Integrand size = 24, antiderivative size = 155 \[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {x}{2 b}-\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{3/2} d}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{3/2} d}-\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d} \] Output:
1/2*x/b-1/2*a^(3/4)*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))/( a^(1/2)-b^(1/2))^(1/2)/b^(3/2)/d+1/2*a^(3/4)*arctanh((a^(1/2)+b^(1/2))^(1/ 2)*tanh(d*x+c)/a^(1/4))/(a^(1/2)+b^(1/2))^(1/2)/b^(3/2)/d-1/2*cosh(d*x+c)* sinh(d*x+c)/b/d
Time = 1.55 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.02 \[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {2 \sqrt {b} (c+d x)+\frac {2 a \arctan \left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {2 a \text {arctanh}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-\sqrt {b} \sinh (2 (c+d x))}{4 b^{3/2} d} \] Input:
Integrate[Sinh[c + d*x]^6/(a - b*Sinh[c + d*x]^4),x]
Output:
(2*Sqrt[b]*(c + d*x) + (2*a*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqr t[-a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] + (2*a*ArcTanh[((Sqrt [a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a] *Sqrt[b]] - Sqrt[b]*Sinh[2*(c + d*x)])/(4*b^(3/2)*d)
Time = 0.46 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 25, 3696, 1610, 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.
| \(\displaystyle \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i c+i d x)^6}{a-b \sin (i c+i d x)^4}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i c+i d x)^6}{a-b \sin (i c+i d x)^4}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tanh ^6(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 1610 |
| \(\displaystyle \frac {\int \left (\frac {a \tanh ^2(c+d x)}{b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {1}{2 b \left (\tanh ^2(c+d x)-1\right )}-\frac {1}{4 b (\tanh (c+d x)-1)^2}-\frac {1}{4 b (\tanh (c+d x)+1)^2}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} \sqrt {\sqrt {a}+\sqrt {b}}}+\frac {\text {arctanh}(\tanh (c+d x))}{2 b}-\frac {1}{4 b (1-\tanh (c+d x))}+\frac {1}{4 b (\tanh (c+d x)+1)}}{d}\) |
Input:
Int[Sinh[c + d*x]^6/(a - b*Sinh[c + d*x]^4),x]
Output:
(ArcTanh[Tanh[c + d*x]]/(2*b) - (a^(3/4)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]* Tanh[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/2)) + (a^(3/4)*Ar cTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/2)) - 1/(4*b*(1 - Tanh[c + d*x])) + 1/(4*b*(1 + Tanh[c + d*x ])))/d
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a + b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 *a*c, 0] && IntegerQ[q] && IntegerQ[m]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) ^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & & IntegerQ[m/2] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.37 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05
| method | result | size |
| risch | \(\frac {x}{2 b}-\frac {{\mathrm e}^{2 d x +2 c}}{8 b d}+\frac {{\mathrm e}^{-2 d x -2 c}}{8 b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a \,b^{6} d^{4}-256 b^{7} d^{4}\right ) \textit {\_Z}^{4}-32 a^{2} b^{3} d^{2} \textit {\_Z}^{2}+a^{3}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (-\frac {128 b^{4} d^{3}}{a}+\frac {128 b^{5} d^{3}}{a^{2}}\right ) \textit {\_R}^{3}+\left (32 b^{2} d^{2}-\frac {32 d^{2} b^{3}}{a}\right ) \textit {\_R}^{2}+16 b d \textit {\_R} -\frac {2 a}{b}-1\right )\right )\) | \(162\) |
| derivativedivides | \(\frac {\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b}-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}\right )}{b}-\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b}}{d}\) | \(206\) |
| default | \(\frac {\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b}-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}\right )}{b}-\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b}}{d}\) | \(206\) |
Input:
int(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/2*x/b-1/8/b/d*exp(2*d*x+2*c)+1/8/b/d*exp(-2*d*x-2*c)+sum(_R*ln(exp(2*d*x +2*c)+(-128/a*b^4*d^3+128/a^2*b^5*d^3)*_R^3+(32*b^2*d^2-32/a*d^2*b^3)*_R^2 +16*b*d*_R-2/b*a-1),_R=RootOf((256*a*b^6*d^4-256*b^7*d^4)*_Z^4-32*a^2*b^3* d^2*_Z^2+a^3))
Leaf count of result is larger than twice the leaf count of optimal. 1441 vs. \(2 (111) = 222\).
