Integrand size = 25, antiderivative size = 163 \[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{2 (a-b)^{7/2} f}+\frac {2 a+3 b}{6 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\text {sech}^2(e+f x)}{2 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {2 a+3 b}{2 (a-b)^3 f \sqrt {a+b \sinh ^2(e+f x)}} \]
-1/2*(2*a+3*b)*arctanh((a+b*sinh(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(7/2)/ f+1/6*(2*a+3*b)/(a-b)^2/f/(a+b*sinh(f*x+e)^2)^(3/2)+1/2*sech(f*x+e)^2/(a-b )/f/(a+b*sinh(f*x+e)^2)^(3/2)+1/2*(2*a+3*b)/(a-b)^3/f/(a+b*sinh(f*x+e)^2)^ (1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.50 \[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a+3 b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sinh ^2(e+f x)}{a-b}\right )+3 (a-b) \text {sech}^2(e+f x)}{6 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]
((2*a + 3*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sinh[e + f*x]^2)/(a - b)] + 3*(a - b)*Sech[e + f*x]^2)/(6*(a - b)^2*f*(a + b*Sinh[e + f*x]^2)^( 3/2))
Time = 0.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 26, 3673, 87, 61, 61, 73, 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.
\(\displaystyle \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \tan (i e+i f x)^3}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\tan (i e+i f x)^3}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\sinh ^2(e+f x)}{\left (\sinh ^2(e+f x)+1\right )^2 \left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\frac {(2 a+3 b) \int \frac {1}{\left (\sinh ^2(e+f x)+1\right ) \left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh ^2(e+f x)}{2 (a-b)}+\frac {1}{(a-b) \left (\sinh ^2(e+f x)+1\right ) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (\frac {\int \frac {1}{\left (\sinh ^2(e+f x)+1\right ) \left (b \sinh ^2(e+f x)+a\right )^{3/2}}d\sinh ^2(e+f x)}{a-b}+\frac {2}{3 (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 (a-b)}+\frac {1}{(a-b) \left (\sinh ^2(e+f x)+1\right ) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (\frac {\frac {\int \frac {1}{\left (\sinh ^2(e+f x)+1\right ) \sqrt {b \sinh ^2(e+f x)+a}}d\sinh ^2(e+f x)}{a-b}+\frac {2}{(a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{a-b}+\frac {2}{3 (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 (a-b)}+\frac {1}{(a-b) \left (\sinh ^2(e+f x)+1\right ) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (\frac {\frac {2 \int \frac {1}{\frac {\sinh ^4(e+f x)}{b}-\frac {a}{b}+1}d\sqrt {b \sinh ^2(e+f x)+a}}{b (a-b)}+\frac {2}{(a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{a-b}+\frac {2}{3 (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 (a-b)}+\frac {1}{(a-b) \left (\sinh ^2(e+f x)+1\right ) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (\frac {\frac {2}{(a-b) \sqrt {a+b \sinh ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}}{a-b}+\frac {2}{3 (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 (a-b)}+\frac {1}{(a-b) \left (\sinh ^2(e+f x)+1\right ) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
(1/((a - b)*(1 + Sinh[e + f*x]^2)*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((2*a + 3*b)*(2/(3*(a - b)*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((-2*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(a - b)^(3/2) + 2/((a - b)*Sqrt[a + b*Si nh[e + f*x]^2]))/(a - b)))/(2*(a - b)))/(2*f)
3.6.2.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.78 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.31
method | result | size |
default | \(\frac {\operatorname {`\,int/indef0`\,}\left (-\frac {\sinh \left (f x +e \right )^{3} \left (b^{2} \sinh \left (f x +e \right )^{4}+2 \sinh \left (f x +e \right )^{2} a b +a^{2}\right ) \cosh \left (f x +e \right )^{2}}{\left (-b^{4} \cosh \left (f x +e \right )^{14}+\left (-4 a \,b^{3}+4 b^{4}\right ) \cosh \left (f x +e \right )^{12}+\left (-6 a^{2} b^{2}+12 a \,b^{3}-6 b^{4}\right ) \cosh \left (f x +e \right )^{10}+\left (-4 a^{3} b +12 a^{2} b^{2}-12 a \,b^{3}+4 b^{4}\right ) \cosh \left (f x +e \right )^{8}+\left (-a^{4}+4 a^{3} b -6 a^{2} b^{2}+4 a \,b^{3}-b^{4}\right ) \cosh \left (f x +e \right )^{6}\right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}}, \sinh \left (f x +e \right )\right )}{f}\) | \(213\) |
risch | \(\text {Expression too large to display}\) | \(309501\) |
`int/indef0`(-sinh(f*x+e)^3*(b^2*sinh(f*x+e)^4+2*sinh(f*x+e)^2*a*b+a^2)*co sh(f*x+e)^2/(-b^4*cosh(f*x+e)^14+(-4*a*b^3+4*b^4)*cosh(f*x+e)^12+(-6*a^2*b ^2+12*a*b^3-6*b^4)*cosh(f*x+e)^10+(-4*a^3*b+12*a^2*b^2-12*a*b^3+4*b^4)*cos h(f*x+e)^8+(-a^4+4*a^3*b-6*a^2*b^2+4*a*b^3-b^4)*cosh(f*x+e)^6)/(a+b*sinh(f *x+e)^2)^(1/2),sinh(f*x+e))/f
Leaf count of result is larger than twice the leaf count of optimal. 5155 vs. \(2 (143) = 286\).
