3.26 \(\int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\)

Optimal. Leaf size=75 \[ \frac {\cosh ^3(c+d x)}{3 d (a+b)}-\frac {a \cosh (c+d x)}{d (a+b)^2}+\frac {a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{d (a+b)^{5/2}} \]

[Out]

-a*cosh(d*x+c)/(a+b)^2/d+1/3*cosh(d*x+c)^3/(a+b)/d+a*arctanh(sech(d*x+c)*b^(1/2)/(a+b)^(1/2))*b^(1/2)/(a+b)^(5
/2)/d

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Rubi [A]  time = 0.13, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3664, 453, 325, 208} \[ \frac {\cosh ^3(c+d x)}{3 d (a+b)}-\frac {a \cosh (c+d x)}{d (a+b)^2}+\frac {a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[b]*ArcTanh[(Sqrt[b]*Sech[c + d*x])/Sqrt[a + b]])/((a + b)^(5/2)*d) - (a*Cosh[c + d*x])/((a + b)^2*d) +
Cosh[c + d*x]^3/(3*(a + b)*d)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {-1+x^2}{x^4 \left (a+b-b x^2\right )} \, dx,x,\text {sech}(c+d x)\right )}{d}\\ &=\frac {\cosh ^3(c+d x)}{3 (a+b) d}+\frac {a \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b-b x^2\right )} \, dx,x,\text {sech}(c+d x)\right )}{(a+b) d}\\ &=-\frac {a \cosh (c+d x)}{(a+b)^2 d}+\frac {\cosh ^3(c+d x)}{3 (a+b) d}+\frac {(a b) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\text {sech}(c+d x)\right )}{(a+b)^2 d}\\ &=\frac {a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2} d}-\frac {a \cosh (c+d x)}{(a+b)^2 d}+\frac {\cosh ^3(c+d x)}{3 (a+b) d}\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 135, normalized size = 1.80 \[ \frac {(a+b)^{3/2} \cosh (3 (c+d x))-3 (3 a-b) \sqrt {a+b} \cosh (c+d x)+12 i a \sqrt {b} \left (\tan ^{-1}\left (\frac {-\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )-i \sqrt {a+b}}{\sqrt {b}}\right )+\tan ^{-1}\left (\frac {\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )-i \sqrt {a+b}}{\sqrt {b}}\right )\right )}{12 d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*I)*a*Sqrt[b]*(ArcTan[((-I)*Sqrt[a + b] - Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[b]] + ArcTan[((-I)*Sqrt[a + b] +
Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[b]]) - 3*(3*a - b)*Sqrt[a + b]*Cosh[c + d*x] + (a + b)^(3/2)*Cosh[3*(c + d*x)
])/(12*(a + b)^(5/2)*d)

