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

Optimal. Leaf size=136 \[ \frac{2 a^2 \sinh ^{\frac{3}{2}}(c+d x)}{3 b^3 d}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{2 \left (a^4+b^4\right ) \sqrt{\sinh (c+d x)}}{b^5 d}-\frac{2 a \left (a^4+b^4\right ) \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{b^6 d}-\frac{a \sinh ^2(c+d x)}{2 b^2 d}+\frac{2 \sinh ^{\frac{5}{2}}(c+d x)}{5 b d} \]

[Out]

(-2*a*(a^4 + b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]])/(b^6*d) + (2*(a^4 + b^4)*Sqrt[Sinh[c + d*x]])/(b^5*d) - (a^3
*Sinh[c + d*x])/(b^4*d) + (2*a^2*Sinh[c + d*x]^(3/2))/(3*b^3*d) - (a*Sinh[c + d*x]^2)/(2*b^2*d) + (2*Sinh[c +
d*x]^(5/2))/(5*b*d)

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Rubi [A]  time = 0.156701, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3223, 1890, 1620} \[ \frac{2 a^2 \sinh ^{\frac{3}{2}}(c+d x)}{3 b^3 d}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{2 \left (a^4+b^4\right ) \sqrt{\sinh (c+d x)}}{b^5 d}-\frac{2 a \left (a^4+b^4\right ) \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{b^6 d}-\frac{a \sinh ^2(c+d x)}{2 b^2 d}+\frac{2 \sinh ^{\frac{5}{2}}(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(a^4 + b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]])/(b^6*d) + (2*(a^4 + b^4)*Sqrt[Sinh[c + d*x]])/(b^5*d) - (a^3
*Sinh[c + d*x])/(b^4*d) + (2*a^2*Sinh[c + d*x]^(3/2))/(3*b^3*d) - (a*Sinh[c + d*x]^2)/(2*b^2*d) + (2*Sinh[c +
d*x]^(5/2))/(5*b*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 1890

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{g = Denominator[n]}, Dist[g, Subst[Int[x^(g - 1)*(
Pq /. x -> x^g)*(a + b*x^(g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && FractionQ[n]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin{align*} \int \frac{\cosh ^3(c+d x)}{a+b \sqrt{\sinh (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{a+b \sqrt{x}} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x \left (1+x^4\right )}{a+b x} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{a^4+b^4}{b^5}-\frac{a^3 x}{b^4}+\frac{a^2 x^2}{b^3}-\frac{a x^3}{b^2}+\frac{x^4}{b}-\frac{a \left (a^4+b^4\right )}{b^5 (a+b x)}\right ) \, dx,x,\sqrt{\sinh (c+d x)}\right )}{d}\\ &=-\frac{2 a \left (a^4+b^4\right ) \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{b^6 d}+\frac{2 \left (a^4+b^4\right ) \sqrt{\sinh (c+d x)}}{b^5 d}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{2 a^2 \sinh ^{\frac{3}{2}}(c+d x)}{3 b^3 d}-\frac{a \sinh ^2(c+d x)}{2 b^2 d}+\frac{2 \sinh ^{\frac{5}{2}}(c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.150835, size = 117, normalized size = 0.86 \[ \frac{20 a^2 b^3 \sinh ^{\frac{3}{2}}(c+d x)-30 a^3 b^2 \sinh (c+d x)+60 b \left (a^4+b^4\right ) \sqrt{\sinh (c+d x)}-60 a \left (a^4+b^4\right ) \log \left (a+b \sqrt{\sinh (c+d x)}\right )-15 a b^4 \sinh ^2(c+d x)+12 b^5 \sinh ^{\frac{5}{2}}(c+d x)}{30 b^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*a*(a^4 + b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]] + 60*b*(a^4 + b^4)*Sqrt[Sinh[c + d*x]] - 30*a^3*b^2*Sinh[c +
d*x] + 20*a^2*b^3*Sinh[c + d*x]^(3/2) - 15*a*b^4*Sinh[c + d*x]^2 + 12*b^5*Sinh[c + d*x]^(5/2))/(30*b^6*d)

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Maple [C]  time = 0.1, size = 359, normalized size = 2.6 \begin{align*} -{\frac{{a}^{5}}{d{b}^{6}}\ln \left ({a}^{2} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,{b}^{2}\tanh \left ( 1/2\,dx+c/2 \right ) -{a}^{2} \right ) }-{\frac{a}{d{b}^{2}}\ln \left ({a}^{2} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,{b}^{2}\tanh \left ( 1/2\,dx+c/2 \right ) -{a}^{2} \right ) }-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{{a}^{3}}{d{b}^{4}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{a}^{5}}{d{b}^{6}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{{a}^{3}}{d{b}^{4}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{a}^{5}}{d{b}^{6}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{d}\mbox{{\tt ` int/indef0`}} \left ( -{\frac{b \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{-{b}^{2}\sinh \left ( dx+c \right ) +{a}^{2}}\sqrt{\sinh \left ( dx+c \right ) }},\sinh \left ( dx+c \right ) \right ) } \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)^(1/2)),x)