Time = 0.16 (sec) , antiderivative size = 1441, normalized size of antiderivative = 9.30 \[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")
Output:
1/8*(4*d*x*cosh(d*x + c)^2 - cosh(d*x + c)^4 - 4*cosh(d*x + c)*sinh(d*x + c)^3 - sinh(d*x + c)^4 + 2*(2*d*x - 3*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 2 *(b*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x + c)^2)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*s inh(d*x + c) + a^2*sinh(d*x + c)^2 + 2*(a^2*b^2 - a*b^3)*d^2*sqrt(a^3/((a^ 2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2 + 2*((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b ^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*d)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a^ 2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))) + 2*(b*d*cosh(d* x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x + c)^2)*sqrt(( (a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + 2*(a^2*b^2 - a*b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b ^6 + b^7)*d^4)) - a^2 - 2*((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*d)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b ^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))) + 2*(b*d*cosh(d*x + c)^2 + 2* b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x + c)^2)*sqrt(-((a*b^3 - b^4 )*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2)/((a*b^3 - b^4)*d^2) )*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d *x + c)^2 - 2*(a^2*b^2 - a*b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)...
Timed out. \[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**6/(a-b*sinh(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{6}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")
Output:
1/8*(4*d*x*e^(2*d*x + 2*c) - e^(4*d*x + 4*c) + 1)*e^(-2*d*x - 2*c)/(b*d) - 1/64*integrate(256*(a*e^(6*d*x + 6*c) - 2*a*e^(4*d*x + 4*c) + a*e^(2*d*x + 2*c))/(b^2*e^(8*d*x + 8*c) - 4*b^2*e^(6*d*x + 6*c) - 4*b^2*e^(2*d*x + 2* c) + b^2 - 2*(8*a*b*e^(4*c) - 3*b^2*e^(4*c))*e^(4*d*x)), x)
\[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{6}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x, algorithm="giac")
Output:
sage0*x
Time = 12.25 (sec) , antiderivative size = 2191, normalized size of antiderivative = 14.14 \[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
int(sinh(c + d*x)^6/(a - b*sinh(c + d*x)^4),x)
Output:
log((((((4194304*a^6*d^2*(512*a^4 - 1184*a^3*b - 253*a*b^3 - b^4 + 930*a^2 *b^2 + b^4*exp(2*c + 2*d*x) + 627*a*b^3*exp(2*c + 2*d*x) + 768*a^3*b*exp(2 *c + 2*d*x) - 1392*a^2*b^2*exp(2*c + 2*d*x)))/(b^12*(a - b)^2) - (16777216 *a^6*d^3*(((a^3*b^7)^(1/2) + a^2*b^3)/(b^6*d^2*(a - b)))^(1/2)*(40*a*b^2 - 35*b^3 + 512*a^3*exp(2*c + 2*d*x) + 64*b^3*exp(2*c + 2*d*x) + 326*a*b^2*e xp(2*c + 2*d*x) - 896*a^2*b*exp(2*c + 2*d*x)))/(b^11*(a - b)))*(((a^3*b^7) ^(1/2) + a^2*b^3)/(b^6*d^2*(a - b)))^(1/2))/4 - (2097152*a^7*d*(256*a^2*b - 256*a*b^2 - 5*b^3 - 1024*a^3*exp(2*c + 2*d*x) + 6*b^3*exp(2*c + 2*d*x) + 756*a*b^2*exp(2*c + 2*d*x) + 256*a^2*b*exp(2*c + 2*d*x)))/(b^14*(a - b))) *(((a^3*b^7)^(1/2) + a^2*b^3)/(b^6*d^2*(a - b)))^(1/2))/4 - (524288*a^8*(1 85*a*b^2 - 464*a^2*b + 256*a^3 + 24*b^3 - 1024*a^3*exp(2*c + 2*d*x) - 35*b ^3*exp(2*c + 2*d*x) - 988*a*b^2*exp(2*c + 2*d*x) + 2048*a^2*b*exp(2*c + 2* d*x)))/(b^15*(a - b)^2))*(-((a^3*b^7)^(1/2) + a^2*b^3)/(16*(b^7*d^2 - a*b^ 6*d^2)))^(1/2) - log((((((4194304*a^6*d^2*(512*a^4 - 1184*a^3*b - 253*a*b^ 3 - b^4 + 930*a^2*b^2 + b^4*exp(2*c + 2*d*x) + 627*a*b^3*exp(2*c + 2*d*x) + 768*a^3*b*exp(2*c + 2*d*x) - 1392*a^2*b^2*exp(2*c + 2*d*x)))/(b^12*(a - b)^2) + (16777216*a^6*d^3*(((a^3*b^7)^(1/2) + a^2*b^3)/(b^6*d^2*(a - b)))^ (1/2)*(40*a*b^2 - 35*b^3 + 512*a^3*exp(2*c + 2*d*x) + 64*b^3*exp(2*c + 2*d *x) + 326*a*b^2*exp(2*c + 2*d*x) - 896*a^2*b*exp(2*c + 2*d*x)))/(b^11*(a - b)))*(((a^3*b^7)^(1/2) + a^2*b^3)/(b^6*d^2*(a - b)))^(1/2))/4 + (20971...
