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

Optimal. Leaf size=32 \[ i \text{Unintegrable}\left (-\frac{i \sinh ^3(c+d x)}{a+b \tanh ^3(c+d x)},x\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0461342, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., 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*}

Mathematica [A]  time = 0.483965, size = 826, normalized size = 25.81 \[ \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)

________________________________________________________________________________________

Maple [A]  time = 0.11, size = 346, normalized size = 10.8 \begin{align*} -8\,{\frac{1}{d \left ( 16\,a-16\,b \right ) \left ( \tanh \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+{\frac{16}{3\,d \left ( 16\,a-16\,b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{a}{2\,d \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{d \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{16}{3\,d \left ( 16\,a+16\,b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-8\,{\frac{1}{d \left ( 16\,a+16\,b \right ) \left ( \tanh \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}+{\frac{a}{2\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{b}{d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{ab}{3\,d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{ \left ( 2\,{a}^{2}+{b}^{2} \right ){{\it \_R}}^{4}-6\,{{\it \_R}}^{3}ab+2\, \left ( 4\,{a}^{2}+5\,{b}^{2} \right ){{\it \_R}}^{2}-6\,ab{\it \_R}+2\,{a}^{2}+{b}^{2}}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

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]

-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-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
-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))

________________________________________________________________________________________

Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.05765, size = 473, normalized size = 14.78 \begin{align*} -\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 d - b d} - \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 d - b d}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d} \end{align*}

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*d - b*d) - (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*
d - b*d))/((a^4 - 2*a^2*b^2 + b^4)*d)