[Out]

-a^5/d/b^6*ln(a^2*tanh(1/2*d*x+1/2*c)^2+2*b^2*tanh(1/2*d*x+1/2*c)-a^2)-a/d/b^2*ln(a^2*tanh(1/2*d*x+1/2*c)^2+2*
b^2*tanh(1/2*d*x+1/2*c)-a^2)-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^2*a+a^3/d/b^4/(tanh(1/2*d*x+1/2*c)+1)+1/2/d/b^2
/(tanh(1/2*d*x+1/2*c)+1)*a+a^5/d/b^6*ln(tanh(1/2*d*x+1/2*c)+1)+1/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/2/d/b^2/(
tanh(1/2*d*x+1/2*c)-1)^2*a+a^3/d/b^4/(tanh(1/2*d*x+1/2*c)-1)-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)-1)*a+a^5/d/b^6*ln(
tanh(1/2*d*x+1/2*c)-1)+1/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)-1)+`int/indef0`(-cosh(d*x+c)^2*b*sinh(d*x+c)^(1/2)/(-b
^2*sinh(d*x+c)+a^2),sinh(d*x+c))/d

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

[Out]

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

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Fricas [B]  time = 7.32631, size = 2182, normalized size = 16.04 \begin{align*} \text{result too large to display} \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)^(1/2)),x, algorithm="fricas")

[Out]

-1/120*(15*a*b^4*cosh(d*x + c)^4 + 15*a*b^4*sinh(d*x + c)^4 + 60*a^3*b^2*cosh(d*x + c)^3 - 60*a^3*b^2*cosh(d*x
+ c) + 15*a*b^4 + 60*(a*b^4*cosh(d*x + c) + a^3*b^2)*sinh(d*x + c)^3 - 120*((a^5 + a*b^4)*d*x + (a^5 + a*b^4)
*c)*cosh(d*x + c)^2 + 30*(3*a*b^4*cosh(d*x + c)^2 + 6*a^3*b^2*cosh(d*x + c) - 4*(a^5 + a*b^4)*d*x - 4*(a^5 + a
*b^4)*c)*sinh(d*x + c)^2 - 120*((a^5 + a*b^4)*cosh(d*x + c)^2 + 2*(a^5 + a*b^4)*cosh(d*x + c)*sinh(d*x + c) +
(a^5 + a*b^4)*sinh(d*x + c)^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) - b^2 + 2*
(b^2*cosh(d*x + c) + a^2)*sinh(d*x + c) - 4*(a*b*cosh(d*x + c) + a*b*sinh(d*x + c))*sqrt(sinh(d*x + c)))/(b^2*
cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 - 2*a^2*cosh(d*x + c) - b^2 + 2*(b^2*cosh(d*x + c) - a^2)*sinh(d*x + c))
) + 120*((a^5 + a*b^4)*cosh(d*x + c)^2 + 2*(a^5 + a*b^4)*cosh(d*x + c)*sinh(d*x + c) + (a^5 + a*b^4)*sinh(d*x
+ c)^2)*log(2*(b^2*sinh(d*x + c) - a^2)/(cosh(d*x + c) - sinh(d*x + c))) + 60*(a*b^4*cosh(d*x + c)^3 + 3*a^3*b
^2*cosh(d*x + c)^2 - a^3*b^2 - 4*((a^5 + a*b^4)*d*x + (a^5 + a*b^4)*c)*cosh(d*x + c))*sinh(d*x + c) - 4*(3*b^5
*cosh(d*x + c)^4 + 3*b^5*sinh(d*x + c)^4 + 10*a^2*b^3*cosh(d*x + c)^3 - 10*a^2*b^3*cosh(d*x + c) + 3*b^5 + 2*(
6*b^5*cosh(d*x + c) + 5*a^2*b^3)*sinh(d*x + c)^3 + 6*(10*a^4*b + 9*b^5)*cosh(d*x + c)^2 + 6*(3*b^5*cosh(d*x +
c)^2 + 5*a^2*b^3*cosh(d*x + c) + 10*a^4*b + 9*b^5)*sinh(d*x + c)^2 + 2*(6*b^5*cosh(d*x + c)^3 + 15*a^2*b^3*cos
h(d*x + c)^2 - 5*a^2*b^3 + 6*(10*a^4*b + 9*b^5)*cosh(d*x + c))*sinh(d*x + c))*sqrt(sinh(d*x + c)))/(b^6*d*cosh
(d*x + c)^2 + 2*b^6*d*cosh(d*x + c)*sinh(d*x + c) + b^6*d*sinh(d*x + c)^2)

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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)**(1/2)),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}}{b \sqrt{\sinh \left (d x + c\right )} + a}\,{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)^(1/2)),x, algorithm="giac")

[Out]

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