3.1.11 \(\int \sinh ^2(a+b x^2) \, dx\) [11]

Optimal. Leaf size=78 \[ -\frac {x}{2}+\frac {e^{-2 a} \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {b} x\right )}{8 \sqrt {b}}+\frac {e^{2 a} \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {b} x\right )}{8 \sqrt {b}} \]

[Out]

-1/2*x+1/16*erf(x*2^(1/2)*b^(1/2))*2^(1/2)*Pi^(1/2)/exp(2*a)/b^(1/2)+1/16*exp(2*a)*erfi(x*2^(1/2)*b^(1/2))*2^(
1/2)*Pi^(1/2)/b^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5408, 5407, 2235, 2236} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} e^{-2 a} \text {Erf}\left (\sqrt {2} \sqrt {b} x\right )}{8 \sqrt {b}}+\frac {\sqrt {\frac {\pi }{2}} e^{2 a} \text {Erfi}\left (\sqrt {2} \sqrt {b} x\right )}{8 \sqrt {b}}-\frac {x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]^2,x]

[Out]

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

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 5407

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^n), x], x] + Dist[1/2, Int[E^(-c - d*
x^n), x], x] /; FreeQ[{c, d}, x] && IGtQ[n, 1]

Rule 5408

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a + b*Sinh[c + d*x^
n])^p, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 1] && IGtQ[p, 1]

Rubi steps

\begin {align*} \int \sinh ^2\left (a+b x^2\right ) \, dx &=\int \left (-\frac {1}{2}+\frac {1}{2} \cosh \left (2 a+2 b x^2\right )\right ) \, dx\\ &=-\frac {x}{2}+\frac {1}{2} \int \cosh \left (2 a+2 b x^2\right ) \, dx\\ &=-\frac {x}{2}+\frac {1}{4} \int e^{-2 a-2 b x^2} \, dx+\frac {1}{4} \int e^{2 a+2 b x^2} \, dx\\ &=-\frac {x}{2}+\frac {e^{-2 a} \sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {b} x\right )}{8 \sqrt {b}}+\frac {e^{2 a} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {b} x\right )}{8 \sqrt {b}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 86, normalized size = 1.10 \begin {gather*} \frac {-4 \sqrt {2} \sqrt {b} x+\sqrt {\pi } \text {Erf}\left (\sqrt {2} \sqrt {b} x\right ) (\cosh (2 a)-\sinh (2 a))+\sqrt {\pi } \text {Erfi}\left (\sqrt {2} \sqrt {b} x\right ) (\cosh (2 a)+\sinh (2 a))}{8 \sqrt {2} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]^2,x]

[Out]

(-4*Sqrt[2]*Sqrt[b]*x + Sqrt[Pi]*Erf[Sqrt[2]*Sqrt[b]*x]*(Cosh[2*a] - Sinh[2*a]) + Sqrt[Pi]*Erfi[Sqrt[2]*Sqrt[b
]*x]*(Cosh[2*a] + Sinh[2*a]))/(8*Sqrt[2]*Sqrt[b])

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Maple [A]
time = 0.81, size = 51, normalized size = 0.65

method	result	size
risch	\(-\frac {x}{2}+\frac {{\mathrm e}^{-2 a} \sqrt {\pi }\, \sqrt {2}\, \erf \left (x \sqrt {2}\, \sqrt {b}\right )}{16 \sqrt {b}}+\frac {{\mathrm e}^{2 a} \sqrt {\pi }\, \erf \left (\sqrt {-2 b}\, x \right )}{8 \sqrt {-2 b}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 56, normalized size = 0.72 \begin {gather*} \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} \sqrt {-b} x\right ) e^{\left (2 \, a\right )}}{16 \, \sqrt {-b}} + \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} \sqrt {b} x\right ) e^{\left (-2 \, a\right )}}{16 \, \sqrt {b}} - \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(2)*sqrt(pi)*erf(sqrt(2)*sqrt(-b)*x)*e^(2*a)/sqrt(-b) + 1/16*sqrt(2)*sqrt(pi)*erf(sqrt(2)*sqrt(b)*x)*
e^(-2*a)/sqrt(b) - 1/2*x

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Fricas [A]
time = 0.38, size = 73, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {2} \sqrt {\pi } \sqrt {-b} {\left (\cosh \left (2 \, a\right ) + \sinh \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {2} \sqrt {-b} x\right ) - \sqrt {2} \sqrt {\pi } \sqrt {b} {\left (\cosh \left (2 \, a\right ) - \sinh \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {2} \sqrt {b} x\right ) + 8 \, b x}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(2)*sqrt(pi)*sqrt(-b)*(cosh(2*a) + sinh(2*a))*erf(sqrt(2)*sqrt(-b)*x) - sqrt(2)*sqrt(pi)*sqrt(b)*(c
osh(2*a) - sinh(2*a))*erf(sqrt(2)*sqrt(b)*x) + 8*b*x)/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sinh ^{2}{\left (a + b x^{2} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x**2+a)**2,x)

[Out]

Integral(sinh(a + b*x**2)**2, x)

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Giac [A]
time = 0.41, size = 58, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\sqrt {2} \sqrt {-b} x\right ) e^{\left (2 \, a\right )}}{16 \, \sqrt {-b}} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\sqrt {2} \sqrt {b} x\right ) e^{\left (-2 \, a\right )}}{16 \, \sqrt {b}} - \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*sqrt(-b)*x)*e^(2*a)/sqrt(-b) - 1/16*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*sqrt(b)*
x)*e^(-2*a)/sqrt(b) - 1/2*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {sinh}\left (b\,x^2+a\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^2)^2,x)

[Out]

int(sinh(a + b*x^2)^2, x)

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