3.423 \(\int \frac{\cosh ^3(c+d x)}{(a+b \sinh ^n(c+d x))^2} \, dx\)

Optimal. Leaf size=84 \[ \frac{\sinh ^3(c+d x) \, _2F_1\left (2,\frac{3}{n};\frac{n+3}{n};-\frac{b \sinh ^n(c+d x)}{a}\right )}{3 a^2 d}+\frac{\sinh (c+d x) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{b \sinh ^n(c+d x)}{a}\right )}{a^2 d} \]

[Out]

(Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x])/(a^2*d) + (Hypergeometric2F
1[2, 3/n, (3 + n)/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^3)/(3*a^2*d)

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Rubi [A]  time = 0.102242, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3223, 1893, 245, 364} \[ \frac{\sinh ^3(c+d x) \, _2F_1\left (2,\frac{3}{n};\frac{n+3}{n};-\frac{b \sinh ^n(c+d x)}{a}\right )}{3 a^2 d}+\frac{\sinh (c+d x) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{b \sinh ^n(c+d x)}{a}\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^3/(a + b*Sinh[c + d*x]^n)^2,x]

[Out]

(Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x])/(a^2*d) + (Hypergeometric2F
1[2, 3/n, (3 + n)/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^3)/(3*a^2*d)

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rule 1893

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\cosh ^3(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{\left (a+b x^n\right )^2}+\frac{x^2}{\left (a+b x^n\right )^2}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}+\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{b \sinh ^n(c+d x)}{a}\right ) \sinh (c+d x)}{a^2 d}+\frac{\, _2F_1\left (2,\frac{3}{n};\frac{3+n}{n};-\frac{b \sinh ^n(c+d x)}{a}\right ) \sinh ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0516983, size = 82, normalized size = 0.98 \[ \frac{\frac{\sinh ^3(c+d x) \, _2F_1\left (2,\frac{3}{n};1+\frac{3}{n};-\frac{b \sinh ^n(c+d x)}{a}\right )}{3 a^2}+\frac{\sinh (c+d x) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{b \sinh ^n(c+d x)}{a}\right )}{a^2}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^3/(a + b*Sinh[c + d*x]^n)^2,x]

[Out]

((Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x])/a^2 + (Hypergeometric2F1[2
, 3/n, 1 + 3/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^3)/(3*a^2))/d

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Maple [F]  time = 5.335, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{ \left ( a+b \left ( \sinh \left ( dx+c \right ) \right ) ^{n} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^n)^2,x)

[Out]

int(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^n)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2^{n} e^{\left (c n + 6 \, d x + 6 \, c\right )} + 2^{n} e^{\left (c n + 4 \, d x + 4 \, c\right )} - 2^{n} e^{\left (c n + 2 \, d x + 2 \, c\right )} - 2^{n} e^{\left (c n\right )}\right )} e^{\left (d n x\right )}}{8 \,{\left (2^{n} a^{2} d n e^{\left (d n x + c n + 3 \, d x + 3 \, c\right )} + a b d n e^{\left (3 \, d x + n \log \left (e^{\left (d x + c\right )} + 1\right ) + n \log \left (e^{\left (d x + c\right )} - 1\right ) + 3 \, c\right )}\right )}} + \frac{1}{8} \, \int \frac{{\left (2^{n} n e^{\left (c n\right )} - 3 \cdot 2^{n} e^{\left (c n\right )} +{\left (2^{n} n e^{\left (c n\right )} - 3 \cdot 2^{n} e^{\left (c n\right )}\right )} e^{\left (6 \, d x + 6 \, c\right )} +{\left (3 \cdot 2^{n} n e^{\left (c n\right )} - 2^{n} e^{\left (c n\right )}\right )} e^{\left (4 \, d x + 4 \, c\right )} +{\left (3 \cdot 2^{n} n e^{\left (c n\right )} - 2^{n} e^{\left (c n\right )}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} e^{\left (d n x\right )}}{2^{n} a^{2} n e^{\left (d n x + c n + 3 \, d x + 3 \, c\right )} + a b n e^{\left (3 \, d x + n \log \left (e^{\left (d x + c\right )} + 1\right ) + n \log \left (e^{\left (d x + c\right )} - 1\right ) + 3 \, c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^n)^2,x, algorithm="maxima")

[Out]

1/8*(2^n*e^(c*n + 6*d*x + 6*c) + 2^n*e^(c*n + 4*d*x + 4*c) - 2^n*e^(c*n + 2*d*x + 2*c) - 2^n*e^(c*n))*e^(d*n*x
)/(2^n*a^2*d*n*e^(d*n*x + c*n + 3*d*x + 3*c) + a*b*d*n*e^(3*d*x + n*log(e^(d*x + c) + 1) + n*log(e^(d*x + c) -
1) + 3*c)) + 1/8*integrate((2^n*n*e^(c*n) - 3*2^n*e^(c*n) + (2^n*n*e^(c*n) - 3*2^n*e^(c*n))*e^(6*d*x + 6*c) +
(3*2^n*n*e^(c*n) - 2^n*e^(c*n))*e^(4*d*x + 4*c) + (3*2^n*n*e^(c*n) - 2^n*e^(c*n))*e^(2*d*x + 2*c))*e^(d*n*x)/
(2^n*a^2*n*e^(d*n*x + c*n + 3*d*x + 3*c) + a*b*n*e^(3*d*x + n*log(e^(d*x + c) + 1) + n*log(e^(d*x + c) - 1) +
3*c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cosh \left (d x + c\right )^{3}}{b^{2} \sinh \left (d x + c\right )^{2 \, n} + 2 \, a b \sinh \left (d x + c\right )^{n} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^n)^2,x, algorithm="fricas")

[Out]

integral(cosh(d*x + c)^3/(b^2*sinh(d*x + c)^(2*n) + 2*a*b*sinh(d*x + c)^n + a^2), x)

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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(cosh(d*x+c)**3/(a+b*sinh(d*x+c)**n)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^n)^2,x, algorithm="giac")

[Out]

integrate(cosh(d*x + c)^3/(b*sinh(d*x + c)^n + a)^2, x)