Integrand size = 16, antiderivative size = 36 \[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\frac {(c x)^{1+m} \operatorname {AppellF1}\left (1+m,\frac {1}{4},-\frac {1}{4},2+m,a x,-a x\right )}{c (1+m)} \] Output:
(c*x)^(1+m)*AppellF1(1+m,1/4,-1/4,2+m,a*x,-a*x)/c/(1+m)
\[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx \] Input:
Integrate[E^(ArcTanh[a*x]/2)*(c*x)^m,x]
Output:
Integrate[E^(ArcTanh[a*x]/2)*(c*x)^m, x]
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6676, 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.
\(\displaystyle \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx\) |
\(\Big \downarrow \) 6676 |
\(\displaystyle \int \frac {\sqrt [4]{a x+1} (c x)^m}{\sqrt [4]{1-a x}}dx\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {(c x)^{m+1} \operatorname {AppellF1}\left (m+1,\frac {1}{4},-\frac {1}{4},m+2,a x,-a x\right )}{c (m+1)}\) |
Input:
Int[E^(ArcTanh[a*x]/2)*(c*x)^m,x]
Output:
((c*x)^(1 + m)*AppellF1[1 + m, 1/4, -1/4, 2 + m, a*x, -(a*x)])/(c*(1 + m))
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[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x) ^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c, m, n}, x] && !Int egerQ[(n - 1)/2]
\[\int \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, \left (x c \right )^{m}d x\]
Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(x*c)^m,x)
Output:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(x*c)^m,x)
\[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\int { \left (c x\right )^{m} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(c*x)^m,x, algorithm="fricas" )
Output:
integral((c*x)^m*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)), x)
\[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\int \left (c x\right )^{m} \sqrt {\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}}\, dx \] Input:
integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(1/2)*(c*x)**m,x)
Output:
Integral((c*x)**m*sqrt((a*x + 1)/sqrt(-a**2*x**2 + 1)), x)
\[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\int { \left (c x\right )^{m} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(c*x)^m,x, algorithm="maxima" )
Output:
integrate((c*x)^m*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1)), x)
\[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\int { \left (c x\right )^{m} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(c*x)^m,x, algorithm="giac")
Output:
integrate((c*x)^m*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1)), x)
Timed out. \[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=\int {\left (c\,x\right )}^m\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}} \,d x \] Input:
int((c*x)^m*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2),x)
Output:
int((c*x)^m*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2), x)
\[ \int e^{\frac {1}{2} \text {arctanh}(a x)} (c x)^m \, dx=-c^{m} \left (\int \frac {x^{m} \sqrt {a x +1}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{4}}}{a^{2} x^{2}-1}d x \right ) \] Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(c*x)^m,x)
Output:
- c**m*int((x**m*sqrt(a*x + 1)*( - a**2*x**2 + 1)**(3/4))/(a**2*x**2 - 1) ,x)