Integrand size = 25, antiderivative size = 120 \[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=-\frac {15 \arctan (\sinh (e+f x)) \sqrt {a \cosh ^2(e+f x)} \text {sech}(e+f x)}{8 f}+\frac {15 \sqrt {a \cosh ^2(e+f x)} \tanh (e+f x)}{8 f}-\frac {5 \sqrt {a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{8 f}-\frac {\sqrt {a \cosh ^2(e+f x)} \tanh ^5(e+f x)}{4 f} \]
-15/8*arctan(sinh(f*x+e))*sech(f*x+e)*(a*cosh(f*x+e)^2)^(1/2)/f+15/8*(a*co sh(f*x+e)^2)^(1/2)*tanh(f*x+e)/f-5/8*(a*cosh(f*x+e)^2)^(1/2)*tanh(f*x+e)^3 /f-1/4*(a*cosh(f*x+e)^2)^(1/2)*tanh(f*x+e)^5/f
Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.62 \[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=-\frac {\sqrt {a \cosh ^2(e+f x)} \text {sech}^5(e+f x) \left (60 \arctan (\sinh (e+f x)) \cosh ^4(e+f x)-5 \sinh (e+f x)-15 \sinh (3 (e+f x))-2 \sinh (5 (e+f x))\right )}{32 f} \]
-1/32*(Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x]^5*(60*ArcTan[Sinh[e + f*x]]*C osh[e + f*x]^4 - 5*Sinh[e + f*x] - 15*Sinh[3*(e + f*x)] - 2*Sinh[5*(e + f* x)]))/f
Time = 0.46 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 25, 3655, 25, 3042, 25, 3686, 25, 3042, 25, 3072, 25, 252, 252, 262, 216}
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.
| \(\displaystyle \int \tanh ^6(e+f x) \sqrt {a \sinh ^2(e+f x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (i e+i f x)^6 \left (-\sqrt {a-a \sin (i e+i f x)^2}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \sqrt {a-a \sin (i e+i f x)^2} \tan (i e+i f x)^6dx\) |
| \(\Big \downarrow \) 3655 |
| \(\displaystyle -\int -\sqrt {a \cosh ^2(e+f x)} \tanh ^6(e+f x)dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \tanh ^6(e+f x) \sqrt {a \cosh ^2(e+f x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sqrt {a \sin \left (i e+i f x+\frac {\pi }{2}\right )^2}}{\tan \left (i e+i f x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sqrt {a \sin \left (\frac {1}{2} (2 i e+\pi )+i f x\right )^2}}{\tan \left (\frac {1}{2} (2 i e+\pi )+i f x\right )^6}dx\) |
| \(\Big \downarrow \) 3686 |
| \(\displaystyle \text {sech}(e+f x) \left (-\sqrt {a \cosh ^2(e+f x)}\right ) \int -\sinh (e+f x) \tanh ^5(e+f x)dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \int \sinh (e+f x) \tanh ^5(e+f x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \int -\sin (i e+i f x) \tan (i e+i f x)^5dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \text {sech}(e+f x) \left (-\sqrt {a \cosh ^2(e+f x)}\right ) \int \sin (i e+i f x) \tan (i e+i f x)^5dx\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle -\frac {\text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \int -\frac {\sinh ^6(e+f x)}{\left (\sinh ^2(e+f x)+1\right )^3}d\sinh (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \int \frac {\sinh ^6(e+f x)}{\left (\sinh ^2(e+f x)+1\right )^3}d\sinh (e+f x)}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \left (\frac {\sinh ^5(e+f x)}{4 \left (\sinh ^2(e+f x)+1\right )^2}-\frac {5}{4} \int \frac {\sinh ^4(e+f x)}{\left (\sinh ^2(e+f x)+1\right )^2}d\sinh (e+f x)\right )}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \left (\frac {\sinh ^5(e+f x)}{4 \left (\sinh ^2(e+f x)+1\right )^2}-\frac {5}{4} \left (\frac {3}{2} \int \frac {\sinh ^2(e+f x)}{\sinh ^2(e+f x)+1}d\sinh (e+f x)-\frac {\sinh ^3(e+f x)}{2 \left (\sinh ^2(e+f x)+1\right )}\right )\right )}{f}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle -\frac {\text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \left (\frac {\sinh ^5(e+f x)}{4 \left (\sinh ^2(e+f x)+1\right )^2}-\frac {5}{4} \left (\frac {3}{2} \left (\sinh (e+f x)-\int \frac {1}{\sinh ^2(e+f x)+1}d\sinh (e+f x)\right )-\frac {\sinh ^3(e+f x)}{2 \left (\sinh ^2(e+f x)+1\right )}\right )\right )}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \left (\frac {\sinh ^5(e+f x)}{4 \left (\sinh ^2(e+f x)+1\right )^2}-\frac {5}{4} \left (\frac {3}{2} (\sinh (e+f x)-\arctan (\sinh (e+f x)))-\frac {\sinh ^3(e+f x)}{2 \left (\sinh ^2(e+f x)+1\right )}\right )\right )}{f}\) |
