3.1174 \(\int e^{3 \tanh ^{-1}(a x)} x^m (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=251 \[ \frac {(4 m+2 p+3) x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+1}{2},\frac {3}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{(m+1) (m+2 p)}+\frac {a (4 m+6 p+5) x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+2}{2},\frac {3}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{(m+2) (m+2 p+1)}-\frac {3 x^{m+1} \left (c-a^2 c x^2\right )^p}{\sqrt {1-a^2 x^2} (m+2 p)}-\frac {a x^{m+2} \left (c-a^2 c x^2\right )^p}{\sqrt {1-a^2 x^2} (m+2 p+1)} \]

[Out]

(3+4*m+2*p)*x^(1+m)*(-a^2*c*x^2+c)^p*hypergeom([3/2-p, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)/(m+2*p)/((-a^2*x^
2+1)^p)+a*(5+4*m+6*p)*x^(2+m)*(-a^2*c*x^2+c)^p*hypergeom([1+1/2*m, 3/2-p],[2+1/2*m],a^2*x^2)/(2+m)/(1+m+2*p)/(
(-a^2*x^2+1)^p)-3*x^(1+m)*(-a^2*c*x^2+c)^p/(m+2*p)/(-a^2*x^2+1)^(1/2)-a*x^(2+m)*(-a^2*c*x^2+c)^p/(1+m+2*p)/(-a
^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6153, 6148, 1809, 808, 364} \[ \frac {(4 m+2 p+3) x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+1}{2},\frac {3}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{(m+1) (m+2 p)}+\frac {a (4 m+6 p+5) x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+2}{2},\frac {3}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{(m+2) (m+2 p+1)}-\frac {3 x^{m+1} \left (c-a^2 c x^2\right )^p}{\sqrt {1-a^2 x^2} (m+2 p)}-\frac {a x^{m+2} \left (c-a^2 c x^2\right )^p}{\sqrt {1-a^2 x^2} (m+2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*x^m*(c - a^2*c*x^2)^p,x]

[Out]

(-3*x^(1 + m)*(c - a^2*c*x^2)^p)/((m + 2*p)*Sqrt[1 - a^2*x^2]) - (a*x^(2 + m)*(c - a^2*c*x^2)^p)/((1 + m + 2*p
)*Sqrt[1 - a^2*x^2]) + ((3 + 4*m + 2*p)*x^(1 + m)*(c - a^2*c*x^2)^p*Hypergeometric2F1[(1 + m)/2, 3/2 - p, (3 +
m)/2, a^2*x^2])/((1 + m)*(m + 2*p)*(1 - a^2*x^2)^p) + (a*(5 + 4*m + 6*p)*x^(2 + m)*(c - a^2*c*x^2)^p*Hypergeo
metric2F1[(2 + m)/2, 3/2 - p, (4 + m)/2, a^2*x^2])/((2 + m)*(1 + m + 2*p)*(1 - a^2*x^2)^p)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6148

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{3 \tanh ^{-1}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int e^{3 \tanh ^{-1}(a x)} x^m \left (1-a^2 x^2\right )^p \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^m (1+a x)^3 \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \, dx\\ &=-\frac {a x^{2+m} \left (c-a^2 c x^2\right )^p}{(1+m+2 p) \sqrt {1-a^2 x^2}}-\frac {\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^m \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \left (-a^2 (1+m+2 p)-a^3 (5+4 m+6 p) x-3 a^4 (1+m+2 p) x^2\right ) \, dx}{a^2 (1+m+2 p)}\\ &=-\frac {3 x^{1+m} \left (c-a^2 c x^2\right )^p}{(m+2 p) \sqrt {1-a^2 x^2}}-\frac {a x^{2+m} \left (c-a^2 c x^2\right )^p}{(1+m+2 p) \sqrt {1-a^2 x^2}}+\frac {\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^m \left (a^4 (1+m+2 p) (3+4 m+2 p)+a^5 (m+2 p) (5+4 m+6 p) x\right ) \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \, dx}{a^4 (m+2 p) (1+m+2 p)}\\ &=-\frac {3 x^{1+m} \left (c-a^2 c x^2\right )^p}{(m+2 p) \sqrt {1-a^2 x^2}}-\frac {a x^{2+m} \left (c-a^2 c x^2\right )^p}{(1+m+2 p) \sqrt {1-a^2 x^2}}+\frac {\left ((3+4 m+2 p) \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^m \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \, dx}{m+2 p}+\frac {\left (a (5+4 m+6 p) \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^{1+m} \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \, dx}{1+m+2 p}\\ &=-\frac {3 x^{1+m} \left (c-a^2 c x^2\right )^p}{(m+2 p) \sqrt {1-a^2 x^2}}-\frac {a x^{2+m} \left (c-a^2 c x^2\right )^p}{(1+m+2 p) \sqrt {1-a^2 x^2}}+\frac {(3+4 m+2 p) x^{1+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {1+m}{2},\frac {3}{2}-p;\frac {3+m}{2};a^2 x^2\right )}{(1+m) (m+2 p)}+\frac {a (5+4 m+6 p) x^{2+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {2+m}{2},\frac {3}{2}-p;\frac {4+m}{2};a^2 x^2\right )}{(2+m) (1+m+2 p)}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 186, normalized size = 0.74 \[ x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\, _2F_1\left (\frac {m+1}{2},\frac {3}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{m+1}+a x \left (\frac {3 \, _2F_1\left (\frac {m+2}{2},\frac {3}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{m+2}+a x \left (\frac {3 \, _2F_1\left (\frac {m+3}{2},\frac {3}{2}-p;\frac {m+5}{2};a^2 x^2\right )}{m+3}+\frac {a x \, _2F_1\left (\frac {m+4}{2},\frac {3}{2}-p;\frac {m+6}{2};a^2 x^2\right )}{m+4}\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a*x])*x^m*(c - a^2*c*x^2)^p,x]

[Out]

(x^(1 + m)*(c - a^2*c*x^2)^p*(Hypergeometric2F1[(1 + m)/2, 3/2 - p, (3 + m)/2, a^2*x^2]/(1 + m) + a*x*((3*Hype
rgeometric2F1[(2 + m)/2, 3/2 - p, (4 + m)/2, a^2*x^2])/(2 + m) + a*x*((3*Hypergeometric2F1[(3 + m)/2, 3/2 - p,
(5 + m)/2, a^2*x^2])/(3 + m) + (a*x*Hypergeometric2F1[(4 + m)/2, 3/2 - p, (6 + m)/2, a^2*x^2])/(4 + m)))))/(1
- a^2*x^2)^p

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x + 1\right )} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{a^{2} x^{2} - 2 \, a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m*(-a^2*c*x^2+c)^p,x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*(a*x + 1)*(-a^2*c*x^2 + c)^p*x^m/(a^2*x^2 - 2*a*x + 1), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m*(-a^2*c*x^2+c)^p,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right )^{3} x^{m} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m*(-a^2*c*x^2+c)^p,x)

[Out]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m*(-a^2*c*x^2+c)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m*(-a^2*c*x^2+c)^p,x, algorithm="maxima")

[Out]

integrate((a*x + 1)^3*(-a^2*c*x^2 + c)^p*x^m/(-a^2*x^2 + 1)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^m\,{\left (c-a^2\,c\,x^2\right )}^p\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(c - a^2*c*x^2)^p*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)

[Out]

int((x^m*(c - a^2*c*x^2)^p*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*x**m*(-a**2*c*x**2+c)**p,x)

[Out]

Timed out

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