Time = 0.75 (sec) , antiderivative size = 10506, normalized size of antiderivative = 64.45 \[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{3}{\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (f x + e\right )^{3}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1981 vs. \(2 (143) = 286\).
Time = 8.26 (sec) , antiderivative size = 1981, normalized size of antiderivative = 12.15 \[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
2/3*((3*(a^18*b^3*e^(21*e) - 12*a^17*b^4*e^(21*e) + 65*a^16*b^5*e^(21*e) - 208*a^15*b^6*e^(21*e) + 429*a^14*b^7*e^(21*e) - 572*a^13*b^8*e^(21*e) + 4 29*a^12*b^9*e^(21*e) - 429*a^10*b^11*e^(21*e) + 572*a^9*b^12*e^(21*e) - 42 9*a^8*b^13*e^(21*e) + 208*a^7*b^14*e^(21*e) - 65*a^6*b^15*e^(21*e) + 12*a^ 5*b^16*e^(21*e) - a^4*b^17*e^(21*e))*e^(2*f*x)/(a^20*b^2*e^(16*e) - 16*a^1 9*b^3*e^(16*e) + 120*a^18*b^4*e^(16*e) - 560*a^17*b^5*e^(16*e) + 1820*a^16 *b^6*e^(16*e) - 4368*a^15*b^7*e^(16*e) + 8008*a^14*b^8*e^(16*e) - 11440*a^ 13*b^9*e^(16*e) + 12870*a^12*b^10*e^(16*e) - 11440*a^11*b^11*e^(16*e) + 80 08*a^10*b^12*e^(16*e) - 4368*a^9*b^13*e^(16*e) + 1820*a^8*b^14*e^(16*e) - 560*a^7*b^15*e^(16*e) + 120*a^6*b^16*e^(16*e) - 16*a^5*b^17*e^(16*e) + a^4 *b^18*e^(16*e)) + 2*(8*a^19*b^2*e^(19*e) - 103*a^18*b^3*e^(19*e) + 608*a^1 7*b^4*e^(19*e) - 2171*a^16*b^5*e^(19*e) + 5200*a^15*b^6*e^(19*e) - 8723*a^ 14*b^7*e^(19*e) + 10296*a^13*b^8*e^(19*e) - 8151*a^12*b^9*e^(19*e) + 3432* a^11*b^10*e^(19*e) + 715*a^10*b^11*e^(19*e) - 2288*a^9*b^12*e^(19*e) + 180 7*a^8*b^13*e^(19*e) - 832*a^7*b^14*e^(19*e) + 239*a^6*b^15*e^(19*e) - 40*a ^5*b^16*e^(19*e) + 3*a^4*b^17*e^(19*e))/(a^20*b^2*e^(16*e) - 16*a^19*b^3*e ^(16*e) + 120*a^18*b^4*e^(16*e) - 560*a^17*b^5*e^(16*e) + 1820*a^16*b^6*e^ (16*e) - 4368*a^15*b^7*e^(16*e) + 8008*a^14*b^8*e^(16*e) - 11440*a^13*b^9* e^(16*e) + 12870*a^12*b^10*e^(16*e) - 11440*a^11*b^11*e^(16*e) + 8008*a^10 *b^12*e^(16*e) - 4368*a^9*b^13*e^(16*e) + 1820*a^8*b^14*e^(16*e) - 560*...
Timed out. \[ \int \frac {\tanh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tanh}\left (e+f\,x\right )}^3}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]