfi}\left (\frac{\sqrt{\log (f)} (b+2 c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c} \sqrt{\log (f)}} \]

[Out]

-(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(4*Sqrt[c]*Sqrt[Log[f]]) - (E^(-2*d
+ (b^2*Log[f]^2)/(8*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(b*Log[f] - 2*x*(2*f - c*Log[f]))/(2*Sqrt[2*f - c*Log[f
]])])/(8*Sqrt[2*f - c*Log[f]]) + (E^(2*d - (b^2*Log[f]^2)/(8*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(b*Log[f] + 2*
x*(2*f + c*Log[f]))/(2*Sqrt[2*f + c*Log[f]])])/(8*Sqrt[2*f + c*Log[f]])

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Rubi [A]  time = 0.394981, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {5512, 2234, 2204, 2287, 2205} \[ -\frac{\sqrt{\pi } f^a e^{\frac{b^2 \log ^2(f)}{8 f-4 c \log (f)}-2 d} \text{Erf}\left (\frac{b \log (f)-2 x (2 f-c \log (f))}{2 \sqrt{2 f-c \log (f)}}\right )}{8 \sqrt{2 f-c \log (f)}}+\frac{\sqrt{\pi } f^a e^{2 d-\frac{b^2 \log ^2(f)}{4 c \log (f)+8 f}} \text{Erfi}\left (\frac{b \log (f)+2 x (c \log (f)+2 f)}{2 \sqrt{c \log (f)+2 f}}\right )}{8 \sqrt{c \log (f)+2 f}}-\frac{\sqrt{\pi } f^{a-\frac{b^2}{4 c}} \text{Erfi}\left (\frac{\sqrt{\log (f)} (b+2 c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c} \sqrt{\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.19205, size = 257, normalized size = 1.14 \[ \frac{1}{8} \sqrt{\pi } f^a \left (-\frac{e^{-\frac{b^2 \log ^2(f)}{4 c \log (f)+8 f}} \left (\sqrt{2 f-c \log (f)} (c \log (f)+2 f) (\cosh (2 d)-\sinh (2 d)) e^{\frac{b^2 f \log ^2(f)}{4 f^2-c^2 \log ^2(f)}} \text{Erf}\left (\frac{4 f x-\log (f) (b+2 c x)}{2 \sqrt{2 f-c \log (f)}}\right )+(2 f-c \log (f)) \sqrt{c \log (f)+2 f} (\sinh (2 d)+\cosh (2 d)) \text{Erfi}\left (\frac{\log (f) (b+2 c x)+4 f x}{2 \sqrt{c \log (f)+2 f}}\right )\right )}{c^2 \log ^2(f)-4 f^2}-\frac{2 f^{-\frac{b^2}{4 c}} \text{Erfi}\left (\frac{\sqrt{\log (f)} (b+2 c x)}{2 \sqrt{c}}\right )}{\sqrt{c} \sqrt{\log (f)}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^a*Sqrt[Pi]*((-2*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(Sqrt[c]*f^(b^2/(4*c))*Sqrt[Log[f]]) - (E^((b
^2*f*Log[f]^2)/(4*f^2 - c^2*Log[f]^2))*Erf[(4*f*x - (b + 2*c*x)*Log[f])/(2*Sqrt[2*f - c*Log[f]])]*Sqrt[2*f - c
*Log[f]]*(2*f + c*Log[f])*(Cosh[2*d] - Sinh[2*d]) + Erfi[(4*f*x + (b + 2*c*x)*Log[f])/(2*Sqrt[2*f + c*Log[f]])
]*(2*f - c*Log[f])*Sqrt[2*f + c*Log[f]]*(Cosh[2*d] + Sinh[2*d]))/(E^((b^2*Log[f]^2)/(8*f + 4*c*Log[f]))*(-4*f^
2 + c^2*Log[f]^2))))/8

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Maple [A]  time = 0.208, size = 217, normalized size = 1. \begin{align*} -{\frac{\sqrt{\pi }{f}^{a}}{8}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+8\,d\ln \left ( f \right ) c-16\,df}{4\,c\ln \left ( f \right ) -8\,f}}}}{\it Erf} \left ( -x\sqrt{2\,f-c\ln \left ( f \right ) }+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{2\,f-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{2\,f-c\ln \left ( f \right ) }}}}-{\frac{\sqrt{\pi }{f}^{a}}{8}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-8\,d\ln \left ( f \right ) c-16\,df}{8\,f+4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) -2\,f}x+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) -2\,f}}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) -2\,f}}}}+{\frac{\sqrt{\pi }{f}^{a}}{4}{f}^{-{\frac{{b}^{2}}{4\,c}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{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(f*x^2+d)^2,x)

[Out]

-1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2+8*d*ln(f)*c-16*d*f)/(-2*f+c*ln(f)))/(2*f-c*ln(f))^(1/2)*erf(-x*(2*f-c*
ln(f))^(1/2)+1/2*ln(f)*b/(2*f-c*ln(f))^(1/2))-1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2-8*d*ln(f)*c-16*d*f)/(2*f+
c*ln(f)))/(-c*ln(f)-2*f)^(1/2)*erf(-(-c*ln(f)-2*f)^(1/2)*x+1/2*ln(f)*b/(-c*ln(f)-2*f)^(1/2))+1/4*Pi^(1/2)*f^a*
f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c*ln(f))^(1/2))

