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

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]

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

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Rubi [A]  time = 0.27, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 = {5513, 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)*Cosh[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 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 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]

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 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 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 5513

Int[Cosh[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cosh[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*} \int f^{a+b x} \cosh ^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*}

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Mathematica [A]  time = 0.65, 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)*Cosh[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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fricas [B]  time = 0.64, size = 334, normalized size = 2.07 \[ -\frac {\sqrt {2} \sqrt {\pi } b \sqrt {-f} \cosh \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, e^{2} - 16 \, d f + 4 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{8 \, f}\right ) \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \relax (f) + 2 \, e\right )} \sqrt {-f}}{4 \, f}\right ) \log \relax (f) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \cosh \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, e^{2} - 16 \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{8 \, f}\right ) \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \relax (f) + 2 \, e\right )}}{4 \, \sqrt {f}}\right ) \log \relax (f) - \sqrt {2} \sqrt {\pi } b \sqrt {-f} \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \relax (f) + 2 \, e\right )} \sqrt {-f}}{4 \, f}\right ) \log \relax (f) \sinh \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, e^{2} - 16 \, d f + 4 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{8 \, f}\right ) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \relax (f) + 2 \, e\right )}}{4 \, \sqrt {f}}\right ) \log \relax (f) \sinh \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, e^{2} - 16 \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{8 \, f}\right ) - 8 \, f \cosh \left ({\left (b x + a\right )} \log \relax (f)\right ) - 8 \, f \sinh \left ({\left (b x + a\right )} \log \relax (f)\right )}{16 \, b f \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cosh(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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giac [C]  time = 0.20, size = 389, normalized size = 2.42 \[ -\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} \sqrt {f} {\left (4 \, x - \frac {b \log \relax (f) - 2 \, e}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \relax (f)^{2} + 8 \, a f \log \relax (f) - 4 \, b e \log \relax (f) - 16 \, d f + 4 \, e^{2}}{8 \, f}\right )}}{16 \, \sqrt {f}} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} \sqrt {-f} {\left (4 \, x + \frac {b \log \relax (f) + 2 \, e}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \relax (f)^{2} - 8 \, a f \log \relax (f) + 4 \, b e \log \relax (f) - 16 \, d f + 4 \, e^{2}}{8 \, f}\right )}}{16 \, \sqrt {-f}} + {\left (\frac {2 \, b \cos \left (-\frac {1}{2} \, \pi b x \mathrm {sgn}\relax (f) + \frac {1}{2} \, \pi b x - \frac {1}{2} \, \pi a \mathrm {sgn}\relax (f) + \frac {1}{2} \, \pi a\right ) \log \left ({\left | f \right |}\right )}{4 \, b^{2} \log \left ({\left | f \right |}\right )^{2} + {\left (\pi b \mathrm {sgn}\relax (f) - \pi b\right )}^{2}} - \frac {{\left (\pi b \mathrm {sgn}\relax (f) - \pi b\right )} \sin \left (-\frac {1}{2} \, \pi b x \mathrm {sgn}\relax (f) + \frac {1}{2} \, \pi b x - \frac {1}{2} \, \pi a \mathrm {sgn}\relax (f) + \frac {1}{2} \, \pi a\right )}{4 \, b^{2} \log \left ({\left | f \right |}\right )^{2} + {\left (\pi b \mathrm {sgn}\relax (f) - \pi b\right )}^{2}}\right )} e^{\left (b x \log \left ({\left | f \right |}\right ) + a \log \left ({\left | f \right |}\right )\right )} - \frac {1}{2} i \, {\left (-\frac {2 i \, e^{\left (\frac {1}{2} i \, \pi b x \mathrm {sgn}\relax (f) - \frac {1}{2} i \, \pi b x + \frac {1}{2} i \, \pi a \mathrm {sgn}\relax (f) - \frac {1}{2} i \, \pi a\right )}}{2 i \, \pi b \mathrm {sgn}\relax (f) - 2 i \, \pi b + 4 \, b \log \left ({\left | f \right |}\right )} + \frac {2 i \, e^{\left (-\frac {1}{2} i \, \pi b x \mathrm {sgn}\relax (f) + \frac {1}{2} i \, \pi b x - \frac {1}{2} i \, \pi a \mathrm {sgn}\relax (f) + \frac {1}{2} i \, \pi a\right )}}{-2 i \, \pi b \mathrm {sgn}\relax (f) + 2 i \, \pi b + 4 \, b \log \left ({\left | f \right |}\right )}\right )} e^{\left (b x \log \left ({\left | f \right |}\right ) + a \log \left ({\left | f \right |}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cosh(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*
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)))

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maple [A]  time = 0.28, size = 158, normalized size = 0.98 \[ -\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {\ln \relax (f )^{2} b^{2}-4 \ln \relax (f ) b e -16 d f +4 e^{2}}{8 f}} \sqrt {2}\, \erf \left (-\sqrt {2}\, \sqrt {f}\, x +\frac {\left (b \ln \relax (f )-2 e \right ) \sqrt {2}}{4 \sqrt {f}}\right )}{16 \sqrt {f}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}+4 \ln \relax (f ) b e -16 d f +4 e^{2}}{8 f}} \erf \left (-\sqrt {-2 f}\, x +\frac {2 e +b \ln \relax (f )}{2 \sqrt {-2 f}}\right )}{8 \sqrt {-2 f}}+\frac {f^{a} f^{b x}}{2 \ln \relax (f ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x+a)*cosh(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 = 0.43, size = 143, normalized size = 0.89 \[ \frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {-f} x - \frac {\sqrt {2} {\left (b \log \relax (f) + 2 \, e\right )}}{4 \, \sqrt {-f}}\right ) e^{\left (2 \, d - \frac {{\left (b \log \relax (f) + 2 \, e\right )}^{2}}{8 \, f}\right )}}{16 \, \sqrt {-f}} + \frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {f} x - \frac {\sqrt {2} {\left (b \log \relax (f) - 2 \, e\right )}}{4 \, \sqrt {f}}\right ) e^{\left (-2 \, d + \frac {{\left (b \log \relax (f) - 2 \, e\right )}^{2}}{8 \, f}\right )}}{16 \, \sqrt {f}} + \frac {f^{b x + a}}{2 \, b \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cosh(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int f^{a+b\,x}\,{\mathrm {cosh}\left (f\,x^2+e\,x+d\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x} \cosh ^{2}{\left (d + e x + f x^{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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