### 3.773 $$\int \frac{1}{(\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x))^{5/2}} \, dx$$

Optimal. Leaf size=205 $\frac{3 \tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{16 \sqrt{2} \left (b^2-c^2\right )^{5/4}}+\frac{3 (b \sinh (x)+c \cosh (x))}{16 \left (b^2-c^2\right ) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{b \sinh (x)+c \cosh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}$

[Out]

(3*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]*Co
sh[x + I*ArcTan[b, (-I)*c]]])])/(16*Sqrt[2]*(b^2 - c^2)^(5/4)) + (c*Cosh[x] + b*Sinh[x])/(4*Sqrt[b^2 - c^2]*(S
qrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(5/2)) + (3*(c*Cosh[x] + b*Sinh[x]))/(16*(b^2 - c^2)*(Sqrt[b^2 - c^2]
+ b*Cosh[x] + c*Sinh[x])^(3/2))

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Rubi [A]  time = 0.175753, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {3116, 3115, 2649, 206} $\frac{3 \tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{16 \sqrt{2} \left (b^2-c^2\right )^{5/4}}+\frac{3 (b \sinh (x)+c \cosh (x))}{16 \left (b^2-c^2\right ) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{b \sinh (x)+c \cosh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-5/2),x]

[Out]

(3*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]*Co
sh[x + I*ArcTan[b, (-I)*c]]])])/(16*Sqrt[2]*(b^2 - c^2)^(5/4)) + (c*Cosh[x] + b*Sinh[x])/(4*Sqrt[b^2 - c^2]*(S
qrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(5/2)) + (3*(c*Cosh[x] + b*Sinh[x]))/(16*(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 )^{5/2}} \, dx &=\frac{c \cosh (x)+b \sinh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}+\frac{3 \int \frac{1}{\left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx}{8 \sqrt{b^2-c^2}}\\ &=\frac{c \cosh (x)+b \sinh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}+\frac{3 (c \cosh (x)+b \sinh (x))}{16 \left (b^2-c^2\right ) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{3 \int \frac{1}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx}{32 \left (b^2-c^2\right )}\\ &=\frac{c \cosh (x)+b \sinh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}+\frac{3 (c \cosh (x)+b \sinh (x))}{16 \left (b^2-c^2\right ) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{3 \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}{32 \left (b^2-c^2\right )}\\ &=\frac{c \cosh (x)+b \sinh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}+\frac{3 (c \cosh (x)+b \sinh (x))}{16 \left (b^2-c^2\right ) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac{(3 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 )}{16 \left (b^2-c^2\right )}\\ &=\frac{3 \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 )}{16 \sqrt{2} \left (b^2-c^2\right )^{5/4}}+\frac{c \cosh (x)+b \sinh (x)}{4 \sqrt{b^2-c^2} \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}+\frac{3 (c \cosh (x)+b \sinh (x))}{16 \left (b^2-c^2\right ) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}\\ \end{align*}

Mathematica [F]  time = 180.009, 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])^(-5/2),x]

[Out]

\$Aborted

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Maple [B]  time = 1.046, size = 954, normalized size = 4.7 \begin{align*} \text{result too large to display} \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))^(5/2),x)

[Out]

1/8/(sinh(x)-1)/sinh(x)/((-b^2+c^2)/(b^2-c^2)^(1/2)*sinh(x)+(b^2-c^2)^(1/2))^(1/2)/(-sinh(x)*(b^2-c^2)^(1/2)+(
b^2-c^2)^(1/2))^(1/2)/(b^2-c^2)*(2*2^(1/2)*(-sinh(x)*(b^2-c^2)^(1/2)+(b^2-c^2)^(1/2))^(1/2)*arctanh(1/2*cosh(x
)*2^(1/2))*sinh(x)^2+(-(b^2-c^2)^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)*ln(-2/(-cosh(x)+2^(1/2))*(-c
osh(x)*(b^2-c^2)^(1/2)*2^(1/2)*sinh(x)+cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)+(-sinh(x)*(b^2-c^2)^(1/2)+(b^2-c^2)^(1/
2))^(1/2)*(-(b^2-c^2)^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)+sinh(x)*(b^2-c^2)^(1/2)-(b^2-c^2)^(1/2)
))*2^(1/2)*sinh(x)-(-(b^2-c^2)^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)*ln(2/(cosh(x)+2^(1/2))*(cosh(x
)*(b^2-c^2)^(1/2)*2^(1/2)*sinh(x)-cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)+(-sinh(x)*(b^2-c^2)^(1/2)+(b^2-c^2)^(1/2))^(
1/2)*(-(b^2-c^2)^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)+sinh(x)*(b^2-c^2)^(1/2)-(b^2-c^2)^(1/2)))*2^
(1/2)*sinh(x)-2*2^(1/2)*(-sinh(x)*(b^2-c^2)^(1/2)+(b^2-c^2)^(1/2))^(1/2)*arctanh(1/2*cosh(x)*2^(1/2))*sinh(x)-
(-(b^2-c^2)^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)*ln(-2/(-cosh(x)+2^(1/2))*(-cosh(x)*(b^2-c^2)^(1/2
)*2^(1/2)*sinh(x)+cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)+(-sinh(x)*(b^2-c^2)^(1/2)+(b^2-c^2)^(1/2))^(1/2)*(-(b^2-c^2)
^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)+sinh(x)*(b^2-c^2)^(1/2)-(b^2-c^2)^(1/2)))*2^(1/2)+(-(b^2-c^2
)^(1/2)*sinh(x)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)*ln(2/(cosh(x)+2^(1/2))*(cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)*sin
h(x)-cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)+(-sinh(x)*(b^2-c^2)^(1/2)+(b^2-c^2)^(1/2))^(1/2)*(-(b^2-c^2)^(1/2)*sinh(x
)^3+(b^2-c^2)^(1/2)*sinh(x)^2)^(1/2)+sinh(x)*(b^2-c^2)^(1/2)-(b^2-c^2)^(1/2)))*2^(1/2)-4*(-sinh(x)*(b^2-c^2)^(
1/2)+(b^2-c^2)^(1/2))^(1/2)*cosh(x)*sinh(x))

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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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^(-5/2), 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))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(5/2),x)

[Out]

Timed out

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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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError