3.74 \(\int \frac {\sinh ^3(c+d x)}{a+b \tanh ^3(c+d x)} \, dx\)

Optimal. Leaf size=33 \[ i \text {Int}\left (-\frac {i \sinh ^3(c+d x)}{a+b \tanh ^3(c+d x)},x\right ) \]

[Out]

I*Unintegrable(-I*sinh(d*x+c)^3/(a+b*tanh(d*x+c)^3),x)

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Rubi [A]  time = 0.05, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^3(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

I*Defer[Int][((-I)*Sinh[c + d*x]^3)/(a + b*Tanh[c + d*x]^3), x]

Rubi steps

\begin {align*} \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^3(c+d x)} \, dx &=i \int -\frac {i \sinh ^3(c+d x)}{a+b \tanh ^3(c+d x)} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 826, normalized size = 25.03 \[ \frac {\cosh (3 (c+d x)) a^3+27 b \sinh (c+d x) a^2-b \sinh (3 (c+d x)) a^2-9 \left (a^2+3 b^2\right ) \cosh (c+d x) a-b^2 \cosh (3 (c+d x)) a-2 b \text {RootSum}\left [a \text {$\#$1}^6+b \text {$\#$1}^6+3 a \text {$\#$1}^4-3 b \text {$\#$1}^4+3 a \text {$\#$1}^2+3 b \text {$\#$1}^2+a-b\& ,\frac {3 a^2 c \text {$\#$1}^4+3 b^2 c \text {$\#$1}^4-3 a b c \text {$\#$1}^4+3 a^2 d x \text {$\#$1}^4+3 b^2 d x \text {$\#$1}^4-3 a b d x \text {$\#$1}^4+6 a^2 \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^4+6 b^2 \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^4-6 a b \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^4+2 a^2 c \text {$\#$1}^2-2 b^2 c \text {$\#$1}^2+2 a^2 d x \text {$\#$1}^2-2 b^2 d x \text {$\#$1}^2+4 a^2 \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^2-4 b^2 \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^2+3 a^2 c+3 b^2 c+3 a b c+3 a^2 d x+3 b^2 d x+3 a b d x+6 a^2 \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right )+6 b^2 \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right )+6 a b \log \left (\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right )}{a \text {$\#$1}^5+b \text {$\#$1}^5+2 a \text {$\#$1}^3-2 b \text {$\#$1}^3+a \text {$\#$1}+b \text {$\#$1}}\& \right ] a+9 b^3 \sinh (c+d x)+b^3 \sinh (3 (c+d x))}{12 (a-b)^2 (a+b)^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a*(a^2 + 3*b^2)*Cosh[c + d*x] + a^3*Cosh[3*(c + d*x)] - a*b^2*Cosh[3*(c + d*x)] - 2*a*b*RootSum[a - b + 3*
a*#1^2 + 3*b*#1^2 + 3*a*#1^4 - 3*b*#1^4 + a*#1^6 + b*#1^6 & , (3*a^2*c + 3*a*b*c + 3*b^2*c + 3*a^2*d*x + 3*a*b
*d*x + 3*b^2*d*x + 6*a^2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]
*#1] + 6*a*b*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1] + 6*b^2
*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1] + 2*a^2*c*#1^2 - 2*
b^2*c*#1^2 + 2*a^2*d*x*#1^2 - 2*b^2*d*x*#1^2 + 4*a^2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*
x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^2 - 4*b^2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#
1 - Sinh[(c + d*x)/2]*#1]*#1^2 + 3*a^2*c*#1^4 - 3*a*b*c*#1^4 + 3*b^2*c*#1^4 + 3*a^2*d*x*#1^4 - 3*a*b*d*x*#1^4
+ 3*b^2*d*x*#1^4 + 6*a^2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]
*#1]*#1^4 - 6*a*b*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1
^4 + 6*b^2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^4)/(a*
#1 + b*#1 + 2*a*#1^3 - 2*b*#1^3 + a*#1^5 + b*#1^5) & ] + 27*a^2*b*Sinh[c + d*x] + 9*b^3*Sinh[c + d*x] - a^2*b*
Sinh[3*(c + d*x)] + b^3*Sinh[3*(c + d*x)])/(12*(a - b)^2*(a + b)^2*d)

