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

Optimal. Leaf size=161 \[ \frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{\frac{(2 e-b \log (f))^2}{8 f}-2 d} \text{Erf}\left (\frac{-b \log (f)+2 e+4 f x}{2 \sqrt{2} \sqrt{f}}\right )+\frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{2 d-\frac{(b \log (f)+2 e)^2}{8 f}} \text{Erfi}\left (\frac{b \log (f)+2 e+4 f x}{2 \sqrt{2} \sqrt{f}}\right )-\frac{f^{a+b x}}{2 b \log (f)} \]

[Out]

(E^(-2*d + (2*e - b*Log[f])^2/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erf[(2*e + 4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])]
)/8 + (E^(2*d - (2*e + b*Log[f])^2/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erfi[(2*e + 4*f*x + b*Log[f])/(2*Sqrt[2]*Sqr
t[f])])/8 - f^(a + b*x)/(2*b*Log[f])

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Rubi [A]  time = 0.283848, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5512, 2194, 2287, 2234, 2205, 2204} \[ \frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{\frac{(2 e-b \log (f))^2}{8 f}-2 d} \text{Erf}\left (\frac{-b \log (f)+2 e+4 f x}{2 \sqrt{2} \sqrt{f}}\right )+\frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{2 d-\frac{(b \log (f)+2 e)^2}{8 f}} \text{Erfi}\left (\frac{b \log (f)+2 e+4 f x}{2 \sqrt{2} \sqrt{f}}\right )-\frac{f^{a+b x}}{2 b \log (f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-2*d + (2*e - b*Log[f])^2/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erf[(2*e + 4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])]
)/8 + (E^(2*d - (2*e + b*Log[f])^2/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erfi[(2*e + 4*f*x + b*Log[f])/(2*Sqrt[2]*Sqr
t[f])])/8 - f^(a + b*x)/(2*b*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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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 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} \sinh ^2\left (d+e x+f x^2\right ) \, dx &=\int \left (-\frac{1}{2} f^{a+b x}+\frac{1}{4} e^{-2 d-2 e x-2 f x^2} f^{a+b x}+\frac{1}{4} e^{2 d+2 e x+2 f x^2} f^{a+b x}\right ) \, dx\\ &=\frac{1}{4} \int e^{-2 d-2 e x-2 f x^2} f^{a+b x} \, dx+\frac{1}{4} \int e^{2 d+2 e x+2 f x^2} f^{a+b x} \, dx-\frac{1}{2} \int f^{a+b x} \, dx\\ &=-\frac{f^{a+b x}}{2 b \log (f)}+\frac{1}{4} \int \exp \left (-2 d-2 f x^2+a \log (f)-x (2 e-b \log (f))\right ) \, dx+\frac{1}{4} \int \exp \left (2 d+2 f x^2+a \log (f)+x (2 e+b \log (f))\right ) \, dx\\ &=-\frac{f^{a+b x}}{2 b \log (f)}+\frac{1}{4} \left (e^{-2 d+\frac{(2 e-b \log (f))^2}{8 f}} f^a\right ) \int e^{-\frac{(-2 e-4 f x+b \log (f))^2}{8 f}} \, dx+\frac{1}{4} \left (e^{2 d-\frac{(2 e+b \log (f))^2}{8 f}} f^a\right ) \int e^{\frac{(2 e+4 f x+b \log (f))^2}{8 f}} \, dx\\ &=\frac{1}{8} e^{-2 d+\frac{(2 e-b \log (f))^2}{8 f}} f^{-\frac{1}{2}+a} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{2 e+4 f x-b \log (f)}{2 \sqrt{2} \sqrt{f}}\right )+\frac{1}{8} e^{2 d-\frac{(2 e+b \log (f))^2}{8 f}} f^{-\frac{1}{2}+a} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{2 e+4 f x+b \log (f)}{2 \sqrt{2} \sqrt{f}}\right )-\frac{f^{a+b x}}{2 b \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.625089, size = 220, normalized size = 1.37 \[ \frac{f^{a-\frac{b e+f}{2 f}} e^{-\frac{b^2 \log ^2(f)+4 e^2}{8 f}} \left (\sqrt{\pi } b \log (f) (\cosh (2 d)-\sinh (2 d)) e^{\frac{b^2 \log ^2(f)+4 e^2}{4 f}} \text{Erf}\left (\frac{-b \log (f)+2 e+4 f x}{2 \sqrt{2} \sqrt{f}}\right )-4 \sqrt{2} f^{b \left (\frac{e}{2 f}+x\right )+\frac{1}{2}} e^{\frac{b^2 \log ^2(f)+4 e^2}{8 f}}+\sqrt{\pi } b \log (f) (\sinh (2 d)+\cosh (2 d)) \text{Erfi}\left (\frac{b \log (f)+2 e+4 f x}{2 \sqrt{2} \sqrt{f}}\right )\right )}{8 \sqrt{2} b \log (f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a - (b*e + f)/(2*f))*(-4*Sqrt[2]*E^((4*e^2 + b^2*Log[f]^2)/(8*f))*f^(1/2 + b*(e/(2*f) + x)) + b*E^((4*e^2
+ b^2*Log[f]^2)/(4*f))*Sqrt[Pi]*Erf[(2*e + 4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*Log[f]*(Cosh[2*d] - Sinh[2*d
]) + b*Sqrt[Pi]*Erfi[(2*e + 4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*Log[f]*(Cosh[2*d] + Sinh[2*d])))/(8*Sqrt[2]
*b*E^((4*e^2 + b^2*Log[f]^2)/(8*f))*Log[f])

