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

Optimal. Leaf size=148 \[ \frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{\frac{b^2 \log ^2(f)}{8 f}-2 d} \text{Erf}\left (\frac{4 f x-b \log (f)}{2 \sqrt{2} \sqrt{f}}\right )+\frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{2 d-\frac{b^2 \log ^2(f)}{8 f}} \text{Erfi}\left (\frac{b \log (f)+4 f x}{2 \sqrt{2} \sqrt{f}}\right )-\frac{f^{a+b x}}{2 b \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.193613, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5512, 2194, 2287, 2234, 2205, 2204} \[ \frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{\frac{b^2 \log ^2(f)}{8 f}-2 d} \text{Erf}\left (\frac{4 f x-b \log (f)}{2 \sqrt{2} \sqrt{f}}\right )+\frac{1}{8} \sqrt{\frac{\pi }{2}} f^{a-\frac{1}{2}} e^{2 d-\frac{b^2 \log ^2(f)}{8 f}} \text{Erfi}\left (\frac{b \log (f)+4 f x}{2 \sqrt{2} \sqrt{f}}\right )-\frac{f^{a+b x}}{2 b \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x)*Sinh[d + f*x^2]^2,x]

[Out]

(E^(-2*d + (b^2*Log[f]^2)/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erf[(4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])])/8 + (E^(
2*d - (b^2*Log[f]^2)/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erfi[(4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[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+f x^2\right ) \, dx &=\int \left (-\frac{1}{2} f^{a+b x}+\frac{1}{4} e^{-2 d-2 f x^2} f^{a+b x}+\frac{1}{4} e^{2 d+2 f x^2} f^{a+b x}\right ) \, dx\\ &=\frac{1}{4} \int e^{-2 d-2 f x^2} f^{a+b x} \, dx+\frac{1}{4} \int e^{2 d+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 e^{-2 d-2 f x^2+a \log (f)+b x \log (f)} \, dx+\frac{1}{4} \int e^{2 d+2 f x^2+a \log (f)+b x \log (f)} \, dx\\ &=-\frac{f^{a+b x}}{2 b \log (f)}+\frac{1}{4} \left (e^{2 d-\frac{b^2 \log ^2(f)}{8 f}} f^a\right ) \int e^{\frac{(4 f x+b \log (f))^2}{8 f}} \, dx+\frac{1}{4} \left (e^{-2 d+\frac{b^2 \log ^2(f)}{8 f}} f^a\right ) \int e^{-\frac{(-4 f x+b \log (f))^2}{8 f}} \, dx\\ &=\frac{1}{8} e^{-2 d+\frac{b^2 \log ^2(f)}{8 f}} f^{-\frac{1}{2}+a} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{4 f x-b \log (f)}{2 \sqrt{2} \sqrt{f}}\right )+\frac{1}{8} e^{2 d-\frac{b^2 \log ^2(f)}{8 f}} f^{-\frac{1}{2}+a} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{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 = 0.751649, size = 149, normalized size = 1.01 \[ \frac{1}{16} f^a \left (\frac{\sqrt{2 \pi } e^{\frac{b^2 \log ^2(f)}{8 f}} (\cosh (2 d)-\sinh (2 d)) \text{Erf}\left (\frac{4 f x-b \log (f)}{2 \sqrt{2} \sqrt{f}}\right )}{\sqrt{f}}+\frac{\sqrt{2 \pi } e^{-\frac{b^2 \log ^2(f)}{8 f}} (\sinh (2 d)+\cosh (2 d)) \text{Erfi}\left (\frac{b \log (f)+4 f x}{2 \sqrt{2} \sqrt{f}}\right )}{\sqrt{f}}-\frac{8 f^{b x}}{b \log (f)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^a*((-8*f^(b*x))/(b*Log[f]) + (E^((b^2*Log[f]^2)/(8*f))*Sqrt[2*Pi]*Erf[(4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f]
)]*(Cosh[2*d] - Sinh[2*d]))/Sqrt[f] + (Sqrt[2*Pi]*Erfi[(4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*(Cosh[2*d] + Si
nh[2*d]))/(E^((b^2*Log[f]^2)/(8*f))*Sqrt[f])))/16

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Maple [A]  time = 0.173, size = 126, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{\pi }{f}^{a}\sqrt{2}}{16}{{\rm e}^{{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-16\,df}{8\,f}}}}{\it Erf} \left ( -\sqrt{2}\sqrt{f}x+{\frac{b\ln \left ( f \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}-16\,df}{8\,f}}}}{\it Erf} \left ( -\sqrt{-2\,f}x+{\frac{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+d)^2,x)

[Out]

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

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Maxima [A]  time = 1.62639, size = 171, normalized size = 1.16 \begin{align*} \frac{\sqrt{2} \sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{2} \sqrt{f} x - \frac{\sqrt{2} b \log \left (f\right )}{4 \, \sqrt{f}}\right ) e^{\left (\frac{b^{2} \log \left (f\right )^{2}}{8 \, f} - 2 \, d\right )}}{16 \, \sqrt{f}} + \frac{\sqrt{2} \sqrt{\pi } f^{a} \operatorname{erf}\left (\sqrt{2} \sqrt{-f} x - \frac{\sqrt{2} b \log \left (f\right )}{4 \, \sqrt{-f}}\right ) e^{\left (-\frac{b^{2} \log \left (f\right )^{2}}{8 \, f} + 2 \, d\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+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)/sqrt(f))*e^(1/8*b^2*log(f)^2/f - 2*d)/s
qrt(f) + 1/16*sqrt(2)*sqrt(pi)*f^a*erf(sqrt(2)*sqrt(-f)*x - 1/4*sqrt(2)*b*log(f)/sqrt(-f))*e^(-1/8*b^2*log(f)^
2/f + 2*d)/sqrt(-f) - 1/2*f^(b*x + a)/(b*log(f))

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Fricas [B]  time = 1.83629, size = 826, normalized size = 5.58 \begin{align*} -\frac{\sqrt{2} \sqrt{\pi } b \sqrt{-f} \cosh \left (\frac{b^{2} \log \left (f\right )^{2} - 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right ) \operatorname{erf}\left (\frac{\sqrt{2}{\left (4 \, f x + b \log \left (f\right )\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} + 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right ) \operatorname{erf}\left (-\frac{\sqrt{2}{\left (4 \, f x - b \log \left (f\right )\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 )\right )}}{4 \, \sqrt{f}}\right ) \log \left (f\right ) \sinh \left (\frac{b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) - 16 \, d f}{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 )\right )} \sqrt{-f}}{4 \, f}\right ) \log \left (f\right ) \sinh \left (\frac{b^{2} \log \left (f\right )^{2} - 8 \, a f \log \left (f\right ) - 16 \, d f}{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+d)^2,x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [C]  time = 1.3181, size = 481, normalized size = 3.25 \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+d)^2,x, algorithm="giac")

[Out]

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