-((Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x]*(Sinh[e + f*x]^5/(4*(1 + Sinh[e + f*x]^2)^2) - (5*((3*(-ArcTan[Sinh[e + f*x]] + Sinh[e + f*x]))/2 - Sinh[e + f*x]^3/(2*(1 + Sinh[e + f*x]^2))))/4))/f)
3.5.31.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.41
| method | result | size |
| default | \(\frac {\sqrt {a \sinh \left (f x +e \right )^{2}}\, \left (15 \ln \left (\frac {2 \sqrt {-a}\, \sqrt {a \sinh \left (f x +e \right )^{2}}-2 a}{\cosh \left (f x +e \right )}\right ) a \cosh \left (f x +e \right )^{4}+8 \sqrt {a \sinh \left (f x +e \right )^{2}}\, \cosh \left (f x +e \right )^{4} \sqrt {-a}+9 \sqrt {a \sinh \left (f x +e \right )^{2}}\, \cosh \left (f x +e \right )^{2} \sqrt {-a}-2 \sqrt {-a}\, \sqrt {a \sinh \left (f x +e \right )^{2}}\right )}{8 \cosh \left (f x +e \right )^{3} \sqrt {-a}\, \sinh \left (f x +e \right ) \sqrt {a \cosh \left (f x +e \right )^{2}}\, f}\) | \(169\) |
| risch | \(\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{2 f x +2 e}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}-\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}+\frac {\left (9 \,{\mathrm e}^{6 f x +6 e}+{\mathrm e}^{4 f x +4 e}-{\mathrm e}^{2 f x +2 e}-9\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{2 f x +2 e}}{4 f \left ({\mathrm e}^{2 f x +2 e}+1\right )^{5}}+\frac {15 i \ln \left ({\mathrm e}^{f x}-i {\mathrm e}^{-e}\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{f x +e}}{8 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}-\frac {15 i \ln \left ({\mathrm e}^{f x}+i {\mathrm e}^{-e}\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{f x +e}}{8 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}\) | \(313\) |
1/8/cosh(f*x+e)^3*(a*sinh(f*x+e)^2)^(1/2)*(15*ln(2/cosh(f*x+e)*((-a)^(1/2) *(a*sinh(f*x+e)^2)^(1/2)-a))*a*cosh(f*x+e)^4+8*(a*sinh(f*x+e)^2)^(1/2)*cos h(f*x+e)^4*(-a)^(1/2)+9*(a*sinh(f*x+e)^2)^(1/2)*cosh(f*x+e)^2*(-a)^(1/2)-2 *(-a)^(1/2)*(a*sinh(f*x+e)^2)^(1/2))/(-a)^(1/2)/sinh(f*x+e)/(a*cosh(f*x+e) ^2)^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 1645 vs. \(2 (104) = 208\).
Time = 0.27 (sec) , antiderivative size = 1645, normalized size of antiderivative = 13.71 \[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=\text {Too large to display} \]
1/4*(20*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^9 + 2*e^(f*x + e)*sinh(f*x + e)^10 + 15*(6*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^8 + 120*(2 *cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^7 + 5*(84*cosh (f*x + e)^4 + 84*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^6 + 6*(84* cosh(f*x + e)^5 + 140*cosh(f*x + e)^3 + 5*cosh(f*x + e))*e^(f*x + e)*sinh( f*x + e)^5 + 5*(84*cosh(f*x + e)^6 + 210*cosh(f*x + e)^4 + 15*cosh(f*x + e )^2 - 1)*e^(f*x + e)*sinh(f*x + e)^4 + 20*(12*cosh(f*x + e)^7 + 42*cosh(f* x + e)^5 + 5*cosh(f*x + e)^3 - cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 15*(6*cosh(f*x + e)^8 + 28*cosh(f*x + e)^6 + 5*cosh(f*x + e)^4 - 2*cosh( f*x + e)^2 - 1)*e^(f*x + e)*sinh(f*x + e)^2 + 10*(2*cosh(f*x + e)^9 + 12*c osh(f*x + e)^7 + 3*cosh(f*x + e)^5 - 2*cosh(f*x + e)^3 - 3*cosh(f*x + e))* e^(f*x + e)*sinh(f*x + e) - 15*(9*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^ 8 + e^(f*x + e)*sinh(f*x + e)^9 + 4*(9*cosh(f*x + e)^2 + 1)*e^(f*x + e)*si nh(f*x + e)^7 + 28*(3*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f* x + e)^6 + 6*(21*cosh(f*x + e)^4 + 14*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sin h(f*x + e)^5 + 2*(63*cosh(f*x + e)^5 + 70*cosh(f*x + e)^3 + 15*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^4 + 4*(21*cosh(f*x + e)^6 + 35*cosh(f*x + e) ^4 + 15*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 12*(3*cosh(f*x + e)^7 + 7*cosh(f*x + e)^5 + 5*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e )*sinh(f*x + e)^2 + (9*cosh(f*x + e)^8 + 28*cosh(f*x + e)^6 + 30*cosh(f...
