Integrand size = 23, antiderivative size = 103 \[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{2},-p,\frac {5}{2},-\sinh ^2(c+d x),-\frac {b \sinh ^2(c+d x)}{a}\right ) \sqrt {\cosh ^2(c+d x)} \sinh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (1+\frac {b \sinh ^2(c+d x)}{a}\right )^{-p} \tanh (c+d x)}{3 d} \]
1/3*AppellF1(3/2,3/2,-p,5/2,-sinh(d*x+c)^2,-b*sinh(d*x+c)^2/a)*sinh(d*x+c) ^2*(a+b*sinh(d*x+c)^2)^p*(cosh(d*x+c)^2)^(1/2)*tanh(d*x+c)/d/((1+b*sinh(d* x+c)^2/a)^p)
\[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx \]
Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 25, 3675, 395, 394}
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.
\(\displaystyle \int \tanh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (i c+i d x)^2 \left (-\left (a-b \sin (i c+i d x)^2\right )^p\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \left (a-b \sin (i c+i d x)^2\right )^p \tan (i c+i d x)^2dx\) |
\(\Big \downarrow \) 3675 |
\(\displaystyle \frac {\sqrt {\cosh ^2(c+d x)} \text {sech}(c+d x) \int \frac {\sinh ^2(c+d x) \left (b \sinh ^2(c+d x)+a\right )^p}{\left (\sinh ^2(c+d x)+1\right )^{3/2}}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\sqrt {\cosh ^2(c+d x)} \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (\frac {b \sinh ^2(c+d x)}{a}+1\right )^{-p} \int \frac {\sinh ^2(c+d x) \left (\frac {b \sinh ^2(c+d x)}{a}+1\right )^p}{\left (\sinh ^2(c+d x)+1\right )^{3/2}}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {\sinh ^2(c+d x) \sqrt {\cosh ^2(c+d x)} \tanh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (\frac {b \sinh ^2(c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{2},-p,\frac {5}{2},-\sinh ^2(c+d x),-\frac {b \sinh ^2(c+d x)}{a}\right )}{3 d}\) |
(AppellF1[3/2, 3/2, -p, 5/2, -Sinh[c + d*x]^2, -((b*Sinh[c + d*x]^2)/a)]*S qrt[Cosh[c + d*x]^2]*Sinh[c + d*x]^2*(a + b*Sinh[c + d*x]^2)^p*Tanh[c + d* x])/(3*d*(1 + (b*Sinh[c + d*x]^2)/a)^p)
3.6.18.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff^(m + 1 )*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])) Subst[Int[x^m*((a + b*ff^2*x^2) ^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b , e, f, p}, x] && IntegerQ[m/2] && !IntegerQ[p]
\[\int \left (a +b \sinh \left (d x +c \right )^{2}\right )^{p} \tanh \left (d x +c \right )^{2}d x\]
\[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\int { {\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\text {Timed out} \]
\[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\int { {\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{2} \,d x } \]
\[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\int { {\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx=\int {\mathrm {tanh}\left (c+d\,x\right )}^2\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^p \,d x \]