3.82 \(\int \cosh ^3(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=30 \[ \frac {(a+b) \sinh ^3(c+d x)}{3 d}+\frac {a \sinh (c+d x)}{d} \]

[Out]

a*sinh(d*x+c)/d+1/3*(a+b)*sinh(d*x+c)^3/d

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Rubi [A]  time = 0.04, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3676} \[ \frac {(a+b) \sinh ^3(c+d x)}{3 d}+\frac {a \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sinh[c + d*x])/d + ((a + b)*Sinh[c + d*x]^3)/(3*d)

Rule 3676

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.47 \[ \frac {a \sinh ^3(c+d x)}{3 d}+\frac {a \sinh (c+d x)}{d}+\frac {b \sinh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sinh[c + d*x])/d + (a*Sinh[c + d*x]^3)/(3*d) + (b*Sinh[c + d*x]^3)/(3*d)

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fricas [A]  time = 0.39, size = 45, normalized size = 1.50 \[ \frac {{\left (a + b\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 3 \, a - b\right )} \sinh \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((a + b)*sinh(d*x + c)^3 + 3*((a + b)*cosh(d*x + c)^2 + 3*a - b)*sinh(d*x + c))/d

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giac [B]  time = 0.16, size = 94, normalized size = 3.13 \[ -\frac {{\left (9 \, a e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-3 \, d x - 3 \, c\right )} - {\left (a e^{\left (3 \, d x + 12 \, c\right )} + b e^{\left (3 \, d x + 12 \, c\right )} + 9 \, a e^{\left (d x + 10 \, c\right )} - 3 \, b e^{\left (d x + 10 \, c\right )}\right )} e^{\left (-9 \, c\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((9*a*e^(2*d*x + 2*c) - 3*b*e^(2*d*x + 2*c) + a + b)*e^(-3*d*x - 3*c) - (a*e^(3*d*x + 12*c) + b*e^(3*d*x
+ 12*c) + 9*a*e^(d*x + 10*c) - 3*b*e^(d*x + 10*c))*e^(-9*c))/d

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maple [A]  time = 0.32, size = 37, normalized size = 1.23 \[ \frac {\frac {b \left (\sinh ^{3}\left (d x +c \right )\right )}{3}+a \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (d x +c \right )\right )}{3}\right ) \sinh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^3*(a+b*tanh(d*x+c)^2),x)

[Out]

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

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maxima [B]  time = 0.34, size = 83, normalized size = 2.77 \[ \frac {b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3}}{24 \, d} + \frac {1}{24} \, a {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*b*(e^(d*x + c) - e^(-d*x - c))^3/d + 1/24*a*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d - e^(
-3*d*x - 3*c)/d)

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mupad [B]  time = 0.21, size = 74, normalized size = 2.47 \[ \frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (a+b\right )}{24\,d}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,\left (a+b\right )}{24\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (3\,a-b\right )}{8\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (3\,a-b\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^3*(a + b*tanh(c + d*x)^2),x)

[Out]

(exp(3*c + 3*d*x)*(a + b))/(24*d) - (exp(- 3*c - 3*d*x)*(a + b))/(24*d) + (exp(c + d*x)*(3*a - b))/(8*d) - (ex
p(- c - d*x)*(3*a - b))/(8*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \cosh ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**3*(a+b*tanh(d*x+c)**2),x)

[Out]

Integral((a + b*tanh(c + d*x)**2)*cosh(c + d*x)**3, x)

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