Integrand size = 23, antiderivative size = 104 \[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {3 a \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{8 f}+\frac {\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{4 f} \]
1/4*sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(3/2)/f+3/8*a^2*arctanh(sinh(f*x+e)*b^ (1/2)/(a+b*sinh(f*x+e)^2)^(1/2))/f/b^(1/2)+3/8*a*sinh(f*x+e)*(a+b*sinh(f*x +e)^2)^(1/2)/f
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a+b \sinh ^2(e+f x)} \left (5 a \sinh (e+f x)+2 b \sinh ^3(e+f x)+\frac {3 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}\right )}{8 f} \]
(Sqrt[a + b*Sinh[e + f*x]^2]*(5*a*Sinh[e + f*x] + 2*b*Sinh[e + f*x]^3 + (3 *a^(3/2)*ArcSinh[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*Si nh[e + f*x]^2)/a])))/(8*f)
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3669, 211, 211, 224, 219}
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.
| \(\displaystyle \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (i e+i f x) \left (a-b \sin (i e+i f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \left (b \sinh ^2(e+f x)+a\right )^{3/2}d\sinh (e+f x)}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {3}{4} a \int \sqrt {b \sinh ^2(e+f x)+a}d\sinh (e+f x)+\frac {1}{4} \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)+\frac {1}{2} \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}\right )+\frac {1}{4} \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b \sinh ^2(e+f x)}{b \sinh ^2(e+f x)+a}}d\frac {\sinh (e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}+\frac {1}{2} \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}\right )+\frac {1}{4} \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 \sqrt {b}}+\frac {1}{2} \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}\right )+\frac {1}{4} \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{f}\) |
((Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^(3/2))/4 + (3*a*((a*ArcTanh[(Sqrt[ b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]])/(2*Sqrt[b]) + (Sinh[e + f* x]*Sqrt[a + b*Sinh[e + f*x]^2])/2))/4)/f
3.4.63.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {\sinh \left (f x +e \right ) \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{4 f}+\frac {3 a \sinh \left (f x +e \right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}}{8 f}+\frac {3 a^{2} \ln \left (\sqrt {b}\, \sinh \left (f x +e \right )+\sqrt {a +b \sinh \left (f x +e \right )^{2}}\right )}{8 f \sqrt {b}}\) | \(90\) |
| default | \(\frac {\sinh \left (f x +e \right ) \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{4 f}+\frac {3 a \sinh \left (f x +e \right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}}{8 f}+\frac {3 a^{2} \ln \left (\sqrt {b}\, \sinh \left (f x +e \right )+\sqrt {a +b \sinh \left (f x +e \right )^{2}}\right )}{8 f \sqrt {b}}\) | \(90\) |
1/4*sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(3/2)/f+3/8*a*sinh(f*x+e)*(a+b*sinh(f* x+e)^2)^(1/2)/f+3/8/f*a^2/b^(1/2)*ln(b^(1/2)*sinh(f*x+e)+(a+b*sinh(f*x+e)^ 2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 1126 vs. \(2 (88) = 176\).
Time = 0.34 (sec) , antiderivative size = 3161, normalized size of antiderivative = 30.39 \[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/64*(6*(a^2*cosh(f*x + e)^4 + 4*a^2*cosh(f*x + e)^3*sinh(f*x + e) + 6*a^ 2*cosh(f*x + e)^2*sinh(f*x + e)^2 + 4*a^2*cosh(f*x + e)*sinh(f*x + e)^3 + a^2*sinh(f*x + e)^4)*sqrt(b)*log(-((a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^8 + 8*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)*sinh(f*x + e)^7 + (a^2*b - 2*a* b^2 + b^3)*sinh(f*x + e)^8 + 2*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^6 + 2*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3 + 14*(a^2*b - 2*a*b^2 + b^3)*c osh(f*x + e)^2)*sinh(f*x + e)^6 + 4*(14*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^3 + 3*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e))*sinh(f*x + e)^5 + (9*a^2*b - 14*a*b^2 + 6*b^3)*cosh(f*x + e)^4 + (70*(a^2*b - 2*a*b^2 + b ^3)*cosh(f*x + e)^4 + 9*a^2*b - 14*a*b^2 + 6*b^3 + 30*(a^3 - 4*a^2*b + 5*a *b^2 - 2*b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 4*(14*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^5 + 10*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^ 3 + (9*a^2*b - 14*a*b^2 + 6*b^3)*cosh(f*x + e))*sinh(f*x + e)^3 + b^3 + 2* (3*a*b^2 - 2*b^3)*cosh(f*x + e)^2 + 2*(14*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^6 + 15*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^4 + 3*a*b^2 - 2*b^3 + 3*(9*a^2*b - 14*a*b^2 + 6*b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + sqrt(2)*((a^2 - 2*a*b + b^2)*cosh(f*x + e)^6 + 6*(a^2 - 2*a*b + b^2)*cosh (f*x + e)*sinh(f*x + e)^5 + (a^2 - 2*a*b + b^2)*sinh(f*x + e)^6 - 3*(a^2 - 2*a*b + b^2)*cosh(f*x + e)^4 + 3*(5*(a^2 - 2*a*b + b^2)*cosh(f*x + e)^2 - a^2 + 2*a*b - b^2)*sinh(f*x + e)^4 + 4*(5*(a^2 - 2*a*b + b^2)*cosh(f*x...
