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

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

[Out]

hypergeom([1, 1/n],[1+1/n],-b*sinh(d*x+c)^n/a)*sinh(d*x+c)/a/d+1/3*hypergeom([1, 3/n],[(3+n)/n],-b*sinh(d*x+c)
^n/a)*sinh(d*x+c)^3/a/d

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Rubi [A]  time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, 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} \[ \frac {\sinh ^3(c+d x) \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{3 a d}+\frac {\sinh (c+d x) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.98 \[ \frac {\frac {\sinh (c+d x) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a}+\frac {\sinh ^3(c+d x) \, _2F_1\left (1,\frac {3}{n};1+\frac {3}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{3 a}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cosh \left (d x + c\right )^{3}}{b \sinh \left (d x + c\right )^{n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (d x + c\right )^{3}}{b \sinh \left (d x + c\right )^{n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{3}\left (d x +c \right )}{a +b \left (\sinh ^{n}\left (d x +c \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (d x + c\right )^{3}}{b \sinh \left (d x + c\right )^{n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^3}{a+b\,{\mathrm {sinh}\left (c+d\,x\right )}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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