Integrand size = 23, antiderivative size = 294 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\frac {x}{b}+\frac {2 (-1)^{2/3} \sqrt [3]{a} \arctan \left (\frac {\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}} b d}+\frac {2 (-1)^{2/3} \sqrt [3]{a} \arctan \left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}} b d}+\frac {2 \sqrt [3]{a} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+b^{2/3}} b d} \]
x/b+2/3*(-1)^(2/3)*a^(1/3)*arctan((-1)^(1/6)*((-1)^(5/6)*b^(1/3)+I*a^(1/3) *tanh(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)-b^(2/3))^(1/2))/b/d/((-1)^(1/3)* a^(2/3)-b^(2/3))^(1/2)+2/3*a^(1/3)*arctanh((b^(1/3)-a^(1/3)*tanh(1/2*d*x+1 /2*c))/(a^(2/3)+b^(2/3))^(1/2))/b/d/(a^(2/3)+b^(2/3))^(1/2)+2/3*(-1)^(2/3) *a^(1/3)*arctan((-1)^(1/6)*((-1)^(1/6)*b^(1/3)+I*a^(1/3)*tanh(1/2*d*x+1/2* c))/((-1)^(1/3)*a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/b/d/((-1)^(1/3)*a^(2/3) -(-1)^(2/3)*b^(2/3))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 11.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.49 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\frac {3 c+3 d x-2 a \text {RootSum}\left [-b+3 b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {c \text {$\#$1}+d x \text {$\#$1}+2 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}}{b+4 a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{3 b d} \]
(3*c + 3*d*x - 2*a*RootSum[-b + 3*b*#1^2 + 8*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (c*#1 + d*x*#1 + 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)/(b + 4*a*#1 - 2*b*#1^2 + b*#1^4) & ])/(3*b*d)
Time = 0.66 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 26, 3699, 2009}
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.
\(\displaystyle \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \sin (i c+i d x)^3}{a+i b \sin (i c+i d x)^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\sin (i c+i d x)^3}{i b \sin (i c+i d x)^3+a}dx\) |
\(\Big \downarrow \) 3699 |
\(\displaystyle i \int \left (\frac {i a}{b \left (b \sinh ^3(c+d x)+a\right )}-\frac {i}{b}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle i \left (\frac {2 \sqrt [6]{-1} \sqrt [3]{a} \arctan \left (\frac {\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b d \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {2 \sqrt [6]{-1} \sqrt [3]{a} \arctan \left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}-\frac {2 i \sqrt [3]{a} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}+b^{2/3}}}-\frac {i x}{b}\right )\) |
I*(((-I)*x)/b + (2*(-1)^(1/6)*a^(1/3)*ArcTan[((-1)^(1/6)*((-1)^(1/6)*b^(1/ 3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^ (2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]*b*d) + (2*(-1)^( 1/6)*a^(1/3)*ArcTan[((-1)^(1/6)*((-1)^(5/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]*b*d) - (((2*I)/3)*a^(1/3)*ArcTanh[(b^(1/3) - a^(1/3)*Tanh[(c + d*x)/2])/Sqrt[a^(2/3) + b^(2/3)]])/(Sqrt[a^(2/3) + b^(2/3)]*b*d))
3.2.74.3.1 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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n) ^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.42
method | result | size |
derivativedivides | \(\frac {\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}+1\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\) | \(124\) |
default | \(\frac {\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}+1\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\) | \(124\) |
risch | \(\frac {x}{b}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{6} d^{6}+729 b^{8} d^{6}\right ) \textit {\_Z}^{6}-243 a^{2} b^{4} d^{4} \textit {\_Z}^{4}+27 a^{2} d^{2} \textit {\_Z}^{2} b^{2}-a^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{d x +c}+\left (486 a \,b^{4} d^{5}+\frac {486 b^{6} d^{5}}{a}\right ) \textit {\_R}^{5}+\left (81 a \,b^{3} d^{4}+\frac {81 b^{5} d^{4}}{a}\right ) \textit {\_R}^{4}+\left (-135 a \,b^{2} d^{3}+\frac {27 b^{4} d^{3}}{a}\right ) \textit {\_R}^{3}-27 a b \,d^{2} \textit {\_R}^{2}+9 a d \textit {\_R} +\frac {2 a}{b}\right )\right )\) | \(175\) |
1/d*(1/3*a/b*sum((_R^4-2*_R^2+1)/(_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))-1/b*ln(t anh(1/2*d*x+1/2*c)-1)+1/b*ln(1+tanh(1/2*d*x+1/2*c)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 27931 vs. \(2 (205) = 410\).
Time = 1.03 (sec) , antiderivative size = 27931, normalized size of antiderivative = 95.00 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\int \frac {\sinh ^{3}{\left (c + d x \right )}}{a + b \sinh ^{3}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{3}}{b \sinh \left (d x + c\right )^{3} + a} \,d x } \]
-8*a*integrate(e^(3*d*x + 3*c)/(b^2*e^(6*d*x + 6*c) - 3*b^2*e^(4*d*x + 4*c ) + 8*a*b*e^(3*d*x + 3*c) + 3*b^2*e^(2*d*x + 2*c) - b^2), x) + x/b
\[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{3}}{b \sinh \left (d x + c\right )^{3} + a} \,d x } \]
Time = 10.08 (sec) , antiderivative size = 1498, normalized size of antiderivative = 5.10 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh ^3(c+d x)} \, dx=\text {Too large to display} \]
symsum(log(-(24576*a^3*(405*root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 2 43*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k)^5*b^7*d^5*exp(root(72 9*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2 *z^2 - a^2, z, k) + d*x) - root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 24 3*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k)*a*b^2*d - 27*root(729* a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z ^2 - a^2, z, k)^4*b^6*d^4*exp(root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k) + d*x) - 4*a^2*exp( root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2* b^2*d^2*z^2 - a^2, z, k) + d*x) + 12*root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^ 6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k)^2*a*b^3*d^2 - 54*root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27 *a^2*b^2*d^2*z^2 - a^2, z, k)^3*a*b^4*d^3 + 108*root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k)^4 *a*b^5*d^4 - 81*root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d ^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k)^5*a*b^6*d^5 + 20*root(729*a^2*b^6 *d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^ 2, z, k)*a^2*b*d*exp(root(729*a^2*b^6*d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2* b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 - a^2, z, k) + d*x) + 24*root(729*a^2*b^6 *d^6*z^6 + 729*b^8*d^6*z^6 - 243*a^2*b^4*d^4*z^4 + 27*a^2*b^2*d^2*z^2 -...