3.2.51 \(\int \coth ^3(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [151]

Optimal. Leaf size=52 \[ -\frac {a^2 \coth ^2(c+d x)}{2 d}+\frac {(a+b)^2 \log (\cosh (c+d x))}{d}+\frac {a (a+2 b) \log (\tanh (c+d x))}{d} \]

[Out]

-1/2*a^2*coth(d*x+c)^2/d+(a+b)^2*ln(cosh(d*x+c))/d+a*(a+2*b)*ln(tanh(d*x+c))/d

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Rubi [A]
time = 0.07, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3751, 457, 90} \begin {gather*} -\frac {a^2 \coth ^2(c+d x)}{2 d}+\frac {a (a+2 b) \log (\tanh (c+d x))}{d}+\frac {(a+b)^2 \log (\cosh (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^2*Coth[c + d*x]^2)/d + ((a + b)^2*Log[Cosh[c + d*x]])/d + (a*(a + 2*b)*Log[Tanh[c + d*x]])/d

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^2}{x^3 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {(a+b x)^2}{(1-x) x^2} \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {(a+b)^2}{-1+x}+\frac {a^2}{x^2}+\frac {a (a+2 b)}{x}\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=-\frac {a^2 \coth ^2(c+d x)}{2 d}+\frac {(a+b)^2 \log (\cosh (c+d x))}{d}+\frac {a (a+2 b) \log (\tanh (c+d x))}{d}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 50, normalized size = 0.96 \begin {gather*} \frac {-a^2 \coth ^2(c+d x)+2 (a+b)^2 \log (\cosh (c+d x))+2 a (a+2 b) \log (\tanh (c+d x))}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*Coth[c + d*x]^2) + 2*(a + b)^2*Log[Cosh[c + d*x]] + 2*a*(a + 2*b)*Log[Tanh[c + d*x]])/(2*d)

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Maple [A]
time = 1.83, size = 50, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {a^{2} \left (\ln \left (\sinh \left (d x +c \right )\right )-\frac {\left (\coth ^{2}\left (d x +c \right )\right )}{2}\right )+2 a b \ln \left (\sinh \left (d x +c \right )\right )+b^{2} \ln \left (\cosh \left (d x +c \right )\right )}{d}\)	\(50\)
default	\(\frac {a^{2} \left (\ln \left (\sinh \left (d x +c \right )\right )-\frac {\left (\coth ^{2}\left (d x +c \right )\right )}{2}\right )+2 a b \ln \left (\sinh \left (d x +c \right )\right )+b^{2} \ln \left (\cosh \left (d x +c \right )\right )}{d}\)	\(50\)
risch	\(-a^{2} x -2 a b x -b^{2} x -\frac {2 b^{2} c}{d}-\frac {4 a b c}{d}-\frac {2 a^{2} c}{d}-\frac {2 a^{2} {\mathrm e}^{2 d x +2 c}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}+\frac {\ln \left (1+{\mathrm e}^{2 d x +2 c}\right ) b^{2}}{d}+\frac {2 a \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (50) = 100\).
time = 0.29, size = 134, normalized size = 2.58 \begin {gather*} a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + \frac {b^{2} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}{d} + \frac {2 \, a b \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - c) - 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) -
e^(-4*d*x - 4*c) - 1))) + b^2*log(e^(d*x + c) + e^(-d*x - c))/d + 2*a*b*log(e^(d*x + c) - e^(-d*x - c))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 677 vs. \(2 (50) = 100\).
time = 0.37, size = 677, normalized size = 13.02 \begin {gather*} -\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} d x \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} d x \sinh \left (d x + c\right )^{4} + {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} d x - a^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x \cosh \left (d x + c\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + a^{2}\right )} \sinh \left (d x + c\right )^{2} - {\left (b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} - 2 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} - b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left ({\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} - a^{2} - 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 2 \, a b + 4 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} d x \cosh \left (d x + c\right )^{3} - {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} d x - a^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} - 2 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2
*a*b + b^2)*d*x*sinh(d*x + c)^4 + (a^2 + 2*a*b + b^2)*d*x - 2*((a^2 + 2*a*b + b^2)*d*x - a^2)*cosh(d*x + c)^2
+ 2*(3*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d*x + a^2)*sinh(d*x + c)^2 - (b^2*cosh(d*
x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 - 2*b^2*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d
*x + c)^2 - b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 - b^2*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh
(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*si
nh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 - 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*cosh(d*x
+ c)^2 - a^2 - 2*a*b)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 - (a^2 + 2*a*b)*cosh(d*
x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*((a^2 + 2*a*b + b^2)*d*x*cosh(
d*x + c)^3 - ((a^2 + 2*a*b + b^2)*d*x - a^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x +
c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 +
4*(d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (50) = 100\).
time = 0.49, size = 141, normalized size = 2.71 \begin {gather*} \frac {b^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right ) + {\left (a^{2} + 2 \, a b\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right ) - \frac {a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a^{2} - 4 \, a b}{e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2) + (a^2 + 2*a*b)*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) -
2) - (a^2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a^2 - 4*a*b)/(
e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2))/d

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Mupad [B]
time = 1.41, size = 211, normalized size = 4.06 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )\,\left (d\,\left (a^2+2\,b\,a\right )+b^2\,d\right )}{2\,d^2}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-x\,{\left (a+b\right )}^2-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (a^2\,\sqrt {-d^2}-b^2\,\sqrt {-d^2}+2\,a\,b\,\sqrt {-d^2}\right )}{d\,\sqrt {a^4+4\,a^3\,b+2\,a^2\,b^2-4\,a\,b^3+b^4}}\right )\,\sqrt {a^4+4\,a^3\,b+2\,a^2\,b^2-4\,a\,b^3+b^4}}{\sqrt {-d^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(exp(4*c + 4*d*x) - 1)*(d*(2*a*b + a^2) + b^2*d))/(2*d^2) - (2*a^2)/(d*(exp(2*c + 2*d*x) - 1)) - (2*a^2)/(
d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - x*(a + b)^2 - (atan((exp(2*c)*exp(2*d*x)*(a^2*(-d^2)^(1/2) -
b^2*(-d^2)^(1/2) + 2*a*b*(-d^2)^(1/2)))/(d*(4*a^3*b - 4*a*b^3 + a^4 + b^4 + 2*a^2*b^2)^(1/2)))*(4*a^3*b - 4*a*
b^3 + a^4 + b^4 + 2*a^2*b^2)^(1/2))/(-d^2)^(1/2)

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