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

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

[Out]

-2*a*(a^4+b^4)*ln(a+b*sinh(d*x+c)^(1/2))/b^6/d-a^3*sinh(d*x+c)/b^4/d+2/3*a^2*sinh(d*x+c)^(3/2)/b^3/d-1/2*a*sin
h(d*x+c)^2/b^2/d+2/5*sinh(d*x+c)^(5/2)/b/d+2*(a^4+b^4)*sinh(d*x+c)^(1/2)/b^5/d

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Rubi [A]  time = 0.16, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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 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 \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*}

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Mathematica [A]  time = 0.15, size = 117, normalized size = 0.86 \[ \frac {60 b \left (a^4+b^4\right ) \sqrt {\sinh (c+d x)}-60 a \left (a^4+b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )-30 a^3 b^2 \sinh (c+d x)+20 a^2 b^3 \sinh ^{\frac {3}{2}}(c+d x)-15 a b^4 \sinh ^2(c+d x)+12 b^5 \sinh ^{\frac {5}{2}}(c+d x)}{30 b^6 d} \]

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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fricas [B]  time = 2.07, size = 879, normalized size = 6.46 \[ \text {result too large to display} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^(1/2)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.20, size = 359, normalized size = 2.64 \[ -\frac {a}{2 d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a^{3}}{d \,b^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {a}{2 d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{5} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{6}}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{2}}-\frac {a}{2 d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a^{3}}{d \,b^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {a}{2 d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {a^{5} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{6}}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{2}}-\frac {a^{5} \ln \left (a^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{2}\right )}{d \,b^{6}}-\frac {a \ln \left (a^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{2}\right )}{d \,b^{2}}+\frac {\mathit {`\,int/indef0`\,}\left (-\frac {\left (\cosh ^{2}\left (d x +c \right )\right ) b \left (\sqrt {\sinh }\left (d x +c \right )\right )}{-b^{2} \sinh \left (d x +c \right )+a^{2}}, \sinh \left (d x +c \right )\right )}{d} \]

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]

-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)-1)^2*a+1/d/b^4/(tanh(1/2*d*x+1/2*c)-1)*a^3-1/2/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
+1/d/b^4/(tanh(1/2*d*x+1/2*c)+1)*a^3+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)-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)+`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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (d x + c\right )^{3}}{b \sqrt {\sinh \left (d x + c\right )} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^3/(a+b*sinh(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^3/(a + b*sinh(c + d*x)^(1/2)),x)

[Out]

int(cosh(c + d*x)^3/(a + b*sinh(c + d*x)^(1/2)), 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)**(1/2)),x)

[Out]

Timed out

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