3.1.93 \(\int (1+\sinh ^2(x))^{3/2} \, dx\) [93]

Optimal. Leaf size=29 \[ \frac {2}{3} \sqrt {\cosh ^2(x)} \tanh (x)+\frac {1}{3} \cosh ^2(x)^{3/2} \tanh (x) \]

[Out]

1/3*(cosh(x)^2)^(3/2)*tanh(x)+2/3*(cosh(x)^2)^(1/2)*tanh(x)

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Rubi [A]
time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3255, 3282, 3286, 2717} \begin {gather*} \frac {1}{3} \cosh ^2(x)^{3/2} \tanh (x)+\frac {2}{3} \sqrt {\cosh ^2(x)} \tanh (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Sinh[x]^2)^(3/2),x]

[Out]

(2*Sqrt[Cosh[x]^2]*Tanh[x])/3 + ((Cosh[x]^2)^(3/2)*Tanh[x])/3

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3255

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3282

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-Cot[e + f*x])*((b*Sin[e + f*x]^2)^p/(2*f*p)),
x] + Dist[b*((2*p - 1)/(2*p)), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p]
&& GtQ[p, 1]

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \left (1+\sinh ^2(x)\right )^{3/2} \, dx &=\int \cosh ^2(x)^{3/2} \, dx\\ &=\frac {1}{3} \cosh ^2(x)^{3/2} \tanh (x)+\frac {2}{3} \int \sqrt {\cosh ^2(x)} \, dx\\ &=\frac {1}{3} \cosh ^2(x)^{3/2} \tanh (x)+\frac {1}{3} \left (2 \sqrt {\cosh ^2(x)} \text {sech}(x)\right ) \int \cosh (x) \, dx\\ &=\frac {2}{3} \sqrt {\cosh ^2(x)} \tanh (x)+\frac {1}{3} \cosh ^2(x)^{3/2} \tanh (x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 23, normalized size = 0.79 \begin {gather*} \frac {1}{12} \sqrt {\cosh ^2(x)} \text {sech}(x) (9 \sinh (x)+\sinh (3 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sinh[x]^2)^(3/2),x]

[Out]

(Sqrt[Cosh[x]^2]*Sech[x]*(9*Sinh[x] + Sinh[3*x]))/12

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Maple [A]
time = 0.76, size = 21, normalized size = 0.72

method	result	size
default	\(\frac {\sqrt {\frac {1}{2}+\frac {\cosh \left (2 x \right )}{2}}\, \sinh \left (x \right ) \left (\sinh ^{2}\left (x \right )+3\right )}{3 \cosh \left (x \right )}\)	\(21\)
risch	\(\frac {{\mathrm e}^{4 x} \sqrt {\left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}{24+24 \,{\mathrm e}^{2 x}}+\frac {3 \sqrt {\left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}\, {\mathrm e}^{2 x}}{8 \left (1+{\mathrm e}^{2 x}\right )}-\frac {3 \sqrt {\left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}{8 \left (1+{\mathrm e}^{2 x}\right )}-\frac {{\mathrm e}^{-2 x} \sqrt {\left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}{24 \left (1+{\mathrm e}^{2 x}\right )}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+sinh(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(cosh(x)^2)^(1/2)*sinh(x)*(sinh(x)^2+3)/cosh(x)

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Maxima [A]
time = 0.48, size = 23, normalized size = 0.79 \begin {gather*} \frac {1}{24} \, e^{\left (3 \, x\right )} - \frac {3}{8} \, e^{\left (-x\right )} - \frac {1}{24} \, e^{\left (-3 \, x\right )} + \frac {3}{8} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sinh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

1/24*e^(3*x) - 3/8*e^(-x) - 1/24*e^(-3*x) + 3/8*e^x

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Fricas [A]
time = 0.43, size = 17, normalized size = 0.59 \begin {gather*} \frac {1}{12} \, \sinh \left (x\right )^{3} + \frac {1}{4} \, {\left (\cosh \left (x\right )^{2} + 3\right )} \sinh \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sinh(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/12*sinh(x)^3 + 1/4*(cosh(x)^2 + 3)*sinh(x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (\sinh ^{2}{\left (x \right )} + 1\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sinh(x)**2)**(3/2),x)

[Out]

Integral((sinh(x)**2 + 1)**(3/2), x)

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Giac [A]
time = 0.42, size = 25, normalized size = 0.86 \begin {gather*} -\frac {1}{24} \, {\left (9 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-3 \, x\right )} + \frac {1}{24} \, e^{\left (3 \, x\right )} + \frac {3}{8} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sinh(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/24*(9*e^(2*x) + 1)*e^(-3*x) + 1/24*e^(3*x) + 3/8*e^x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int {\left ({\mathrm {sinh}\left (x\right )}^2+1\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(x)^2 + 1)^(3/2),x)

[Out]

int((sinh(x)^2 + 1)^(3/2), x)

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