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Maxima [A]  time = 1.10386, size = 269, normalized size = 1.2 \begin{align*} \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right ) - 2 \, f} x - \frac{b \log \left (f\right )}{2 \, \sqrt{-c \log \left (f\right ) - 2 \, f}}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2}}{4 \,{\left (c \log \left (f\right ) + 2 \, f\right )}} + 2 \, d\right )}}{8 \, \sqrt{-c \log \left (f\right ) - 2 \, f}} + \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right ) + 2 \, f} x - \frac{b \log \left (f\right )}{2 \, \sqrt{-c \log \left (f\right ) + 2 \, f}}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2}}{4 \,{\left (c \log \left (f\right ) - 2 \, f\right )}} - 2 \, d\right )}}{8 \, \sqrt{-c \log \left (f\right ) + 2 \, f}} - \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right )} x - \frac{b \log \left (f\right )}{2 \, \sqrt{-c \log \left (f\right )}}\right )}{4 \, \sqrt{-c \log \left (f\right )} f^{\frac{b^{2}}{4 \, c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - 2*f)*x - 1/2*b*log(f)/sqrt(-c*log(f) - 2*f))*e^(-1/4*b^2*log(f)^2/(c*log
(f) + 2*f) + 2*d)/sqrt(-c*log(f) - 2*f) + 1/8*sqrt(pi)*f^a*erf(sqrt(-c*log(f) + 2*f)*x - 1/2*b*log(f)/sqrt(-c*
log(f) + 2*f))*e^(-1/4*b^2*log(f)^2/(c*log(f) - 2*f) - 2*d)/sqrt(-c*log(f) + 2*f) - 1/4*sqrt(pi)*f^a*erf(sqrt(
-c*log(f))*x - 1/2*b*log(f)/sqrt(-c*log(f)))/(sqrt(-c*log(f))*f^(1/4*b^2/c))

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Fricas [B]  time = 1.89417, size = 1269, normalized size = 5.64 \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(f*x^2+d)^2,x, algorithm="fricas")

[Out]

-1/8*((sqrt(pi)*(c^2*log(f)^2 + 2*c*f*log(f))*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 16*d*f + 8*(c*d + a*f)*log(f
))/(c*log(f) - 2*f)) + sqrt(pi)*(c^2*log(f)^2 + 2*c*f*log(f))*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 16*d*f + 8*(
c*d + a*f)*log(f))/(c*log(f) - 2*f)))*sqrt(-c*log(f) + 2*f)*erf(-1/2*(4*f*x - (2*c*x + b)*log(f))*sqrt(-c*log(
f) + 2*f)/(c*log(f) - 2*f)) + (sqrt(pi)*(c^2*log(f)^2 - 2*c*f*log(f))*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 16*d
*f - 8*(c*d + a*f)*log(f))/(c*log(f) + 2*f)) + sqrt(pi)*(c^2*log(f)^2 - 2*c*f*log(f))*sinh(-1/4*((b^2 - 4*a*c)
*log(f)^2 - 16*d*f - 8*(c*d + a*f)*log(f))/(c*log(f) + 2*f)))*sqrt(-c*log(f) - 2*f)*erf(1/2*(4*f*x + (2*c*x +
b)*log(f))*sqrt(-c*log(f) - 2*f)/(c*log(f) + 2*f)) - 2*(sqrt(pi)*(c^2*log(f)^2 - 4*f^2)*cosh(-1/4*(b^2 - 4*a*c
)*log(f)/c) + sqrt(pi)*(c^2*log(f)^2 - 4*f^2)*sinh(-1/4*(b^2 - 4*a*c)*log(f)/c))*sqrt(-c*log(f))*erf(1/2*(2*c*
x + b)*sqrt(-c*log(f))/c))/(c^3*log(f)^3 - 4*c*f^2*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(f*x**2+d)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.29495, size = 323, normalized size = 1.44 \begin{align*} -\frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right ) - 2 \, f}{\left (2 \, x + \frac{b \log \left (f\right )}{c \log \left (f\right ) + 2 \, f}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 8 \, c d \log \left (f\right ) - 8 \, a f \log \left (f\right ) - 16 \, d f}{4 \,{\left (c \log \left (f\right ) + 2 \, f\right )}}\right )}}{8 \, \sqrt{-c \log \left (f\right ) - 2 \, f}} - \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right ) + 2 \, f}{\left (2 \, x + \frac{b \log \left (f\right )}{c \log \left (f\right ) - 2 \, f}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 8 \, c d \log \left (f\right ) + 8 \, a f \log \left (f\right ) - 16 \, d f}{4 \,{\left (c \log \left (f\right ) - 2 \, f\right )}}\right )}}{8 \, \sqrt{-c \log \left (f\right ) + 2 \, f}} + \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right )}{\left (2 \, x + \frac{b}{c}\right )}\right ) e^{\left (-\frac{b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{4 \, \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(f*x^2+d)^2,x, algorithm="giac")

[Out]

-1/8*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) - 2*f)*(2*x + b*log(f)/(c*log(f) + 2*f)))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*
log(f)^2 - 8*c*d*log(f) - 8*a*f*log(f) - 16*d*f)/(c*log(f) + 2*f))/sqrt(-c*log(f) - 2*f) - 1/8*sqrt(pi)*erf(-1
/2*sqrt(-c*log(f) + 2*f)*(2*x + b*log(f)/(c*log(f) - 2*f)))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 + 8*c*d*log
(f) + 8*a*f*log(f) - 16*d*f)/(c*log(f) - 2*f))/sqrt(-c*log(f) + 2*f) + 1/4*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(
2*x + b/c))*e^(-1/4*(b^2*log(f) - 4*a*c*log(f))/c)/sqrt(-c*log(f))