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Maple [A]  time = 0.142, size = 158, normalized size = 1. \begin{align*} -{\frac{\sqrt{\pi }{f}^{a}\sqrt{2}}{16}{{\rm e}^{{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-4\,\ln \left ( f \right ) be-16\,df+4\,{e}^{2}}{8\,f}}}}{\it Erf} \left ( -\sqrt{2}\sqrt{f}x+{\frac{ \left ( b\ln \left ( f \right ) -2\,e \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{f}}}} \right ){\frac{1}{\sqrt{f}}}}-{\frac{\sqrt{\pi }{f}^{a}}{8}{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+4\,\ln \left ( f \right ) be-16\,df+4\,{e}^{2}}{8\,f}}}}{\it Erf} \left ( -\sqrt{-2\,f}x+{\frac{2\,e+b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-2\,f}}}} \right ){\frac{1}{\sqrt{-2\,f}}}}-{\frac{{f}^{a}{f}^{bx}}{2\,b\ln \left ( f \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.53298, size = 193, normalized size = 1.2 \begin{align*} \frac{\sqrt{2} \sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{2} \sqrt{-f} x - \frac{\sqrt{2}{\left (b \log \left (f\right ) + 2 \, e\right )}}{4 \, \sqrt{-f}}\right ) e^{\left (2 \, d - \frac{{\left (b \log \left (f\right ) + 2 \, e\right )}^{2}}{8 \, f}\right )}}{16 \, \sqrt{-f}} + \frac{\sqrt{2} \sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{2} \sqrt{f} x - \frac{\sqrt{2}{\left (b \log \left (f\right ) - 2 \, e\right )}}{4 \, \sqrt{f}}\right ) e^{\left (-2 \, d + \frac{{\left (b \log \left (f\right ) - 2 \, e\right )}^{2}}{8 \, f}\right )}}{16 \, \sqrt{f}} - \frac{f^{b x + a}}{2 \, b \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.8799, size = 956, normalized size = 5.94 \begin{align*} -\frac{\sqrt{2} \sqrt{\pi } b \sqrt{-f} \cosh \left (\frac{b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f + 4 \,{\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) \operatorname{erf}\left (\frac{\sqrt{2}{\left (4 \, f x + b \log \left (f\right ) + 2 \, e\right )} \sqrt{-f}}{4 \, f}\right ) \log \left (f\right ) + \sqrt{2} \sqrt{\pi } b \sqrt{f} \cosh \left (\frac{b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f - 4 \,{\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) \operatorname{erf}\left (-\frac{\sqrt{2}{\left (4 \, f x - b \log \left (f\right ) + 2 \, e\right )}}{4 \, \sqrt{f}}\right ) \log \left (f\right ) - \sqrt{2} \sqrt{\pi } b \sqrt{-f} \operatorname{erf}\left (\frac{\sqrt{2}{\left (4 \, f x + b \log \left (f\right ) + 2 \, e\right )} \sqrt{-f}}{4 \, f}\right ) \log \left (f\right ) \sinh \left (\frac{b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f + 4 \,{\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) + \sqrt{2} \sqrt{\pi } b \sqrt{f} \operatorname{erf}\left (-\frac{\sqrt{2}{\left (4 \, f x - b \log \left (f\right ) + 2 \, e\right )}}{4 \, \sqrt{f}}\right ) \log \left (f\right ) \sinh \left (\frac{b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f - 4 \,{\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) + 8 \, f \cosh \left ({\left (b x + a\right )} \log \left (f\right )\right ) + 8 \, f \sinh \left ({\left (b x + a\right )} \log \left (f\right )\right )}{16 \, b f \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.35316, size = 527, normalized size = 3.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*sqrt(f)*(4*x - (b*log(f) - 2*e)/f))*e^(1/8*(b^2*log(f)^2 + 8*a*f*log(f
) - 4*b*e*log(f) - 16*d*f + 4*e^2)/f)/sqrt(f) - 1/16*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*sqrt(-f)*(4*x + (b*log(
f) + 2*e)/f))*e^(-1/8*(b^2*log(f)^2 - 8*a*f*log(f) + 4*b*e*log(f) - 16*d*f + 4*e^2)/f)/sqrt(-f) - (2*b*cos(-1/
2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)*log(abs(f))/(4*b^2*log(abs(f))^2 + (pi*b*sgn(f) - p
i*b)^2) - (pi*b*sgn(f) - pi*b)*sin(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)/(4*b^2*log(ab
s(f))^2 + (pi*b*sgn(f) - pi*b)^2))*e^(b*x*log(abs(f)) + a*log(abs(f))) - 1/2*I*(2*I*e^(1/2*I*pi*b*x*sgn(f) - 1
/2*I*pi*b*x + 1/2*I*pi*a*sgn(f) - 1/2*I*pi*a)/(2*I*pi*b*sgn(f) - 2*I*pi*b + 4*b*log(abs(f))) - 2*I*e^(-1/2*I*p
i*b*x*sgn(f) + 1/2*I*pi*b*x - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a)/(-2*I*pi*b*sgn(f) + 2*I*pi*b + 4*b*log(abs(f))))
*e^(b*x*log(abs(f)) + a*log(abs(f)))