\(\int f^{a+c x^2} \sinh ^2(d+e x) \, dx\) [349]

Optimal result

Integrand size = 18, antiderivative size = 161 \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=-\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 d-\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e-c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{2 d-\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

1/8*exp(-2*d-e^2/c/ln(f))*f^a*erfi((-e+c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)+1/8*exp(2*
d-e^2/c/ln(f))*f^a*erfi((e+c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)-1/4*f^a*erfi(x*c^(1/2)
*ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5623, 2235, 2325, 2266} \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=-\frac {\sqrt {\pi } f^a e^{-\frac {e^2}{c \log (f)}-2 d} \text {erfi}\left (\frac {e-c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{2 d-\frac {e^2}{c \log (f)}} \text {erfi}\left (\frac {c x \log (f)+e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(a + c*x^2)*Sinh[d + e*x]^2,x]

[Out]

-1/4*(f^a*Sqrt[Pi]*Erfi[Sqrt[c]*x*Sqrt[Log[f]]])/(Sqrt[c]*Sqrt[Log[f]]) - (E^(-2*d - e^2/(c*Log[f]))*f^a*Sqrt[
Pi]*Erfi[(e - c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])])/(8*Sqrt[c]*Sqrt[Log[f]]) + (E^(2*d - e^2/(c*Log[f]))*f^a*Sq
rt[Pi]*Erfi[(e + c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])])/(8*Sqrt[c]*Sqrt[Log[f]])

Rule 2235

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 2266

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 2325

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 5623

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]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.81 \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=\frac {e^{-\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \left (-2 e^{\frac {e^2}{c \log (f)}} \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )+\text {erfi}\left (\frac {-e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (2 d)-\sinh (2 d))+\text {erfi}\left (\frac {e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (2 d)+\sinh (2 d))\right )}{8 \sqrt {c} \sqrt {\log (f)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {\operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x +\frac {e}{\sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {2 d \ln \left (f \right ) c +e^{2}}{\ln \left (f \right ) c}}}{8 \sqrt {-c \ln \left (f \right )}}-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {e}{\sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {2 d \ln \left (f \right ) c -e^{2}}{\ln \left (f \right ) c}}}{8 \sqrt {-c \ln \left (f \right )}}-\frac {f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x \right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(139\)

[In]

int(f^(c*x^2+a)*sinh(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.52 \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=\frac {2 \, \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (a \log \left (f\right )\right ) + \sqrt {\pi } \sinh \left (a \log \left (f\right )\right )\right )} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x\right ) - \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (\frac {a c \log \left (f\right )^{2} + 2 \, c d \log \left (f\right ) - e^{2}}{c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (\frac {a c \log \left (f\right )^{2} + 2 \, c d \log \left (f\right ) - e^{2}}{c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left (c x \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right )}}{c \log \left (f\right )}\right ) - \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (\frac {a c \log \left (f\right )^{2} - 2 \, c d \log \left (f\right ) - e^{2}}{c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (\frac {a c \log \left (f\right )^{2} - 2 \, c d \log \left (f\right ) - e^{2}}{c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left (c x \log \left (f\right ) - e\right )} \sqrt {-c \log \left (f\right )}}{c \log \left (f\right )}\right )}{8 \, c \log \left (f\right )} \]

[In]

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

[Out]

1/8*(2*sqrt(-c*log(f))*(sqrt(pi)*cosh(a*log(f)) + sqrt(pi)*sinh(a*log(f)))*erf(sqrt(-c*log(f))*x) - sqrt(-c*lo
g(f))*(sqrt(pi)*cosh((a*c*log(f)^2 + 2*c*d*log(f) - e^2)/(c*log(f))) + sqrt(pi)*sinh((a*c*log(f)^2 + 2*c*d*log
(f) - e^2)/(c*log(f))))*erf((c*x*log(f) + e)*sqrt(-c*log(f))/(c*log(f))) - sqrt(-c*log(f))*(sqrt(pi)*cosh((a*c
*log(f)^2 - 2*c*d*log(f) - e^2)/(c*log(f))) + sqrt(pi)*sinh((a*c*log(f)^2 - 2*c*d*log(f) - e^2)/(c*log(f))))*e
rf((c*x*log(f) - e)*sqrt(-c*log(f))/(c*log(f))))/(c*log(f))

Sympy [F]

\[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=\int f^{a + c x^{2}} \sinh ^{2}{\left (d + e x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.81 \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {e}{\sqrt {-c \log \left (f\right )}}\right ) e^{\left (2 \, d - \frac {e^{2}}{c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x + \frac {e}{\sqrt {-c \log \left (f\right )}}\right ) e^{\left (-2 \, d - \frac {e^{2}}{c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x\right )}{4 \, \sqrt {-c \log \left (f\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93 \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-c \log \left (f\right )} x\right )}{4 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {-c \log \left (f\right )} {\left (x + \frac {e}{c \log \left (f\right )}\right )}\right ) e^{\left (\frac {a c \log \left (f\right )^{2} + 2 \, c d \log \left (f\right ) - e^{2}}{c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {-c \log \left (f\right )} {\left (x - \frac {e}{c \log \left (f\right )}\right )}\right ) e^{\left (\frac {a c \log \left (f\right )^{2} - 2 \, c d \log \left (f\right ) - e^{2}}{c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int f^{a+c x^2} \sinh ^2(d+e x) \, dx=\int f^{c\,x^2+a}\,{\mathrm {sinh}\left (d+e\,x\right )}^2 \,d x \]

[In]

int(f^(a + c*x^2)*sinh(d + e*x)^2,x)

[Out]

int(f^(a + c*x^2)*sinh(d + e*x)^2, x)