\[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=\int \sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )} \tanh ^{6}{\left (e + f x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 891 vs. \(2 (104) = 208\).
Time = 0.32 (sec) , antiderivative size = 891, normalized size of antiderivative = 7.42 \[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=\text {Too large to display} \]
315/128*sqrt(a)*arctan(e^(-f*x - e))/f + 1/128*(105*sqrt(a)*arctan(e^(-f*x - e)) + (279*sqrt(a)*e^(-f*x - e) + 511*sqrt(a)*e^(-3*f*x - 3*e) + 385*sq rt(a)*e^(-5*f*x - 5*e) + 105*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f + 1/ 128*(105*sqrt(a)*arctan(e^(-f*x - e)) - (105*sqrt(a)*e^(-f*x - e) + 385*sq rt(a)*e^(-3*f*x - 3*e) + 511*sqrt(a)*e^(-5*f*x - 5*e) + 279*sqrt(a)*e^(-7* f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f - 5/256*(15*sqrt(a)*arctan(e^(-f*x - e)) - (15* sqrt(a)*e^(-f*x - e) + 55*sqrt(a)*e^(-3*f*x - 3*e) + 73*sqrt(a)*e^(-5*f*x - 5*e) - 15*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f - 5/256*(15*sqrt(a)*a rctan(e^(-f*x - e)) - (15*sqrt(a)*e^(-f*x - e) - 73*sqrt(a)*e^(-3*f*x - 3* e) - 55*sqrt(a)*e^(-5*f*x - 5*e) - 15*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f *x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1 ))/f + 5/64*(3*sqrt(a)*arctan(e^(-f*x - e)) - (3*sqrt(a)*e^(-f*x - e) + 11 *sqrt(a)*e^(-3*f*x - 3*e) - 11*sqrt(a)*e^(-5*f*x - 5*e) - 3*sqrt(a)*e^(-7* f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f + 1/256*(837*sqrt(a)*e^(-2*f*x - 2*e) + 1533*sq rt(a)*e^(-4*f*x - 4*e) + 1155*sqrt(a)*e^(-6*f*x - 6*e) + 315*sqrt(a)*e^(-8 *f*x - 8*e) + 128*sqrt(a))/(f*(e^(-f*x - e) + 4*e^(-3*f*x - 3*e) + 6*e^...
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03 \[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=-\frac {{\left (15 \, \pi - \frac {4 \, {\left (9 \, {\left (e^{\left (f x + e\right )} - e^{\left (-f x - e\right )}\right )}^{3} + 28 \, e^{\left (f x + e\right )} - 28 \, e^{\left (-f x - e\right )}\right )}}{{\left ({\left (e^{\left (f x + e\right )} - e^{\left (-f x - e\right )}\right )}^{2} + 4\right )}^{2}} + 30 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, f x + 2 \, e\right )} - 1\right )} e^{\left (-f x - e\right )}\right ) - 8 \, e^{\left (f x + e\right )} + 8 \, e^{\left (-f x - e\right )}\right )} \sqrt {a}}{16 \, f} \]
-1/16*(15*pi - 4*(9*(e^(f*x + e) - e^(-f*x - e))^3 + 28*e^(f*x + e) - 28*e ^(-f*x - e))/((e^(f*x + e) - e^(-f*x - e))^2 + 4)^2 + 30*arctan(1/2*(e^(2* f*x + 2*e) - 1)*e^(-f*x - e)) - 8*e^(f*x + e) + 8*e^(-f*x - e))*sqrt(a)/f
Timed out. \[ \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx=\int {\mathrm {tanh}\left (e+f\,x\right )}^6\,\sqrt {a\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a} \,d x \]