Integrand size = 23, antiderivative size = 75 \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2} d}-\frac {a \cosh (c+d x)}{(a+b)^2 d}+\frac {\cosh ^3(c+d x)}{3 (a+b) d} \] Output:
a*b^(1/2)*arctanh(b^(1/2)*sech(d*x+c)/(a+b)^(1/2))/(a+b)^(5/2)/d-a*cosh(d* x+c)/(a+b)^2/d+1/3*cosh(d*x+c)^3/(a+b)/d
Result contains complex when optimal does not.
Time = 0.73 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.80 \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {12 i a \sqrt {b} \left (\arctan \left (\frac {-i \sqrt {a+b}-\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+\arctan \left (\frac {-i \sqrt {a+b}+\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )\right )-3 (3 a-b) \sqrt {a+b} \cosh (c+d x)+(a+b)^{3/2} \cosh (3 (c+d x))}{12 (a+b)^{5/2} d} \] Input:
Integrate[Sinh[c + d*x]^3/(a + b*Tanh[c + d*x]^2),x]
Output:
((12*I)*a*Sqrt[b]*(ArcTan[((-I)*Sqrt[a + b] - Sqrt[a]*Tanh[(c + d*x)/2])/S qrt[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)
Time = 0.52 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4147, 25, 359, 264, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \sin (i c+i d x)^3}{a-b \tan (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\sin (i c+i d x)^3}{a-b \tan (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\cosh ^4(c+d x) \left (1-\text {sech}^2(c+d x)\right )}{-b \text {sech}^2(c+d x)+a+b}d\text {sech}(c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cosh ^4(c+d x) \left (1-\text {sech}^2(c+d x)\right )}{-b \text {sech}^2(c+d x)+a+b}d\text {sech}(c+d x)}{d}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {a \int \frac {\cosh ^2(c+d x)}{-b \text {sech}^2(c+d x)+a+b}d\text {sech}(c+d x)}{a+b}+\frac {\cosh ^3(c+d x)}{3 (a+b)}}{d}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {\frac {a \left (\frac {b \int \frac {1}{-b \text {sech}^2(c+d x)+a+b}d\text {sech}(c+d x)}{a+b}-\frac {\cosh (c+d x)}{a+b}\right )}{a+b}+\frac {\cosh ^3(c+d x)}{3 (a+b)}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {a \left (\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {\cosh (c+d x)}{a+b}\right )}{a+b}+\frac {\cosh ^3(c+d x)}{3 (a+b)}}{d}\) |
Input:
Int[Sinh[c + d*x]^3/(a + b*Tanh[c + d*x]^2),x]
Output:
(Cosh[c + d*x]^3/(3*(a + b)) + (a*((Sqrt[b]*ArcTanh[(Sqrt[b]*Sech[c + d*x] )/Sqrt[a + b]])/(a + b)^(3/2) - Cosh[c + d*x]/(a + b)))/(a + b))/d
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(201\) vs. \(2(65)=130\).
Time = 7.68 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.69
| method | result | size |
| derivativedivides | \(\frac {-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{3 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (16 a +16 b \right )}-\frac {a -b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {16}{3 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a +b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a b \,\operatorname {arctanh}\left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{\left (a +b \right )^{2} \sqrt {a b +b^{2}}}}{d}\) | \(202\) |
| default | \(\frac {-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{3 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (16 a +16 b \right )}-\frac {a -b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {16}{3 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a +b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a b \,\operatorname {arctanh}\left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{\left (a +b \right )^{2} \sqrt {a b +b^{2}}}}{d}\) | \(202\) |
| risch | \(\frac {{\mathrm e}^{3 d x +3 c}}{24 \left (a +b \right ) d}-\frac {3 \,{\mathrm e}^{d x +c} a}{8 \left (a +b \right )^{2} d}+\frac {{\mathrm e}^{d x +c} b}{8 \left (a +b \right )^{2} d}-\frac {3 \,{\mathrm e}^{-d x -c} a}{8 \left (a +b \right )^{2} d}+\frac {{\mathrm e}^{-d x -c} b}{8 \left (a +b \right )^{2} d}+\frac {{\mathrm e}^{-3 d x -3 c}}{24 \left (a +b \right ) d}+\frac {\sqrt {\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right )}{2 \left (a +b \right )^{3} d}-\frac {\sqrt {\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right )}{2 \left (a +b \right )^{3} d}\) | \(214\) |
Input:
int(sinh(d*x+c)^3/(a+tanh(d*x+c)^2*b),x,method=_RETURNVERBOSE)
Output:
1/d*(-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)-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)/(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)))
Leaf count of result is larger than twice the leaf count of optimal. 616 vs. \(2 (65) = 130\).
Time = 0.13 (sec) , antiderivative size = 1367, normalized size of antiderivative = 18.23 \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")
Output:
[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 +...
\[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\sinh ^{3}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(sinh(d*x+c)**3/(a+b*tanh(d*x+c)**2),x)
Output:
Integral(sinh(c + d*x)**3/(a + b*tanh(c + d*x)**2), x)
\[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{3}}{b \tanh \left (d x + c\right )^{2} + a} \,d x } \] Input:
integrate(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")
Output:
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)
Exception generated. \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 2.39 (sec) , antiderivative size = 955, normalized size of antiderivative = 12.73 \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx =\text {Too large to display} \] Input:
int(sinh(c + d*x)^3/(a + b*tanh(c + d*x)^2),x)
Output:
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)) - (e...
Time = 0.27 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.75 \[ \int \frac {\sinh ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {-12 e^{3 d x +3 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (e^{2 d x +2 c} \sqrt {a +b}+\sqrt {a +b}-2 e^{d x +c} \sqrt {b}\right ) a +12 e^{3 d x +3 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (e^{2 d x +2 c} \sqrt {a +b}+\sqrt {a +b}+2 e^{d x +c} \sqrt {b}\right ) a +e^{6 d x +6 c} a^{2}+2 e^{6 d x +6 c} a b +e^{6 d x +6 c} b^{2}-9 e^{4 d x +4 c} a^{2}-6 e^{4 d x +4 c} a b +3 e^{4 d x +4 c} b^{2}-9 e^{2 d x +2 c} a^{2}-6 e^{2 d x +2 c} a b +3 e^{2 d x +2 c} b^{2}+a^{2}+2 a b +b^{2}}{24 e^{3 d x +3 c} d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )} \] Input:
int(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^2),x)
Output:
( - 12*e**(3*c + 3*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a + 12*e**(3*c + 3*d*x)*sqrt(b) *sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d* x)*sqrt(b))*a + e**(6*c + 6*d*x)*a**2 + 2*e**(6*c + 6*d*x)*a*b + e**(6*c + 6*d*x)*b**2 - 9*e**(4*c + 4*d*x)*a**2 - 6*e**(4*c + 4*d*x)*a*b + 3*e**(4* c + 4*d*x)*b**2 - 9*e**(2*c + 2*d*x)*a**2 - 6*e**(2*c + 2*d*x)*a*b + 3*e** (2*c + 2*d*x)*b**2 + a**2 + 2*a*b + b**2)/(24*e**(3*c + 3*d*x)*d*(a**3 + 3 *a**2*b + 3*a*b**2 + b**3))