Integrand size = 25, antiderivative size = 122 \[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1+m}{2},-p,\frac {3+m}{2},-\sinh ^2(e+f x),-\frac {b \sinh ^2(e+f x)}{a}\right ) \cosh ^2(e+f x)^{\frac {1+m}{2}} \left (a+b \sinh ^2(e+f x)\right )^p \left (1+\frac {b \sinh ^2(e+f x)}{a}\right )^{-p} (d \tanh (e+f x))^{1+m}}{d f (1+m)} \]
AppellF1(1/2+1/2*m,1/2+1/2*m,-p,3/2+1/2*m,-sinh(f*x+e)^2,-b*sinh(f*x+e)^2/ a)*(cosh(f*x+e)^2)^(1/2+1/2*m)*(a+b*sinh(f*x+e)^2)^p*(d*tanh(f*x+e))^(1+m) /d/f/(1+m)/((1+b*sinh(f*x+e)^2/a)^p)
\[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx \]
Time = 0.39 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3676, 393, 152, 150}
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.
\(\displaystyle \int (d \tanh (e+f x))^m \left (a+b \sinh ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i d \tan (i e+i f x))^m \left (a-b \sin (i e+i f x)^2\right )^pdx\) |
\(\Big \downarrow \) 3676 |
\(\displaystyle \frac {i (i \sinh (e+f x))^{-m-1} \cosh ^2(e+f x)^{\frac {m+1}{2}} (d \tanh (e+f x))^{m+1} \int (i \sinh (e+f x))^m \left (\sinh ^2(e+f x)+1\right )^{\frac {1}{2} (-m-1)} \left (b \sinh ^2(e+f x)+a\right )^pd\sinh (e+f x)}{d f}\) |
\(\Big \downarrow \) 393 |
\(\displaystyle \frac {\sinh ^2(e+f x)^{\frac {1-m}{2}-1} \cosh ^2(e+f x)^{\frac {m+1}{2}} (d \tanh (e+f x))^{m+1} \int \sinh ^2(e+f x)^{\frac {m-1}{2}} \left (\sinh ^2(e+f x)+1\right )^{\frac {1}{2} (-m-1)} \left (b \sinh ^2(e+f x)+a\right )^pd\sinh ^2(e+f x)}{2 d f}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle \frac {\sinh ^2(e+f x)^{\frac {1-m}{2}-1} \cosh ^2(e+f x)^{\frac {m+1}{2}} (d \tanh (e+f x))^{m+1} \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac {b \sinh ^2(e+f x)}{a}+1\right )^{-p} \int \sinh ^2(e+f x)^{\frac {m-1}{2}} \left (\sinh ^2(e+f x)+1\right )^{\frac {1}{2} (-m-1)} \left (\frac {b \sinh ^2(e+f x)}{a}+1\right )^pd\sinh ^2(e+f x)}{2 d f}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {\sinh ^2(e+f x)^{\frac {1-m}{2}+\frac {m+1}{2}-1} \cosh ^2(e+f x)^{\frac {m+1}{2}} (d \tanh (e+f x))^{m+1} \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac {b \sinh ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {m+1}{2},\frac {m+1}{2},-p,\frac {m+3}{2},-\sinh ^2(e+f x),-\frac {b \sinh ^2(e+f x)}{a}\right )}{d f (m+1)}\) |
(AppellF1[(1 + m)/2, (1 + m)/2, -p, (3 + m)/2, -Sinh[e + f*x]^2, -((b*Sinh [e + f*x]^2)/a)]*(Cosh[e + f*x]^2)^((1 + m)/2)*(Sinh[e + f*x]^2)^(-1 + (1 - m)/2 + (1 + m)/2)*(a + b*Sinh[e + f*x]^2)^p*(d*Tanh[e + f*x])^(1 + m))/( d*f*(1 + m)*(1 + (b*Sinh[e + f*x]^2)/a)^p)
3.6.12.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Simp[(e*x)^m/(2*x*(x^2)^(Simplify[(m + 1)/2] - 1)) Subs t[Int[x^(Simplify[(m + 1)/2] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ[Simp lify[m + 2*p]] && !IntegerQ[m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*((d_.)*tan[(e_.) + (f_.) *(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[f f*(d*Tan[e + f*x])^(m + 1)*((Cos[e + f*x]^2)^((m + 1)/2)/(d*f*Sin[e + f*x]^ (m + 1))) Subst[Int[(ff*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, d, e, f, m, p}, x] && !I ntegerQ[m]
\[\int \left (a +b \sinh \left (f x +e \right )^{2}\right )^{p} \left (d \tanh \left (f x +e \right )\right )^{m}d x\]
\[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tanh \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\text {Timed out} \]
\[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tanh \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tanh \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int \left (a+b \sinh ^2(e+f x)\right )^p (d \tanh (e+f x))^m \, dx=\int {\left (d\,\mathrm {tanh}\left (e+f\,x\right )\right )}^m\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]