Integrand size = 23, antiderivative size = 79 \[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=-\frac {(2 a+b) x}{2 b^2}+\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a-b} b^2 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d} \]
-1/2*(2*a+b)*x/b^2+1/2*cosh(d*x+c)*sinh(d*x+c)/b/d+a^(3/2)*arctanh((a-b)^( 1/2)*tanh(d*x+c)/a^(1/2))/b^2/d/(a-b)^(1/2)
Time = 0.54 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {-2 (2 a+b) (c+d x)+\frac {4 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a-b}}+b \sinh (2 (c+d x))}{4 b^2 d} \]
(-2*(2*a + b)*(c + d*x) + (4*a^(3/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/S qrt[a]])/Sqrt[a - b] + b*Sinh[2*(c + d*x)])/(4*b^2*d)
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3666, 372, 397, 219, 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.
\(\displaystyle \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (i c+i d x)^4}{a-b \sin (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 3666 |
\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2 \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\int \frac {(a+b) \tanh ^2(c+d x)+a}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{2 b}}{d}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(2 a+b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{b}-\frac {2 a^2 \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{2 b}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(2 a+b) \text {arctanh}(\tanh (c+d x))}{b}-\frac {2 a^2 \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{2 b}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(2 a+b) \text {arctanh}(\tanh (c+d x))}{b}-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{b \sqrt {a-b}}}{2 b}}{d}\) |
(-1/2*(((2*a + b)*ArcTanh[Tanh[c + d*x]])/b - (2*a^(3/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a - b]*b))/b + Tanh[c + d*x]/(2*b*(1 - Tanh[c + d*x]^2)))/d
3.1.31.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(163\) vs. \(2(67)=134\).
Time = 0.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.08
method | result | size |
risch | \(-\frac {a x}{b^{2}}-\frac {x}{2 b}+\frac {{\mathrm e}^{2 d x +2 c}}{8 b d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 b d}+\frac {\sqrt {\left (a -b \right ) a}\, a \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-2 a +2 \sqrt {\left (a -b \right ) a}+b}{b}\right )}{2 \left (a -b \right ) d \,b^{2}}-\frac {\sqrt {\left (a -b \right ) a}\, a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a +2 \sqrt {\left (a -b \right ) a}-b}{b}\right )}{2 \left (a -b \right ) d \,b^{2}}\) | \(164\) |
derivativedivides | \(\frac {\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{2}}-\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (-2 a -b \right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b^{2}}-\frac {2 a^{3} \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{b^{2}}}{d}\) | \(301\) |
default | \(\frac {\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{2}}-\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (-2 a -b \right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b^{2}}-\frac {2 a^{3} \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{b^{2}}}{d}\) | \(301\) |
-a*x/b^2-1/2*x/b+1/8/b/d*exp(2*d*x+2*c)-1/8/b/d*exp(-2*d*x-2*c)+1/2*((a-b) *a)^(1/2)/(a-b)*a/d/b^2*ln(exp(2*d*x+2*c)-(-2*a+2*((a-b)*a)^(1/2)+b)/b)-1/ 2*((a-b)*a)^(1/2)/(a-b)*a/d/b^2*ln(exp(2*d*x+2*c)+(2*a+2*((a-b)*a)^(1/2)-b )/b)
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (67) = 134\).
Time = 0.36 (sec) , antiderivative size = 859, normalized size of antiderivative = 10.87 \[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Too large to display} \]
[-1/8*(4*(2*a + b)*d*x*cosh(d*x + c)^2 - 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*(2*a + b)*d*x - 3*b*cosh(d *x + c)^2)*sinh(d*x + c)^2 - 4*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh (d*x + c) + a*sinh(d*x + c)^2)*sqrt(a/(a - b))*log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*(2*a*b - b^2 )*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b - b^2)*sinh(d*x + c)^ 2 + 8*a^2 - 8*a*b + b^2 + 4*(b^2*cosh(d*x + c)^3 + (2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a*b - b^2)*cosh(d*x + c)^2 + 2*(a*b - b^2)*cosh( d*x + c)*sinh(d*x + c) + (a*b - b^2)*sinh(d*x + c)^2 + 2*a^2 - 3*a*b + b^2 )*sqrt(a/(a - 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*(2*(2*a + b)*d*x*cosh(d*x + c) - b*cosh(d*x + c)^3)*sinh(d*x + c) + b)/(b^2*d*cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x + c)*si nh(d*x + c) + b^2*d*sinh(d*x + c)^2), -1/8*(4*(2*a + b)*d*x*cosh(d*x + c)^ 2 - 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*(2*a + b)*d*x - 3*b*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 8*(a*cos h(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2)*sqrt(- a/(a - b))*arctan(1/2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*a - b)*sqrt(-a/(a - b))/a) + 4*(2*(2*a + b)*d*...
Timed out. \[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
\[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{4}}{b \sinh \left (d x + c\right )^{2} + a} \,d x } \]
Time = 1.69 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.73 \[ \int \frac {\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}-\frac {x\,\left (2\,a+b\right )}{2\,b^2}+\frac {a^{3/2}\,\ln \left (-\frac {4\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{b^3}-\frac {2\,a^{3/2}\,\left (b\,d+2\,a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^3\,d\,\sqrt {a-b}}\right )}{2\,b^2\,d\,\sqrt {a-b}}-\frac {a^{3/2}\,\ln \left (\frac {2\,a^{3/2}\,\left (b\,d+2\,a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^3\,d\,\sqrt {a-b}}-\frac {4\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{b^3}\right )}{2\,b^2\,d\,\sqrt {a-b}} \]
exp(2*c + 2*d*x)/(8*b*d) - exp(- 2*c - 2*d*x)/(8*b*d) - (x*(2*a + b))/(2*b ^2) + (a^(3/2)*log(- (4*a^2*exp(2*c + 2*d*x))/b^3 - (2*a^(3/2)*(b*d + 2*a* d*exp(2*c + 2*d*x) - b*d*exp(2*c + 2*d*x)))/(b^3*d*(a - b)^(1/2))))/(2*b^2 *d*(a - b)^(1/2)) - (a^(3/2)*log((2*a^(3/2)*(b*d + 2*a*d*exp(2*c + 2*d*x) - b*d*exp(2*c + 2*d*x)))/(b^3*d*(a - b)^(1/2)) - (4*a^2*exp(2*c + 2*d*x))/ b^3))/(2*b^2*d*(a - b)^(1/2))