Integrand size = 23, antiderivative size = 90 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {(a-2 b) \arctan \left (\frac {\sqrt {b} \cosh (c+d x)}{\sqrt {a-b}}\right )}{2 (a-b)^{3/2} b^{3/2} d}-\frac {a \cosh (c+d x)}{2 (a-b) b d \left (a-b+b \cosh ^2(c+d x)\right )} \]
1/2*(a-2*b)*arctan(cosh(d*x+c)*b^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/b^(3/2)/d- 1/2*a*cosh(d*x+c)/(a-b)/b/d/(a-b+b*cosh(d*x+c)^2)
Result contains complex when optimal does not.
Time = 2.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.57 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {\frac {(a-2 b) \left (\arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a-b}}\right )+\arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a-b}}\right )\right )}{(a-b)^{3/2}}-\frac {2 a \sqrt {b} \cosh (c+d x)}{(a-b) (2 a-b+b \cosh (2 (c+d x)))}}{2 b^{3/2} d} \]
(((a - 2*b)*(ArcTan[(Sqrt[b] - I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]] + ArcTan[(Sqrt[b] + I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]]))/(a - b)^(3/ 2) - (2*a*Sqrt[b]*Cosh[c + d*x])/((a - b)*(2*a - b + b*Cosh[2*(c + d*x)])) )/(2*b^(3/2)*d)
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 26, 3665, 298, 218}
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.
\(\displaystyle \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \sin (i c+i d x)^3}{\left (a-b \sin (i c+i d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\sin (i c+i d x)^3}{\left (a-b \sin (i c+i d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1-\cosh ^2(c+d x)}{\left (b \cosh ^2(c+d x)+a-b\right )^2}d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle -\frac {\frac {a \cosh (c+d x)}{2 b (a-b) \left (a+b \cosh ^2(c+d x)-b\right )}-\frac {(a-2 b) \int \frac {1}{b \cosh ^2(c+d x)+a-b}d\cosh (c+d x)}{2 b (a-b)}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\frac {a \cosh (c+d x)}{2 b (a-b) \left (a+b \cosh ^2(c+d x)-b\right )}-\frac {(a-2 b) \arctan \left (\frac {\sqrt {b} \cosh (c+d x)}{\sqrt {a-b}}\right )}{2 b^{3/2} (a-b)^{3/2}}}{d}\) |
-((-1/2*((a - 2*b)*ArcTan[(Sqrt[b]*Cosh[c + d*x])/Sqrt[a - b]])/((a - b)^( 3/2)*b^(3/2)) + (a*Cosh[c + d*x])/(2*(a - b)*b*(a - b + b*Cosh[c + d*x]^2) ))/d)
3.1.43.3.1 Defintions of rubi rules used
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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.72
method | result | size |
derivativedivides | \(\frac {\frac {8 \left (2 a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-16 a}{\left (16 a b -16 b^{2}\right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )}+\frac {4 \left (2 a -4 b \right ) \arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 a +4 b}{4 \sqrt {a b -b^{2}}}\right )}{\left (16 a b -16 b^{2}\right ) \sqrt {a b -b^{2}}}}{d}\) | \(155\) |
default | \(\frac {\frac {8 \left (2 a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-16 a}{\left (16 a b -16 b^{2}\right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )}+\frac {4 \left (2 a -4 b \right ) \arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 a +4 b}{4 \sqrt {a b -b^{2}}}\right )}{\left (16 a b -16 b^{2}\right ) \sqrt {a b -b^{2}}}}{d}\) | \(155\) |
risch | \(-\frac {a \,{\mathrm e}^{d x +c} \left ({\mathrm e}^{2 d x +2 c}+1\right )}{b d \left (a -b \right ) \left (b \,{\mathrm e}^{4 d x +4 c}+4 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+b \right )}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \left (a -b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a b +b^{2}}}+1\right ) a}{4 \sqrt {-a b +b^{2}}\, \left (a -b \right ) d b}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \left (a -b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a b +b^{2}}}+1\right )}{2 \sqrt {-a b +b^{2}}\, \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a b +b^{2}}}+1\right ) a}{4 \sqrt {-a b +b^{2}}\, \left (a -b \right ) d b}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a b +b^{2}}}+1\right )}{2 \sqrt {-a b +b^{2}}\, \left (a -b \right ) d}\) | \(310\) |
1/d*(8*((2*a-4*b)*tanh(1/2*d*x+1/2*c)^2-2*a)/(16*a*b-16*b^2)/(tanh(1/2*d*x +1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tanh(1/2*d*x+1/2*c)^2+a)+4*(2*a- 4*b)/(16*a*b-16*b^2)/(a*b-b^2)^(1/2)*arctan(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. 953 vs. \(2 (78) = 156\).