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fricas [B]  time = 0.65, size = 1367, normalized size = 18.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((a + b)*cosh(d*x + c)^6 + 6*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 - 3*(3*a -
b)*cosh(d*x + c)^4 + 3*(5*(a + b)*cosh(d*x + c)^2 - 3*a + b)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 -
3*(3*a - b)*cosh(d*x + c))*sinh(d*x + c)^3 - 3*(3*a - b)*cosh(d*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 - 6*(3
*a - b)*cosh(d*x + c)^2 - 3*a + b)*sinh(d*x + c)^2 + 12*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)^2*sinh(d*x + c)
+ 3*a*cosh(d*x + c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3)*sqrt(b/(a + b))*log(((a + b)*cosh(d*x + c)^4 + 4*(a
+ b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a + 3*b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh
(d*x + c)^2 + a + 3*b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a + 3*b)*cosh(d*x + c))*sinh(d*x + c) +
4*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3 + (a + b)*cosh
(d*x + c) + (3*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c))*sqrt(b/(a + b)) + a + b)/((a + b)*cosh(d*x + c)
^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a +
b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x +
c) + a + b)) + 6*((a + b)*cosh(d*x + c)^5 - 2*(3*a - b)*cosh(d*x + c)^3 - (3*a - b)*cosh(d*x + c))*sinh(d*x +
c) + a + b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2*sinh(d*x + c) + 3
*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c)^2 + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^3), 1/24*((a + b)*c
osh(d*x + c)^6 + 6*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 - 3*(3*a - b)*cosh(d*x + c)
^4 + 3*(5*(a + b)*cosh(d*x + c)^2 - 3*a + b)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 - 3*(3*a - b)*cosh
(d*x + c))*sinh(d*x + c)^3 - 3*(3*a - b)*cosh(d*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 - 6*(3*a - b)*cosh(d*x
+ c)^2 - 3*a + b)*sinh(d*x + c)^2 + 24*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)^2*sinh(d*x + c) + 3*a*cosh(d*x
+ c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3)*sqrt(-b/(a + b))*arctan(1/2*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cos
h(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3 + (a - 3*b)*cosh(d*x + c) + (3*(a + b)*cosh(d*x + c)^2 +
a - 3*b)*sinh(d*x + c))*sqrt(-b/(a + b))/b) - 24*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)^2*sinh(d*x + c) + 3*a*
cosh(d*x + c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3)*sqrt(-b/(a + b))*arctan(1/2*((a + b)*cosh(d*x + c) + (a + b
)*sinh(d*x + c))*sqrt(-b/(a + b))/b) + 6*((a + b)*cosh(d*x + c)^5 - 2*(3*a - b)*cosh(d*x + c)^3 - (3*a - b)*co
sh(d*x + c))*sinh(d*x + c) + a + b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*d*cosh(d*x
+ c)^2*sinh(d*x + c) + 3*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c)^2 + (a^2 + 2*a*b + b^2)*d*sinh(d*x
+ c)^3)]

