3.956 \(\int e^{\tanh ^{-1}(a x)} (c-a^2 c x^2)^{7/2} \, dx\)

Optimal. Leaf size=183 \[ -\frac {c^3 (a x+1)^8 \sqrt {c-a^2 c x^2}}{8 a \sqrt {1-a^2 x^2}}+\frac {6 c^3 (a x+1)^7 \sqrt {c-a^2 c x^2}}{7 a \sqrt {1-a^2 x^2}}-\frac {2 c^3 (a x+1)^6 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}+\frac {8 c^3 (a x+1)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}} \]

[Out]

8/5*c^3*(a*x+1)^5*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-2*c^3*(a*x+1)^6*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1
)^(1/2)+6/7*c^3*(a*x+1)^7*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-1/8*c^3*(a*x+1)^8*(-a^2*c*x^2+c)^(1/2)/a/(
-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6143, 6140, 43} \[ -\frac {c^3 (a x+1)^8 \sqrt {c-a^2 c x^2}}{8 a \sqrt {1-a^2 x^2}}+\frac {6 c^3 (a x+1)^7 \sqrt {c-a^2 c x^2}}{7 a \sqrt {1-a^2 x^2}}-\frac {2 c^3 (a x+1)^6 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}+\frac {8 c^3 (a x+1)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(7/2),x]

[Out]

(8*c^3*(1 + a*x)^5*Sqrt[c - a^2*c*x^2])/(5*a*Sqrt[1 - a^2*x^2]) - (2*c^3*(1 + a*x)^6*Sqrt[c - a^2*c*x^2])/(a*S
qrt[1 - a^2*x^2]) + (6*c^3*(1 + a*x)^7*Sqrt[c - a^2*c*x^2])/(7*a*Sqrt[1 - a^2*x^2]) - (c^3*(1 + a*x)^8*Sqrt[c
- a^2*c*x^2])/(8*a*Sqrt[1 - a^2*x^2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6140

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

Rule 6143

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx &=\frac {\left (c^3 \sqrt {c-a^2 c x^2}\right ) \int e^{\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{7/2} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\left (c^3 \sqrt {c-a^2 c x^2}\right ) \int (1-a x)^3 (1+a x)^4 \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\left (c^3 \sqrt {c-a^2 c x^2}\right ) \int \left (8 (1+a x)^4-12 (1+a x)^5+6 (1+a x)^6-(1+a x)^7\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {8 c^3 (1+a x)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}}-\frac {2 c^3 (1+a x)^6 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}+\frac {6 c^3 (1+a x)^7 \sqrt {c-a^2 c x^2}}{7 a \sqrt {1-a^2 x^2}}-\frac {c^3 (1+a x)^8 \sqrt {c-a^2 c x^2}}{8 a \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 68, normalized size = 0.37 \[ -\frac {c^3 (a x+1)^5 \left (35 a^3 x^3-135 a^2 x^2+185 a x-93\right ) \sqrt {c-a^2 c x^2}}{280 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(7/2),x]

[Out]

-1/280*(c^3*(1 + a*x)^5*Sqrt[c - a^2*c*x^2]*(-93 + 185*a*x - 135*a^2*x^2 + 35*a^3*x^3))/(a*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.69, size = 120, normalized size = 0.66 \[ \frac {{\left (35 \, a^{7} c^{3} x^{8} + 40 \, a^{6} c^{3} x^{7} - 140 \, a^{5} c^{3} x^{6} - 168 \, a^{4} c^{3} x^{5} + 210 \, a^{3} c^{3} x^{4} + 280 \, a^{2} c^{3} x^{3} - 140 \, a c^{3} x^{2} - 280 \, c^{3} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{280 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

1/280*(35*a^7*c^3*x^8 + 40*a^6*c^3*x^7 - 140*a^5*c^3*x^6 - 168*a^4*c^3*x^5 + 210*a^3*c^3*x^4 + 280*a^2*c^3*x^3
- 140*a*c^3*x^2 - 280*c^3*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (a x + 1\right )}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(7/2)*(a*x + 1)/sqrt(-a^2*x^2 + 1), x)

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maple [A]  time = 0.03, size = 97, normalized size = 0.53 \[ \frac {x \left (35 a^{7} x^{7}+40 x^{6} a^{6}-140 x^{5} a^{5}-168 x^{4} a^{4}+210 x^{3} a^{3}+280 a^{2} x^{2}-140 a x -280\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{280 \left (a x -1\right )^{3} \left (a x +1\right )^{3} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x)

[Out]

1/280*x*(35*a^7*x^7+40*a^6*x^6-140*a^5*x^5-168*a^4*x^4+210*a^3*x^3+280*a^2*x^2-140*a*x-280)*(-a^2*c*x^2+c)^(7/
2)/(a*x-1)^3/(a*x+1)^3/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.41, size = 154, normalized size = 0.84 \[ -\frac {1}{7} \, a^{6} c^{\frac {7}{2}} x^{7} + \frac {3}{5} \, a^{4} c^{\frac {7}{2}} x^{5} - a^{2} c^{\frac {7}{2}} x^{3} + c^{\frac {7}{2}} x + \frac {1}{8} \, {\left (\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{4} c^{3} x^{6} - 3 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2} c^{3} x^{4} + 3 \, a^{2} c^{\frac {7}{2}} x^{4} + 6 \, c^{\frac {7}{2}} x^{2} - \frac {10 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} c^{3}}{a^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

-1/7*a^6*c^(7/2)*x^7 + 3/5*a^4*c^(7/2)*x^5 - a^2*c^(7/2)*x^3 + c^(7/2)*x + 1/8*(sqrt(a^4*c*x^4 - 2*a^2*c*x^2 +
c)*a^4*c^3*x^6 - 3*sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*a^2*c^3*x^4 + 3*a^2*c^(7/2)*x^4 + 6*c^(7/2)*x^2 - 10*sqr
t(a^4*c*x^4 - 2*a^2*c*x^2 + c)*c^3/a^2)*a

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mupad [B]  time = 1.06, size = 107, normalized size = 0.58 \[ \frac {\sqrt {c-a^2\,c\,x^2}\,\left (-\frac {a^7\,c^3\,x^8}{8}-\frac {a^6\,c^3\,x^7}{7}+\frac {a^5\,c^3\,x^6}{2}+\frac {3\,a^4\,c^3\,x^5}{5}-\frac {3\,a^3\,c^3\,x^4}{4}-a^2\,c^3\,x^3+\frac {a\,c^3\,x^2}{2}+c^3\,x\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a^2*c*x^2)^(7/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)

[Out]

((c - a^2*c*x^2)^(1/2)*(c^3*x + (a*c^3*x^2)/2 - a^2*c^3*x^3 - (3*a^3*c^3*x^4)/4 + (3*a^4*c^3*x^5)/5 + (a^5*c^3
*x^6)/2 - (a^6*c^3*x^7)/7 - (a^7*c^3*x^8)/8))/(1 - a^2*x^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a**2*c*x**2+c)**(7/2),x)

[Out]

Integral((-c*(a*x - 1)*(a*x + 1))**(7/2)*(a*x + 1)/sqrt(-(a*x - 1)*(a*x + 1)), x)

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