Time = 0.31 (sec) , antiderivative size = 1889, normalized size of antiderivative = 20.99 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
[-1/4*(4*(a^2*b - a*b^2)*cosh(d*x + c)^3 + 12*(a^2*b - a*b^2)*cosh(d*x + c )*sinh(d*x + c)^2 + 4*(a^2*b - a*b^2)*sinh(d*x + c)^3 + ((a*b - 2*b^2)*cos h(d*x + c)^4 + 4*(a*b - 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a*b - 2*b^ 2)*sinh(d*x + c)^4 + 2*(2*a^2 - 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(a*b - 2*b^2)*cosh(d*x + c)^2 + 2*a^2 - 5*a*b + 2*b^2)*sinh(d*x + c)^2 + a*b - 2*b^2 + 4*((a*b - 2*b^2)*cosh(d*x + c)^3 + (2*a^2 - 5*a*b + 2*b^2)*cosh(d *x + c))*sinh(d*x + c))*sqrt(-a*b + b^2)*log((b*cosh(d*x + c)^4 + 4*b*cosh (d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(2*a - 3*b)*cosh(d*x + c )^2 + 2*(3*b*cosh(d*x + c)^2 - 2*a + 3*b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - (2*a - 3*b)*cosh(d*x + c))*sinh(d*x + c) - 4*(cosh(d*x + c)^3 + 3 *cosh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + (3*cosh(d*x + c)^2 + 1) *sinh(d*x + c) + cosh(d*x + c))*sqrt(-a*b + b^2) + b)/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(2*a - b)*cosh( d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh (d*x + c)^3 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + b)) + 4*(a^2*b - a* b^2)*cosh(d*x + c) + 4*(a^2*b - a*b^2 + 3*(a^2*b - a*b^2)*cosh(d*x + c)^2) *sinh(d*x + c))/((a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)^4 + 4*(a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2*b^3 - 2*a*b^4 + b^ 5)*d*sinh(d*x + c)^4 + 2*(2*a^3*b^2 - 5*a^2*b^3 + 4*a*b^4 - b^5)*d*cosh(d* x + c)^2 + 2*(3*(a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)^2 + (2*a^3*b^...
Timed out. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int { \frac {\sinh \left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
-(a*e^(3*d*x + 3*c) + a*e^(d*x + c))/(a*b^2*d - b^3*d + (a*b^2*d*e^(4*c) - b^3*d*e^(4*c))*e^(4*d*x) + 2*(2*a^2*b*d*e^(2*c) - 3*a*b^2*d*e^(2*c) + b^3 *d*e^(2*c))*e^(2*d*x)) + 1/8*integrate(8*((a*e^(3*c) - 2*b*e^(3*c))*e^(3*d *x) - (a*e^c - 2*b*e^c)*e^(d*x))/(a*b^2 - b^3 + (a*b^2*e^(4*c) - b^3*e^(4* c))*e^(4*d*x) + 2*(2*a^2*b*e^(2*c) - 3*a*b^2*e^(2*c) + b^3*e^(2*c))*e^(2*d *x)), x)
\[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int { \frac {\sinh \left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^3}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]