Timed out. \[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cosh \left (f x + e\right ) \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (88) = 176\).
Time = 0.70 (sec) , antiderivative size = 791, normalized size of antiderivative = 7.61 \[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\frac {{\left (\frac {24 \, a^{2} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}}{\sqrt {-b}}\right ) e^{\left (4 \, e\right )}}{\sqrt {-b}} - \frac {12 \, a^{2} e^{\left (4 \, e\right )} \log \left ({\left | -{\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} b - 2 \, a \sqrt {b} + b^{\frac {3}{2}} \right |}\right )}{\sqrt {b}} + \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b} {\left (b e^{\left (2 \, f x + 6 \, e\right )} + \frac {{\left (10 \, a b e^{\left (6 \, e\right )} - 3 \, b^{2} e^{\left (6 \, e\right )}\right )} e^{\left (-2 \, e\right )}}{b}\right )} + \frac {4 \, {\left (10 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{3} a^{2} e^{\left (4 \, e\right )} - 8 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{3} a b e^{\left (4 \, e\right )} + 2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{3} b^{2} e^{\left (4 \, e\right )} + 8 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} a b^{\frac {3}{2}} e^{\left (4 \, e\right )} - 3 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} b^{\frac {5}{2}} e^{\left (4 \, e\right )} - 6 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} a^{2} b e^{\left (4 \, e\right )} + 4 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} a b^{2} e^{\left (4 \, e\right )} - 4 \, a b^{\frac {5}{2}} e^{\left (4 \, e\right )} + b^{\frac {7}{2}} e^{\left (4 \, e\right )}\right )}}{{\left ({\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} - b\right )}^{2}}\right )} e^{\left (-4 \, e\right )}}{64 \, f} \]
1/64*(24*a^2*arctan(-(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4 *a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))/sqrt(-b))*e^(4*e)/sqrt(-b) - 12*a^2*e^(4*e)*log(abs(-(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e ) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*b - 2*a*sqrt(b) + b^(3 /2)))/sqrt(b) + sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f* x + 2*e) + b)*(b*e^(2*f*x + 6*e) + (10*a*b*e^(6*e) - 3*b^2*e^(6*e))*e^(-2* e)/b) + 4*(10*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2 *f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^3*a^2*e^(4*e) - 8*(sqrt(b)*e^(2*f* x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2 *e) + b))^3*a*b*e^(4*e) + 2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4 *e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^3*b^2*e^(4*e) + 8*(s qrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2* b*e^(2*f*x + 2*e) + b))^2*a*b^(3/2)*e^(4*e) - 3*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^ 2*b^(5/2)*e^(4*e) - 6*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*a^2*b*e^(4*e) + 4*(sqrt(b) *e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2 *f*x + 2*e) + b))*a*b^2*e^(4*e) - 4*a*b^(5/2)*e^(4*e) + b^(7/2)*e^(4*e))/( (sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^2 - b)^2)*e^(-4*e)/f
Time = 1.93 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.58 \[ \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\frac {\mathrm {sinh}\left (e+f\,x\right )\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2}{a}\right )}{f\,{\left (\frac {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2}{a}+1\right )}^{3/2}} \]