3.772 \(\int \frac{1}{(\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x))^{3/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac{b \sinh (x)+c \cosh (x)}{2 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt{2} \left (b^2-c^2\right )^{3/4}} \]

[Out]

ArcTan[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[Sqrt[b^2 - c^2] + Sqrt[b^2 - c^2]*Cosh[
x + I*ArcTan[b, (-I)*c]]])]/(2*Sqrt[2]*(b^2 - c^2)^(3/4)) + (c*Cosh[x] + b*Sinh[x])/(2*Sqrt[b^2 - c^2]*(Sqrt[b
^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(3/2))

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Rubi [A]  time = 0.132054, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3116, 3115, 2649, 206} \[ \frac{b \sinh (x)+c \cosh (x)}{2 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt{2} \left (b^2-c^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-3/2),x]

[Out]

ArcTan[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[Sqrt[b^2 - c^2] + Sqrt[b^2 - c^2]*Cosh[
x + I*ArcTan[b, (-I)*c]]])]/(2*Sqrt[2]*(b^2 - c^2)^(3/4)) + (c*Cosh[x] + b*Sinh[x])/(2*Sqrt[b^2 - c^2]*(Sqrt[b
^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(3/2))

Rule 3116

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((c*Cos[d +
 e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^n)/(a*e*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n +
1)), Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 -
 c^2, 0] && LtQ[n, -1]

Rule 3115

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Int[1/Sqrt[a +
Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx &=\frac{c \cosh (x)+b \sinh (x)}{2 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{\int \frac{1}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx}{4 \sqrt{b^2-c^2}}\\ &=\frac{c \cosh (x)+b \sinh (x)}{2 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{\int \frac{1}{\sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}} \, dx}{4 \sqrt{b^2-c^2}}\\ &=\frac{c \cosh (x)+b \sinh (x)}{2 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{i \operatorname{Subst}\left (\int \frac{1}{2 \sqrt{b^2-c^2}-x^2} \, dx,x,-\frac{i \sqrt{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt{b^2-c^2}}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt{2} \left (b^2-c^2\right )^{3/4}}+\frac{c \cosh (x)+b \sinh (x)}{2 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}\\ \end{align*}

Mathematica [F]  time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-3/2),x]

[Out]

$Aborted

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Maple [B]  time = 0.785, size = 417, normalized size = 2.7 \begin{align*}{\frac{\sqrt{2}}{2}{\it Artanh} \left ({\frac{\cosh \left ( x \right ) \sqrt{2}}{2}} \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}{\frac{1}{\sqrt{-{(\sinh \left ( x \right ){b}^{2}-\sinh \left ( x \right ){c}^{2}-{b}^{2}+{c}^{2}){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}}}}+{\frac{\sqrt{2}}{ \left ( 4\,b-4\,c \right ) \left ( b+c \right ) \sinh \left ( x \right ) }\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}\sqrt{{b}^{2}-{c}^{2}} \left ( \ln \left ( -2\,{\frac{\cosh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}\sqrt{2}\sinh \left ( x \right ) -\sinh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}-\cosh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}\sqrt{2}+\sqrt{{b}^{2}-{c}^{2}}-\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) }\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}}{\cosh \left ( x \right ) -\sqrt{2}}} \right ) -\ln \left ( 2\,{\frac{\cosh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}\sqrt{2}\sinh \left ( x \right ) +\sinh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}-\cosh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}\sqrt{2}-\sqrt{{b}^{2}-{c}^{2}}+\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) }\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}}{\cosh \left ( x \right ) +\sqrt{2}}} \right ) \right ){\frac{1}{\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) }}}{\frac{1}{\sqrt{-{(\sinh \left ( x \right ){b}^{2}-\sinh \left ( x \right ){c}^{2}-{b}^{2}+{c}^{2}){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*cosh(x)+c*sinh(x)+(b^2-c^2)^(1/2))^(3/2),x)

[Out]

1/2/(b^2-c^2)^(1/2)/(-(sinh(x)*b^2-sinh(x)*c^2-b^2+c^2)/(b^2-c^2)^(1/2))^(1/2)*2^(1/2)*arctanh(1/2*cosh(x)*2^(
1/2))+1/4*(-(b^2-c^2)^(1/2)*(sinh(x)-1)*sinh(x)^2)^(1/2)*2^(1/2)*(b^2-c^2)^(1/2)*(ln(-2*(cosh(x)*(b^2-c^2)^(1/
2)*2^(1/2)*sinh(x)-sinh(x)*(b^2-c^2)^(1/2)-cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)+(b^2-c^2)^(1/2)-(-(b^2-c^2)^(1/2)*(
sinh(x)-1))^(1/2)*(-(b^2-c^2)^(1/2)*(sinh(x)-1)*sinh(x)^2)^(1/2))/(cosh(x)-2^(1/2)))-ln(2*(cosh(x)*(b^2-c^2)^(
1/2)*2^(1/2)*sinh(x)+sinh(x)*(b^2-c^2)^(1/2)-cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)-(b^2-c^2)^(1/2)+(-(b^2-c^2)^(1/2)
*(sinh(x)-1))^(1/2)*(-(b^2-c^2)^(1/2)*(sinh(x)-1)*sinh(x)^2)^(1/2))/(cosh(x)+2^(1/2))))/(b-c)/(b+c)/(-(b^2-c^2
)^(1/2)*(sinh(x)-1))^(1/2)/sinh(x)/(-(sinh(x)*b^2-sinh(x)*c^2-b^2+c^2)/(b^2-c^2)^(1/2))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + \sqrt{b^{2} - c^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)+(b^2-c^2)^(1/2))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^(-3/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)+(b^2-c^2)^(1/2))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \cosh{\left (x \right )} + c \sinh{\left (x \right )} + \sqrt{b^{2} - c^{2}}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)+(b**2-c**2)**(1/2))**(3/2),x)

[Out]

Integral((b*cosh(x) + c*sinh(x) + sqrt(b**2 - c**2))**(-3/2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)+(b^2-c^2)^(1/2))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError