\(\int f^{a+b x} \sinh (d+e x+f x^2) \, dx\) [345]

Optimal result

Integrand size = 19, antiderivative size = 115 \[ \int f^{a+b x} \sinh \left (d+e x+f x^2\right ) \, dx=-\frac {1}{4} e^{-d+\frac {(e-b \log (f))^2}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {e+2 f x-b \log (f)}{2 \sqrt {f}}\right )+\frac {1}{4} e^{d-\frac {(e+b \log (f))^2}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {e+2 f x+b \log (f)}{2 \sqrt {f}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5623, 2325, 2266, 2236, 2235} \[ \int f^{a+b x} \sinh \left (d+e x+f x^2\right ) \, dx=\frac {1}{4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{d-\frac {(b \log (f)+e)^2}{4 f}} \text {erfi}\left (\frac {b \log (f)+e+2 f x}{2 \sqrt {f}}\right )-\frac {1}{4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {(e-b \log (f))^2}{4 f}-d} \text {erf}\left (\frac {-b \log (f)+e+2 f x}{2 \sqrt {f}}\right ) \]

[In]

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

[Out]

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

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 2236

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.08 \[ \int f^{a+b x} \sinh \left (d+e x+f x^2\right ) \, dx=\frac {1}{4} e^{-\frac {e^2+b^2 \log ^2(f)}{4 f}} f^{a-\frac {b e+f}{2 f}} \sqrt {\pi } \left (-e^{\frac {e^2+b^2 \log ^2(f)}{2 f}} \text {erf}\left (\frac {e+2 f x-b \log (f)}{2 \sqrt {f}}\right ) (\cosh (d)-\sinh (d))+\text {erfi}\left (\frac {e+2 f x+b \log (f)}{2 \sqrt {f}}\right ) (\cosh (d)+\sinh (d))\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.10

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-\sqrt {-f}\, x +\frac {e +b \ln \left (f \right )}{2 \sqrt {-f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+2 \ln \left (f \right ) b e -4 d f +e^{2}}{4 f}}}{4 \sqrt {-f}}+\frac {\operatorname {erf}\left (-\sqrt {f}\, x +\frac {b \ln \left (f \right )-e}{2 \sqrt {f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {b^{2} \ln \left (f \right )^{2}-2 \ln \left (f \right ) b e -4 d f +e^{2}}{4 f}}}{4 \sqrt {f}}\)	\(126\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (90) = 180\).

Time = 0.29 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.20 \[ \int f^{a+b x} \sinh \left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \sqrt {-f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + e^{2} - 4 \, d f + 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right ) \operatorname {erf}\left (\frac {{\left (2 \, f x + b \log \left (f\right ) + e\right )} \sqrt {-f}}{2 \, f}\right ) - \sqrt {\pi } \sqrt {f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + e^{2} - 4 \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right ) \operatorname {erf}\left (-\frac {2 \, f x - b \log \left (f\right ) + e}{2 \, \sqrt {f}}\right ) - \sqrt {\pi } \sqrt {-f} \operatorname {erf}\left (\frac {{\left (2 \, f x + b \log \left (f\right ) + e\right )} \sqrt {-f}}{2 \, f}\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + e^{2} - 4 \, d f + 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right ) - \sqrt {\pi } \sqrt {f} \operatorname {erf}\left (-\frac {2 \, f x - b \log \left (f\right ) + e}{2 \, \sqrt {f}}\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + e^{2} - 4 \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )}{4 \, f} \]

[In]

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

[Out]

-1/4*(sqrt(pi)*sqrt(-f)*cosh(1/4*(b^2*log(f)^2 + e^2 - 4*d*f + 2*(b*e - 2*a*f)*log(f))/f)*erf(1/2*(2*f*x + b*l
og(f) + e)*sqrt(-f)/f) - sqrt(pi)*sqrt(f)*cosh(1/4*(b^2*log(f)^2 + e^2 - 4*d*f - 2*(b*e - 2*a*f)*log(f))/f)*er
f(-1/2*(2*f*x - b*log(f) + e)/sqrt(f)) - sqrt(pi)*sqrt(-f)*erf(1/2*(2*f*x + b*log(f) + e)*sqrt(-f)/f)*sinh(1/4
*(b^2*log(f)^2 + e^2 - 4*d*f + 2*(b*e - 2*a*f)*log(f))/f) - sqrt(pi)*sqrt(f)*erf(-1/2*(2*f*x - b*log(f) + e)/s
qrt(f))*sinh(1/4*(b^2*log(f)^2 + e^2 - 4*d*f - 2*(b*e - 2*a*f)*log(f))/f))/f

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int f^{a+b x} \sinh \left (d+e x+f x^2\right ) \, dx=-\frac {1}{4} \, \sqrt {\pi } f^{a - \frac {1}{2}} \operatorname {erf}\left (\sqrt {f} x - \frac {b \log \left (f\right ) - e}{2 \, \sqrt {f}}\right ) e^{\left (-d + \frac {{\left (b \log \left (f\right ) - e\right )}^{2}}{4 \, f}\right )} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-f} x - \frac {b \log \left (f\right ) + e}{2 \, \sqrt {-f}}\right ) e^{\left (d - \frac {{\left (b \log \left (f\right ) + e\right )}^{2}}{4 \, f}\right )}}{4 \, \sqrt {-f}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.15 \[ \int f^{a+b x} \sinh \left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-f} {\left (2 \, x + \frac {b \log \left (f\right ) + e}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} + 2 \, b e \log \left (f\right ) - 4 \, a f \log \left (f\right ) + e^{2} - 4 \, d f}{4 \, f}\right )}}{4 \, \sqrt {-f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {f} {\left (2 \, x - \frac {b \log \left (f\right ) - e}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2} - 2 \, b e \log \left (f\right ) + 4 \, a f \log \left (f\right ) + e^{2} - 4 \, d f}{4 \, f}\right )}}{4 \, \sqrt {f}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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