3.516 \(\int \coth ^3(c+d x) (a+b \sinh ^2(c+d x))^p \, dx\)

Optimal. Leaf size=94 \[ -\frac{(a+b p) \left (a+b \sinh ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sinh ^2(c+d x)}{a}+1\right )}{2 a^2 d (p+1)}-\frac{\text{csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{p+1}}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.0907626, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3194, 78, 65} \[ -\frac{(a+b p) \left (a+b \sinh ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sinh ^2(c+d x)}{a}+1\right )}{2 a^2 d (p+1)}-\frac{\text{csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{p+1}}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^p,x]

[Out]

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

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

Mathematica [A]  time = 0.379479, size = 71, normalized size = 0.76 \[ -\frac{\left (a+b \sinh ^2(c+d x)\right )^{p+1} \left (\frac{(a+b p) \, _2F_1\left (1,p+1;p+2;\frac{b \sinh ^2(c+d x)}{a}+1\right )}{p+1}+a \text{csch}^2(c+d x)\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^p,x]

[Out]

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

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Maple [F]  time = 0.265, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm coth} \left (dx+c\right ) \right ) ^{3} \left ( a+b \left ( \sinh \left ( dx+c \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x)

[Out]

int(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sinh(d*x + c)^2 + a)^p*coth(d*x + c)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x, algorithm="fricas")

[Out]

integral((b*sinh(d*x + c)^2 + a)^p*coth(d*x + c)^3, 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(coth(d*x+c)**3*(a+b*sinh(d*x+c)**2)**p,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sinh(d*x + c)^2 + a)^p*coth(d*x + c)^3, x)