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

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]

1/2*f^(b*x+a)/b/ln(f)+1/16*exp(-2*d+1/8*b^2*ln(f)^2/f)*f^(-1/2+a)*erf(1/4*(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*b^2*ln(f)^2/f)*f^(-1/2+a)*erfi(1/4*(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 = 0.19, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 = {5513, 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)*Cosh[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 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+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*}

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Mathematica [A]  time = 0.70, 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)*Cosh[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] + Sin
h[2*d]))/(E^((b^2*Log[f]^2)/(8*f))*Sqrt[f])))/16

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fricas [B]  time = 0.62, size = 278, normalized size = 1.88 \[ -\frac {\sqrt {2} \sqrt {\pi } b \sqrt {-f} \cosh \left (\frac {b^{2} \log \relax (f)^{2} - 8 \, a f \log \relax (f) - 16 \, d f}{8 \, f}\right ) \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \relax (f)\right )} \sqrt {-f}}{4 \, f}\right ) \log \relax (f) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \cosh \left (\frac {b^{2} \log \relax (f)^{2} + 8 \, a f \log \relax (f) - 16 \, d f}{8 \, f}\right ) \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \relax (f)\right )}}{4 \, \sqrt {f}}\right ) \log \relax (f) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \relax (f)\right )}}{4 \, \sqrt {f}}\right ) \log \relax (f) \sinh \left (\frac {b^{2} \log \relax (f)^{2} + 8 \, a f \log \relax (f) - 16 \, d f}{8 \, f}\right ) - \sqrt {2} \sqrt {\pi } b \sqrt {-f} \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \relax (f)\right )} \sqrt {-f}}{4 \, f}\right ) \log \relax (f) \sinh \left (\frac {b^{2} \log \relax (f)^{2} - 8 \, a f \log \relax (f) - 16 \, d f}{8 \, f}\right ) - 8 \, f \cosh \left ({\left (b x + a\right )} \log \relax (f)\right ) - 8 \, f \sinh \left ({\left (b x + a\right )} \log \relax (f)\right )}{16 \, b f \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cosh(f*x^2+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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giac [C]  time = 0.18, size = 355, normalized size = 2.40 \[ -\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} \sqrt {f} {\left (4 \, x - \frac {b \log \relax (f)}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \relax (f)^{2} + 8 \, a f \log \relax (f) - 16 \, d f}{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)}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \relax (f)^{2} - 8 \, a f \log \relax (f) - 16 \, d f}{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+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)))

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maple [A]  time = 0.29, size = 126, normalized size = 0.85 \[ -\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {\ln \relax (f )^{2} b^{2}-16 d f}{8 f}} \sqrt {2}\, \erf \left (-\sqrt {2}\, \sqrt {f}\, x +\frac {\ln \relax (f ) b \sqrt {2}}{4 \sqrt {f}}\right )}{16 \sqrt {f}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}-16 d f}{8 f}} \erf \left (-\sqrt {-2 f}\, x +\frac {\ln \relax (f ) b}{2 \sqrt {-2 f}}\right )}{8 \sqrt {-2 f}}+\frac {f^{a} f^{b x}}{2 \ln \relax (f ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x+a)*cosh(f*x^2+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 = 0.42, size = 127, normalized size = 0.86 \[ \frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {f} x - \frac {\sqrt {2} b \log \relax (f)}{4 \, \sqrt {f}}\right ) e^{\left (\frac {b^{2} \log \relax (f)^{2}}{8 \, f} - 2 \, d\right )}}{16 \, \sqrt {f}} + \frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {-f} x - \frac {\sqrt {2} b \log \relax (f)}{4 \, \sqrt {-f}}\right ) e^{\left (-\frac {b^{2} \log \relax (f)^{2}}{8 \, f} + 2 \, d\right )}}{16 \, \sqrt {-f}} + \frac {f^{b x + a}}{2 \, b \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x)*cosh(d + f*x^2)^2,x)

[Out]

int(f^(a + b*x)*cosh(d + f*x^2)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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