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fricas [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(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 3.16, size = 344, normalized size = 10.42 \[ -\frac {\frac {{\left (9 \, a e^{\left (2 \, d x + 2 \, c\right )} + 9 \, b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )} e^{\left (-3 \, d x\right )}}{a^{2} e^{\left (3 \, c\right )} - 2 \, a b e^{\left (3 \, c\right )} + b^{2} e^{\left (3 \, c\right )}} - \frac {a^{2} e^{\left (3 \, d x + 30 \, c\right )} + 2 \, a b e^{\left (3 \, d x + 30 \, c\right )} + b^{2} e^{\left (3 \, d x + 30 \, c\right )} - 9 \, a^{2} e^{\left (d x + 28 \, c\right )} + 9 \, b^{2} e^{\left (d x + 28 \, c\right )}}{a^{3} e^{\left (27 \, c\right )} + 3 \, a^{2} b e^{\left (27 \, c\right )} + 3 \, a b^{2} e^{\left (27 \, c\right )} + b^{3} e^{\left (27 \, c\right )}}}{24 \, d} - \frac {\frac {6 \, {\left (a^{3} b e^{c} + a^{2} b^{2} e^{c} + a b^{3} e^{c}\right )} d x}{a - b} - \frac {{\left (a^{3} b e^{c} + a^{2} b^{2} e^{c} + a b^{3} e^{c}\right )} \log \left ({\left | a e^{\left (6 \, d x + 6 \, c\right )} + b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + a - b \right |}\right )}{a - b}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((9*a*e^(2*d*x + 2*c) + 9*b*e^(2*d*x + 2*c) - a + b)*e^(-3*d*x)/(a^2*e^(3*c) - 2*a*b*e^(3*c) + b^2*e^(3*
c)) - (a^2*e^(3*d*x + 30*c) + 2*a*b*e^(3*d*x + 30*c) + b^2*e^(3*d*x + 30*c) - 9*a^2*e^(d*x + 28*c) + 9*b^2*e^(
d*x + 28*c))/(a^3*e^(27*c) + 3*a^2*b*e^(27*c) + 3*a*b^2*e^(27*c) + b^3*e^(27*c)))/d - (6*(a^3*b*e^c + a^2*b^2*
e^c + a*b^3*e^c)*d*x/(a - b) - (a^3*b*e^c + a^2*b^2*e^c + a*b^3*e^c)*log(abs(a*e^(6*d*x + 6*c) + b*e^(6*d*x +
6*c) + 3*a*e^(4*d*x + 4*c) - 3*b*e^(4*d*x + 4*c) + 3*a*e^(2*d*x + 2*c) + 3*b*e^(2*d*x + 2*c) + a - b))/(a - b)
)/((a^4 - 2*a^2*b^2 + b^4)*d^2)

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maple [A]  time = 0.47, size = 346, normalized size = 10.48 \[ -\frac {a b \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\left (2 a^{2}+b^{2}\right ) \textit {\_R}^{4}-6 \textit {\_R}^{3} a b +2 \left (4 a^{2}+5 b^{2}\right ) \textit {\_R}^{2}-6 a b \textit {\_R} +2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {16}{3 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{d \left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{2 d \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b}{d \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{d \left (16 a -16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{3 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (16 a -16 b \right )}-\frac {a}{2 d \left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b}{d \left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^3),x)

[Out]

-1/3/d*a*b/(a+b)^2/(a-b)^2*sum(((2*a^2+b^2)*_R^4-6*_R^3*a*b+2*(4*a^2+5*b^2)*_R^2-6*a*b*_R+2*a^2+b^2)/(_R^5*a+2
*_R^3*a+4*_R^2*b+_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))-16/3/d/(tanh
(1/2*d*x+1/2*c)-1)^3/(16*a+16*b)-8/d/(16*a+16*b)/(tanh(1/2*d*x+1/2*c)-1)^2+1/2/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-
1)*a-1/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*b-8/d/(16*a-16*b)/(tanh(1/2*d*x+1/2*c)+1)^2+16/3/d/(tanh(1/2*d*x+1/2*
c)+1)^3/(16*a-16*b)-1/2/d/(a-b)^2/(tanh(1/2*d*x+1/2*c)+1)*a-1/d/(a-b)^2/(tanh(1/2*d*x+1/2*c)+1)*b

