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

Optimal. Leaf size=141 \[ \frac {256 c^5 \left (1-a^2 x^2\right )^{3/2}}{315 a (c-a c x)^{3/2}}+\frac {64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a \sqrt {c-a c x}}+\frac {8 c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt {c-a c x}}{21 a}+\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2} (c-a c x)^{3/2}}{9 a} \]

[Out]

256/315*c^5*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(3/2)+2/9*c^2*(-a*c*x+c)^(3/2)*(-a^2*x^2+1)^(3/2)/a+64/105*c^4*(-a
^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(1/2)+8/21*c^3*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)/a

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Rubi [A]  time = 0.10, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6127, 657, 649} \[ \frac {256 c^5 \left (1-a^2 x^2\right )^{3/2}}{315 a (c-a c x)^{3/2}}+\frac {64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a \sqrt {c-a c x}}+\frac {8 c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt {c-a c x}}{21 a}+\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2} (c-a c x)^{3/2}}{9 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(256*c^5*(1 - a^2*x^2)^(3/2))/(315*a*(c - a*c*x)^(3/2)) + (64*c^4*(1 - a^2*x^2)^(3/2))/(105*a*Sqrt[c - a*c*x])
+ (8*c^3*Sqrt[c - a*c*x]*(1 - a^2*x^2)^(3/2))/(21*a) + (2*c^2*(c - a*c*x)^(3/2)*(1 - a^2*x^2)^(3/2))/(9*a)

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 6127

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} (c-a c x)^{7/2} \, dx &=c \int (c-a c x)^{5/2} \sqrt {1-a^2 x^2} \, dx\\ &=\frac {2 c^2 (c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}}{9 a}+\frac {1}{3} \left (4 c^2\right ) \int (c-a c x)^{3/2} \sqrt {1-a^2 x^2} \, dx\\ &=\frac {8 c^3 \sqrt {c-a c x} \left (1-a^2 x^2\right )^{3/2}}{21 a}+\frac {2 c^2 (c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}}{9 a}+\frac {1}{21} \left (32 c^3\right ) \int \sqrt {c-a c x} \sqrt {1-a^2 x^2} \, dx\\ &=\frac {64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a \sqrt {c-a c x}}+\frac {8 c^3 \sqrt {c-a c x} \left (1-a^2 x^2\right )^{3/2}}{21 a}+\frac {2 c^2 (c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}}{9 a}+\frac {1}{105} \left (128 c^4\right ) \int \frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx\\ &=\frac {256 c^5 \left (1-a^2 x^2\right )^{3/2}}{315 a (c-a c x)^{3/2}}+\frac {64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a \sqrt {c-a c x}}+\frac {8 c^3 \sqrt {c-a c x} \left (1-a^2 x^2\right )^{3/2}}{21 a}+\frac {2 c^2 (c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}}{9 a}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 62, normalized size = 0.44 \[ -\frac {2 c^3 (a x+1)^{3/2} \left (35 a^3 x^3-165 a^2 x^2+321 a x-319\right ) \sqrt {c-a c x}}{315 a \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c^3*(1 + a*x)^(3/2)*Sqrt[c - a*c*x]*(-319 + 321*a*x - 165*a^2*x^2 + 35*a^3*x^3))/(315*a*Sqrt[1 - a*x])

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fricas [A]  time = 0.45, size = 80, normalized size = 0.57 \[ \frac {2 \, {\left (35 \, a^{4} c^{3} x^{4} - 130 \, a^{3} c^{3} x^{3} + 156 \, a^{2} c^{3} x^{2} + 2 \, a c^{3} x - 319 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{315 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*a^4*c^3*x^4 - 130*a^3*c^3*x^3 + 156*a^2*c^3*x^2 + 2*a*c^3*x - 319*c^3)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*
x + c)/(a^2*x - a)

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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)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(7/2),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 [A]  time = 0.03, size = 63, normalized size = 0.45 \[ \frac {2 \left (a x +1\right )^{2} \left (35 x^{3} a^{3}-165 a^{2} x^{2}+321 a x -319\right ) \left (-a c x +c \right )^{\frac {7}{2}}}{315 a \left (a x -1\right )^{3} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(a*x+1)^2*(35*a^3*x^3-165*a^2*x^2+321*a*x-319)*(-a*c*x+c)^(7/2)/a/(a*x-1)^3/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.39, size = 128, normalized size = 0.91 \[ -\frac {2 \, {\left (5 \, a^{5} c^{\frac {7}{2}} x^{5} - 20 \, a^{4} c^{\frac {7}{2}} x^{4} + 32 \, a^{3} c^{\frac {7}{2}} x^{3} - 34 \, a^{2} c^{\frac {7}{2}} x^{2} + 91 \, a c^{\frac {7}{2}} x + 182 \, c^{\frac {7}{2}}\right )}}{45 \, \sqrt {a x + 1} a} - \frac {2 \, {\left (5 \, a^{4} c^{\frac {7}{2}} x^{4} - 22 \, a^{3} c^{\frac {7}{2}} x^{3} + 44 \, a^{2} c^{\frac {7}{2}} x^{2} - 106 \, a c^{\frac {7}{2}} x - 177 \, c^{\frac {7}{2}}\right )}}{35 \, \sqrt {a x + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45*(5*a^5*c^(7/2)*x^5 - 20*a^4*c^(7/2)*x^4 + 32*a^3*c^(7/2)*x^3 - 34*a^2*c^(7/2)*x^2 + 91*a*c^(7/2)*x + 182
*c^(7/2))/(sqrt(a*x + 1)*a) - 2/35*(5*a^4*c^(7/2)*x^4 - 22*a^3*c^(7/2)*x^3 + 44*a^2*c^(7/2)*x^2 - 106*a*c^(7/2
)*x - 177*c^(7/2))/(sqrt(a*x + 1)*a)

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mupad [B]  time = 0.99, size = 79, normalized size = 0.56 \[ \frac {\sqrt {c-a\,c\,x}\,\left (\frac {634\,c^3\,x}{315}+\frac {638\,c^3}{315\,a}-\frac {316\,a\,c^3\,x^2}{315}-\frac {52\,a^2\,c^3\,x^3}{315}+\frac {38\,a^3\,c^3\,x^4}{63}-\frac {2\,a^4\,c^3\,x^5}{9}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - a*c*x)^(1/2)*((634*c^3*x)/315 + (638*c^3)/(315*a) - (316*a*c^3*x^2)/315 - (52*a^2*c^3*x^3)/315 + (38*a^3
*c^3*x^4)/63 - (2*a^4*c^3*x^5)/9))/(1 - a^2*x^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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