\[ \int \frac {\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {too large to display} \] Input:
int(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x)
Output:
( - 8*e**(4*c + 4*d*x)*a*b - e**(4*c + 4*d*x)*b**2 - 32768*e**(4*c + 2*d*x )*int(e**(2*d*x)/(8*e**(8*c + 8*d*x)*a*b + e**(8*c + 8*d*x)*b**2 - 32*e**( 6*c + 6*d*x)*a*b - 4*e**(6*c + 6*d*x)*b**2 - 128*e**(4*c + 4*d*x)*a**2 + 3 2*e**(4*c + 4*d*x)*a*b + 6*e**(4*c + 4*d*x)*b**2 - 32*e**(2*c + 2*d*x)*a*b - 4*e**(2*c + 2*d*x)*b**2 + 8*a*b + b**2),x)*a**4*d + 10240*e**(4*c + 2*d *x)*int(e**(2*d*x)/(8*e**(8*c + 8*d*x)*a*b + e**(8*c + 8*d*x)*b**2 - 32*e* *(6*c + 6*d*x)*a*b - 4*e**(6*c + 6*d*x)*b**2 - 128*e**(4*c + 4*d*x)*a**2 + 32*e**(4*c + 4*d*x)*a*b + 6*e**(4*c + 4*d*x)*b**2 - 32*e**(2*c + 2*d*x)*a *b - 4*e**(2*c + 2*d*x)*b**2 + 8*a*b + b**2),x)*a**3*b*d + 1536*e**(4*c + 2*d*x)*int(e**(2*d*x)/(8*e**(8*c + 8*d*x)*a*b + e**(8*c + 8*d*x)*b**2 - 32 *e**(6*c + 6*d*x)*a*b - 4*e**(6*c + 6*d*x)*b**2 - 128*e**(4*c + 4*d*x)*a** 2 + 32*e**(4*c + 4*d*x)*a*b + 6*e**(4*c + 4*d*x)*b**2 - 32*e**(2*c + 2*d*x )*a*b - 4*e**(2*c + 2*d*x)*b**2 + 8*a*b + b**2),x)*a**2*b**2*d - 32*e**(4* c + 2*d*x)*int(e**(2*d*x)/(8*e**(8*c + 8*d*x)*a*b + e**(8*c + 8*d*x)*b**2 - 32*e**(6*c + 6*d*x)*a*b - 4*e**(6*c + 6*d*x)*b**2 - 128*e**(4*c + 4*d*x) *a**2 + 32*e**(4*c + 4*d*x)*a*b + 6*e**(4*c + 4*d*x)*b**2 - 32*e**(2*c + 2 *d*x)*a*b - 4*e**(2*c + 2*d*x)*b**2 + 8*a*b + b**2),x)*a*b**3*d + 2048*e** (2*c + 2*d*x)*int(1/(8*e**(10*c + 10*d*x)*a*b + e**(10*c + 10*d*x)*b**2 - 32*e**(8*c + 8*d*x)*a*b - 4*e**(8*c + 8*d*x)*b**2 - 128*e**(6*c + 6*d*x)*a **2 + 32*e**(6*c + 6*d*x)*a*b + 6*e**(6*c + 6*d*x)*b**2 - 32*e**(4*c + ...