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (a^{3} + a^{2} b - a b^{2} - b^{3} + {\left (a^{3} e^{\left (6 \, c\right )} - a^{2} b e^{\left (6 \, c\right )} - a b^{2} e^{\left (6 \, c\right )} + b^{3} e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} - 9 \, {\left (a^{3} e^{\left (4 \, c\right )} - 3 \, a^{2} b e^{\left (4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, c\right )} - b^{3} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} - 9 \, {\left (a^{3} e^{\left (2 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, c\right )} + b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}\right )} e^{\left (-3 \, d x\right )}}{24 \, {\left (a^{4} d e^{\left (3 \, c\right )} - 2 \, a^{2} b^{2} d e^{\left (3 \, c\right )} + b^{4} d e^{\left (3 \, c\right )}\right )}} - \frac {1}{8} \, \int \frac {16 \, {\left (3 \, {\left (a^{3} b e^{\left (5 \, c\right )} - a^{2} b^{2} e^{\left (5 \, c\right )} + a b^{3} e^{\left (5 \, c\right )}\right )} e^{\left (5 \, d x\right )} + 2 \, {\left (a^{3} b e^{\left (3 \, c\right )} - a b^{3} e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + 3 \, {\left (a^{3} b e^{c} + a^{2} b^{2} e^{c} + a b^{3} e^{c}\right )} e^{\left (d x\right )}\right )}}{a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5} + {\left (a^{5} e^{\left (6 \, c\right )} + a^{4} b e^{\left (6 \, c\right )} - 2 \, a^{3} b^{2} e^{\left (6 \, c\right )} - 2 \, a^{2} b^{3} e^{\left (6 \, c\right )} + a b^{4} e^{\left (6 \, c\right )} + b^{5} e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} + 3 \, {\left (a^{5} e^{\left (4 \, c\right )} - a^{4} b e^{\left (4 \, c\right )} - 2 \, a^{3} b^{2} e^{\left (4 \, c\right )} + 2 \, a^{2} b^{3} e^{\left (4 \, c\right )} + a b^{4} e^{\left (4 \, c\right )} - b^{5} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \, {\left (a^{5} e^{\left (2 \, c\right )} + a^{4} b e^{\left (2 \, c\right )} - 2 \, a^{3} b^{2} e^{\left (2 \, c\right )} - 2 \, a^{2} b^{3} e^{\left (2 \, c\right )} + a b^{4} e^{\left (2 \, c\right )} + b^{5} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3 + a^2*b - a*b^2 - b^3 + (a^3*e^(6*c) - a^2*b*e^(6*c) - a*b^2*e^(6*c) + b^3*e^(6*c))*e^(6*d*x) - 9*(a
^3*e^(4*c) - 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) - b^3*e^(4*c))*e^(4*d*x) - 9*(a^3*e^(2*c) + 3*a^2*b*e^(2*c) + 3
*a*b^2*e^(2*c) + b^3*e^(2*c))*e^(2*d*x))*e^(-3*d*x)/(a^4*d*e^(3*c) - 2*a^2*b^2*d*e^(3*c) + b^4*d*e^(3*c)) - 1/
8*integrate(16*(3*(a^3*b*e^(5*c) - a^2*b^2*e^(5*c) + a*b^3*e^(5*c))*e^(5*d*x) + 2*(a^3*b*e^(3*c) - a*b^3*e^(3*
c))*e^(3*d*x) + 3*(a^3*b*e^c + a^2*b^2*e^c + a*b^3*e^c)*e^(d*x))/(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4
- b^5 + (a^5*e^(6*c) + a^4*b*e^(6*c) - 2*a^3*b^2*e^(6*c) - 2*a^2*b^3*e^(6*c) + a*b^4*e^(6*c) + b^5*e^(6*c))*e^
(6*d*x) + 3*(a^5*e^(4*c) - a^4*b*e^(4*c) - 2*a^3*b^2*e^(4*c) + 2*a^2*b^3*e^(4*c) + a*b^4*e^(4*c) - b^5*e^(4*c)
)*e^(4*d*x) + 3*(a^5*e^(2*c) + a^4*b*e^(2*c) - 2*a^3*b^2*e^(2*c) - 2*a^2*b^3*e^(2*c) + a*b^4*e^(2*c) + b^5*e^(
2*c))*e^(2*d*x)), x)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^3/(a + b*tanh(c + d*x)^3),x)

[Out]

\text{Hanged}

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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(sinh(d*x+c)**3/(a+b*tanh(d*x+c)**3),x)

[Out]

Timed out

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