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

Optimal. Leaf size=154 \[ \frac{\sqrt{\pi } f^a e^{\frac{b^2 \log ^2(f)}{4 f-4 c \log (f)}-d} \text{Erf}\left (\frac{b \log (f)-2 x (f-c \log (f))}{2 \sqrt{f-c \log (f)}}\right )}{4 \sqrt{f-c \log (f)}}+\frac{\sqrt{\pi } f^a e^{d-\frac{b^2 \log ^2(f)}{4 (c \log (f)+f)}} \text{Erfi}\left (\frac{b \log (f)+2 x (c \log (f)+f)}{2 \sqrt{c \log (f)+f}}\right )}{4 \sqrt{c \log (f)+f}} \]

[Out]

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

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Rubi [A]  time = 0.331566, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {5512, 2287, 2234, 2205, 2204} \[ \frac{\sqrt{\pi } f^a e^{\frac{b^2 \log ^2(f)}{4 f-4 c \log (f)}-d} \text{Erf}\left (\frac{b \log (f)-2 x (f-c \log (f))}{2 \sqrt{f-c \log (f)}}\right )}{4 \sqrt{f-c \log (f)}}+\frac{\sqrt{\pi } f^a e^{d-\frac{b^2 \log ^2(f)}{4 (c \log (f)+f)}} \text{Erfi}\left (\frac{b \log (f)+2 x (c \log (f)+f)}{2 \sqrt{c \log (f)+f}}\right )}{4 \sqrt{c \log (f)+f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.785795, size = 179, normalized size = 1.16 \[ \frac{\sqrt{\pi } f^a e^{-\frac{b^2 \log ^2(f)}{4 (c \log (f)+f)}} \left (\sqrt{f-c \log (f)} (\sinh (d)+\cosh (d)) \text{Erfi}\left (\frac{\log (f) (b+2 c x)+2 f x}{2 \sqrt{c \log (f)+f}}\right )-\sqrt{c \log (f)+f} (\cosh (d)-\sinh (d)) e^{\frac{b^2 f \log ^2(f)}{2 f^2-2 c^2 \log ^2(f)}} \text{Erf}\left (\frac{2 f x-\log (f) (b+2 c x)}{2 \sqrt{f-c \log (f)}}\right )\right )}{4 \sqrt{f-c \log (f)} \sqrt{c \log (f)+f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.184, size = 160, normalized size = 1. \begin{align*} -{\frac{\sqrt{\pi }{f}^{a}}{4}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-4\,d\ln \left ( f \right ) c-4\,df}{4\,c\ln \left ( f \right ) +4\,f}}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) -f}x+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) -f}}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) -f}}}}+{\frac{\sqrt{\pi }{f}^{a}}{4}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+4\,d\ln \left ( f \right ) c-4\,df}{4\,c\ln \left ( f \right ) -4\,f}}}}{\it Erf} \left ( -x\sqrt{f-c\ln \left ( f \right ) }+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{f-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{f-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),x)

[Out]

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

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Maxima [A]  time = 1.05674, size = 188, normalized size = 1.22 \begin{align*} \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right ) - f} x - \frac{b \log \left (f\right )}{2 \, \sqrt{-c \log \left (f\right ) - f}}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2}}{4 \,{\left (c \log \left (f\right ) + f\right )}} + d\right )}}{4 \, \sqrt{-c \log \left (f\right ) - f}} - \frac{\sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{-c \log \left (f\right ) + f} x - \frac{b \log \left (f\right )}{2 \, \sqrt{-c \log \left (f\right ) + f}}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2}}{4 \,{\left (c \log \left (f\right ) - f\right )}} - d\right )}}{4 \, \sqrt{-c \log \left (f\right ) + f}} \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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.87224, size = 876, normalized size = 5.69 \begin{align*} \frac{{\left (\sqrt{\pi }{\left (c \log \left (f\right ) + f\right )} \cosh \left (-\frac{{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f + 4 \,{\left (c d + a f\right )} \log \left (f\right )}{4 \,{\left (c \log \left (f\right ) - f\right )}}\right ) + \sqrt{\pi }{\left (c \log \left (f\right ) + f\right )} \sinh \left (-\frac{{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f + 4 \,{\left (c d + a f\right )} \log \left (f\right )}{4 \,{\left (c \log \left (f\right ) - f\right )}}\right )\right )} \sqrt{-c \log \left (f\right ) + f} \operatorname{erf}\left (-\frac{{\left (2 \, f x -{\left (2 \, c x + b\right )} \log \left (f\right )\right )} \sqrt{-c \log \left (f\right ) + f}}{2 \,{\left (c \log \left (f\right ) - f\right )}}\right ) -{\left (\sqrt{\pi }{\left (c \log \left (f\right ) - f\right )} \cosh \left (-\frac{{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f - 4 \,{\left (c d + a f\right )} \log \left (f\right )}{4 \,{\left (c \log \left (f\right ) + f\right )}}\right ) + \sqrt{\pi }{\left (c \log \left (f\right ) - f\right )} \sinh \left (-\frac{{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f - 4 \,{\left (c d + a f\right )} \log \left (f\right )}{4 \,{\left (c \log \left (f\right ) + f\right )}}\right )\right )} \sqrt{-c \log \left (f\right ) - f} \operatorname{erf}\left (\frac{{\left (2 \, f x +{\left (2 \, c x + b\right )} \log \left (f\right )\right )} \sqrt{-c \log \left (f\right ) - f}}{2 \,{\left (c \log \left (f\right ) + f\right )}}\right )}{4 \,{\left (c^{2} \log \left (f\right )^{2} - f^{2}\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),x, algorithm="fricas")

[Out]

1/4*((sqrt(pi)*(c*log(f) + f)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f + 4*(c*d + a*f)*log(f))/(c*log(f) - f)
) + sqrt(pi)*(c*log(f) + f)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f + 4*(c*d + a*f)*log(f))/(c*log(f) - f)))
*sqrt(-c*log(f) + f)*erf(-1/2*(2*f*x - (2*c*x + b)*log(f))*sqrt(-c*log(f) + f)/(c*log(f) - f)) - (sqrt(pi)*(c*
log(f) - f)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f - 4*(c*d + a*f)*log(f))/(c*log(f) + f)) + sqrt(pi)*(c*lo
g(f) - f)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f - 4*(c*d + a*f)*log(f))/(c*log(f) + f)))*sqrt(-c*log(f) -
f)*erf(1/2*(2*f*x + (2*c*x + b)*log(f))*sqrt(-c*log(f) - f)/(c*log(f) + f)))/(c^2*log(f)^2 - f^2)

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

[Out]

Integral(f**(a + b*x + c*x**2)*sinh(d + f*x**2), x)

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Giac [A]  time = 1.27654, size = 244, normalized size = 1.58 \begin{align*} -\frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right ) - f}{\left (2 \, x + \frac{b \log \left (f\right )}{c \log \left (f\right ) + f}\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 ) - 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \,{\left (c \log \left (f\right ) + f\right )}}\right )}}{4 \, \sqrt{-c \log \left (f\right ) - f}} + \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{-c \log \left (f\right ) + f}{\left (2 \, x + \frac{b \log \left (f\right )}{c \log \left (f\right ) - f}\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 ) + 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \,{\left (c \log \left (f\right ) - f\right )}}\right )}}{4 \, \sqrt{-c \log \left (f\right ) + f}} \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),x, algorithm="giac")

[Out]

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