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giac [B]  time = 0.49, size = 810, normalized size = 10.80 \[ -\frac {\frac {24 \, \sqrt {a^{2} - b^{2} + 2 \, \sqrt {-a b} {\left (a + b\right )}} {\left (a b - \sqrt {-a b} a\right )} {\left | a e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )} \right |} \arctan \left (\frac {e^{\left (d x\right )}}{\sqrt {\frac {a^{3} e^{\left (2 \, c\right )} + a^{2} b e^{\left (2 \, c\right )} - a b^{2} e^{\left (2 \, c\right )} - b^{3} e^{\left (2 \, c\right )} + \sqrt {{\left (a^{3} e^{\left (2 \, c\right )} + a^{2} b e^{\left (2 \, c\right )} - a b^{2} e^{\left (2 \, c\right )} - b^{3} e^{\left (2 \, c\right )}\right )}^{2} - {\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 )} {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}}}{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 (-2 \, c\right )}}{a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5} + 2 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {-a b}} + \frac {24 \, {\left (3 \, a^{2} b - a b^{2} + {\left (a^{2} - 3 \, a b\right )} \sqrt {-a b}\right )} {\left | a e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )} \right |} \arctan \left (\frac {e^{\left (d x\right )}}{\sqrt {\frac {a^{3} e^{\left (2 \, c\right )} + a^{2} b e^{\left (2 \, c\right )} - a b^{2} e^{\left (2 \, c\right )} - b^{3} e^{\left (2 \, c\right )} - \sqrt {{\left (a^{3} e^{\left (2 \, c\right )} + a^{2} b e^{\left (2 \, c\right )} - a b^{2} e^{\left (2 \, c\right )} - b^{3} e^{\left (2 \, c\right )}\right )}^{2} - {\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 )} {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}}}{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 (-2 \, c\right )}}{{\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {-a b}\right )} \sqrt {a^{2} - b^{2} - 2 \, \sqrt {-a b} {\left (a + b\right )}}} + \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\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 + 24 \, c\right )} + 2 \, a b e^{\left (3 \, d x + 24 \, c\right )} + b^{2} e^{\left (3 \, d x + 24 \, c\right )} - 9 \, a^{2} e^{\left (d x + 22 \, c\right )} - 6 \, a b e^{\left (d x + 22 \, c\right )} + 3 \, b^{2} e^{\left (d x + 22 \, c\right )}}{a^{3} e^{\left (21 \, c\right )} + 3 \, a^{2} b e^{\left (21 \, c\right )} + 3 \, a b^{2} e^{\left (21 \, c\right )} + b^{3} e^{\left (21 \, c\right )}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(24*sqrt(a^2 - b^2 + 2*sqrt(-a*b)*(a + b))*(a*b - sqrt(-a*b)*a)*abs(a*e^(2*c) + b*e^(2*c))*arctan(e^(d*x
)/sqrt((a^3*e^(2*c) + a^2*b*e^(2*c) - a*b^2*e^(2*c) - b^3*e^(2*c) + sqrt((a^3*e^(2*c) + a^2*b*e^(2*c) - a*b^2*
e^(2*c) - b^3*e^(2*c))^2 - (a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c))*(a^3 + 3*a^2*b + 3*
a*b^2 + b^3)))/(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^(-2*c)/(a^5 + 3*a^4*b + 2*a
^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5 + 2*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(-a*b)) + 24*(3*a^2*b -
a*b^2 + (a^2 - 3*a*b)*sqrt(-a*b))*abs(a*e^(2*c) + b*e^(2*c))*arctan(e^(d*x)/sqrt((a^3*e^(2*c) + a^2*b*e^(2*c)
- a*b^2*e^(2*c) - b^3*e^(2*c) - sqrt((a^3*e^(2*c) + a^2*b*e^(2*c) - a*b^2*e^(2*c) - b^3*e^(2*c))^2 - (a^3*e^(
4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c))*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)))/(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^(-2*c)/((a^4 + 2*a^3*b - 2*a*b^3 - b^4 - 2*(a^3 + 3*a^2*b + 3*
a*b^2 + b^3)*sqrt(-a*b))*sqrt(a^2 - b^2 - 2*sqrt(-a*b)*(a + b))) + (9*a*e^(2*d*x + 2*c) - 3*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 + 24*c) + 2*a*b*e^(3*d*x + 24*
c) + b^2*e^(3*d*x + 24*c) - 9*a^2*e^(d*x + 22*c) - 6*a*b*e^(d*x + 22*c) + 3*b^2*e^(d*x + 22*c))/(a^3*e^(21*c)
+ 3*a^2*b*e^(21*c) + 3*a*b^2*e^(21*c) + b^3*e^(21*c)))/d

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maple [B]  time = 0.30, size = 202, normalized size = 2.69 \[ \frac {-\frac {16}{3 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a +b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{3 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (16 a +16 b \right )}-\frac {a -b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {a b \arctanh \left (\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{\left (a +b \right )^{2} \sqrt {a b +b^{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-16/3/(tanh(1/2*d*x+1/2*c)-1)^3/(16*a+16*b)-8/(16*a+16*b)/(tanh(1/2*d*x+1/2*c)-1)^2-1/2/(a+b)^2*(-a+b)/(t
anh(1/2*d*x+1/2*c)-1)-8/(16*a+16*b)/(tanh(1/2*d*x+1/2*c)+1)^2+16/3/(tanh(1/2*d*x+1/2*c)+1)^3/(16*a+16*b)-1/2*(
a-b)/(a+b)^2/(tanh(1/2*d*x+1/2*c)+1)+a*b/(a+b)^2/(a*b+b^2)^(1/2)*arctanh(1/4*(2*tanh(1/2*d*x+1/2*c)^2*a+2*a+4*
b)/(a*b+b^2)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.65, size = 955, normalized size = 12.73 \[ \frac {{\mathrm {e}}^{-3\,c-3\,d\,x}}{24\,d\,\left (a+b\right )}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}}{24\,d\,\left (a+b\right )}-\frac {\sqrt {a^2\,b}\,\left (2\,\mathrm {atan}\left (\frac {\left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (\frac {4\,\left (2\,a^2\,b^3\,d\,\sqrt {a^2\,b}+4\,a^3\,b^2\,d\,\sqrt {a^2\,b}+2\,a^4\,b\,d\,\sqrt {a^2\,b}\right )}{a\,\left (a+b\right )\,\sqrt {-d^2\,{\left (a+b\right )}^5}\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}}+\frac {2\,a^3\,b}{d\,{\left (a+b\right )}^3\,\sqrt {a^2\,b}\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}\right )+\frac {2\,a^3\,b\,{\mathrm {e}}^{3\,c}\,{\mathrm {e}}^{3\,d\,x}}{d\,{\left (a+b\right )}^3\,\sqrt {a^2\,b}\,\left (a^2+2\,a\,b+b^2\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}\right )\,\left (a^6\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}+b^6\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}+15\,a^2\,b^4\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}+20\,a^3\,b^3\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}+15\,a^4\,b^2\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}+6\,a\,b^5\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}+6\,a^5\,b\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}\right )}{4\,a^2\,b}\right )-2\,\mathrm {atan}\left (\frac {a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-d^2\,{\left (a+b\right )}^5}}{2\,d\,{\left (a+b\right )}^2\,\sqrt {a^2\,b}}\right )\right )}{2\,\sqrt {-a^5\,d^2-5\,a^4\,b\,d^2-10\,a^3\,b^2\,d^2-10\,a^2\,b^3\,d^2-5\,a\,b^4\,d^2-b^5\,d^2}}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (3\,a-b\right )}{8\,d\,{\left (a+b\right )}^2}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (3\,a-b\right )}{8\,d\,{\left (a+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(- 3*c - 3*d*x)/(24*d*(a + b)) + exp(3*c + 3*d*x)/(24*d*(a + b)) - ((a^2*b)^(1/2)*(2*atan(((exp(d*x)*exp(c)
*((4*(2*a^2*b^3*d*(a^2*b)^(1/2) + 4*a^3*b^2*d*(a^2*b)^(1/2) + 2*a^4*b*d*(a^2*b)^(1/2)))/(a*(a + b)*(-d^2*(a +
b)^5)^(1/2)*(2*a*b + a^2 + b^2)*(3*a*b^2 + 3*a^2*b + a^3 + b^3)*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d
^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2)) + (2*a^3*b)/(d*(a + b)^3*(a^2*b)^(1/2)*(2*a*b + a^2 + b^2)*(3*a*b
^2 + 3*a^2*b + a^3 + b^3))) + (2*a^3*b*exp(3*c)*exp(3*d*x))/(d*(a + b)^3*(a^2*b)^(1/2)*(2*a*b + a^2 + b^2)*(3*
a*b^2 + 3*a^2*b + a^3 + b^3)))*(a^6*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3
*b^2*d^2)^(1/2) + b^6*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2
) + 15*a^2*b^4*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2) + 20*
a^3*b^3*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2) + 15*a^4*b^2
*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2) + 6*a*b^5*(- a^5*d^
2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2) + 6*a^5*b*(- a^5*d^2 - b^5*d^
2 - 5*a*b^4*d^2 - 5*a^4*b*d^2 - 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2)))/(4*a^2*b)) - 2*atan((a*exp(d*x)*exp(c
)*(-d^2*(a + b)^5)^(1/2))/(2*d*(a + b)^2*(a^2*b)^(1/2)))))/(2*(- a^5*d^2 - b^5*d^2 - 5*a*b^4*d^2 - 5*a^4*b*d^2
- 10*a^2*b^3*d^2 - 10*a^3*b^2*d^2)^(1/2)) - (exp(- c - d*x)*(3*a - b))/(8*d*(a + b)^2) - (exp(c + d*x)*(3*a -
b))/(8*d*(a + b)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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