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

Optimal. Leaf size=171 \[ \frac{4096 c^4 \sqrt{1-a^2 x^2}}{315 a \sqrt{c-a c x}}+\frac{1024 c^3 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{315 a}+\frac{128 c^2 \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}{105 a}+\frac{32 c \sqrt{1-a^2 x^2} (c-a c x)^{5/2}}{63 a}+\frac{2 \sqrt{1-a^2 x^2} (c-a c x)^{7/2}}{9 a} \]

[Out]

(4096*c^4*Sqrt[1 - a^2*x^2])/(315*a*Sqrt[c - a*c*x]) + (1024*c^3*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(315*a) +
(128*c^2*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(105*a) + (32*c*(c - a*c*x)^(5/2)*Sqrt[1 - a^2*x^2])/(63*a) + (2
*(c - a*c*x)^(7/2)*Sqrt[1 - a^2*x^2])/(9*a)

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Rubi [A]  time = 0.133945, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6127, 657, 649} \[ \frac{4096 c^4 \sqrt{1-a^2 x^2}}{315 a \sqrt{c-a c x}}+\frac{1024 c^3 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{315 a}+\frac{128 c^2 \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}{105 a}+\frac{32 c \sqrt{1-a^2 x^2} (c-a c x)^{5/2}}{63 a}+\frac{2 \sqrt{1-a^2 x^2} (c-a c x)^{7/2}}{9 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4096*c^4*Sqrt[1 - a^2*x^2])/(315*a*Sqrt[c - a*c*x]) + (1024*c^3*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(315*a) +
(128*c^2*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(105*a) + (32*c*(c - a*c*x)^(5/2)*Sqrt[1 - a^2*x^2])/(63*a) + (2
*(c - a*c*x)^(7/2)*Sqrt[1 - a^2*x^2])/(9*a)

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]

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 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]

Rubi steps

\begin{align*} \int e^{-\tanh ^{-1}(a x)} (c-a c x)^{7/2} \, dx &=\frac{\int \frac{(c-a c x)^{9/2}}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=\frac{2 (c-a c x)^{7/2} \sqrt{1-a^2 x^2}}{9 a}+\frac{16}{9} \int \frac{(c-a c x)^{7/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{32 c (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{63 a}+\frac{2 (c-a c x)^{7/2} \sqrt{1-a^2 x^2}}{9 a}+\frac{1}{21} (64 c) \int \frac{(c-a c x)^{5/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{128 c^2 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{105 a}+\frac{32 c (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{63 a}+\frac{2 (c-a c x)^{7/2} \sqrt{1-a^2 x^2}}{9 a}+\frac{1}{105} \left (512 c^2\right ) \int \frac{(c-a c x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{1024 c^3 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{315 a}+\frac{128 c^2 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{105 a}+\frac{32 c (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{63 a}+\frac{2 (c-a c x)^{7/2} \sqrt{1-a^2 x^2}}{9 a}+\frac{1}{315} \left (2048 c^3\right ) \int \frac{\sqrt{c-a c x}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{4096 c^4 \sqrt{1-a^2 x^2}}{315 a \sqrt{c-a c x}}+\frac{1024 c^3 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{315 a}+\frac{128 c^2 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{105 a}+\frac{32 c (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{63 a}+\frac{2 (c-a c x)^{7/2} \sqrt{1-a^2 x^2}}{9 a}\\ \end{align*}

Mathematica [A]  time = 0.0395576, size = 65, normalized size = 0.38 \[ \frac{2 c^4 \sqrt{1-a^2 x^2} \left (35 a^4 x^4-220 a^3 x^3+642 a^2 x^2-1276 a x+2867\right )}{315 a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^4*Sqrt[1 - a^2*x^2]*(2867 - 1276*a*x + 642*a^2*x^2 - 220*a^3*x^3 + 35*a^4*x^4))/(315*a*Sqrt[c - a*c*x])

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Maple [A]  time = 0.034, size = 64, normalized size = 0.4 \begin{align*}{\frac{70\,{x}^{4}{a}^{4}-440\,{x}^{3}{a}^{3}+1284\,{a}^{2}{x}^{2}-2552\,ax+5734}{315\, \left ( ax-1 \right ) ^{4}a}\sqrt{-{a}^{2}{x}^{2}+1} \left ( -acx+c \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(7/2)*(35*a^4*x^4-220*a^3*x^3+642*a^2*x^2-1276*a*x+2867)/(a*x-1)^4/a

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Maxima [A]  time = 1.01109, size = 96, normalized size = 0.56 \begin{align*} \frac{2 \,{\left (35 \, a^{4} c^{\frac{7}{2}} x^{4} - 220 \, a^{3} c^{\frac{7}{2}} x^{3} + 642 \, a^{2} c^{\frac{7}{2}} x^{2} - 1276 \, a c^{\frac{7}{2}} x + 2867 \, c^{\frac{7}{2}}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{315 \,{\left (a^{2} x - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*a^4*c^(7/2)*x^4 - 220*a^3*c^(7/2)*x^3 + 642*a^2*c^(7/2)*x^2 - 1276*a*c^(7/2)*x + 2867*c^(7/2))*sqrt(
a*x + 1)*(a*x - 1)/(a^2*x - a)

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Fricas [A]  time = 2.2511, size = 182, normalized size = 1.06 \begin{align*} -\frac{2 \,{\left (35 \, a^{4} c^{3} x^{4} - 220 \, a^{3} c^{3} x^{3} + 642 \, a^{2} c^{3} x^{2} - 1276 \, a c^{3} x + 2867 \, c^{3}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{315 \,{\left (a^{2} x - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*a^4*c^3*x^4 - 220*a^3*c^3*x^3 + 642*a^2*c^3*x^2 - 1276*a*c^3*x + 2867*c^3)*sqrt(-a^2*x^2 + 1)*sqrt(
-a*c*x + c)/(a^2*x - a)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.30607, size = 115, normalized size = 0.67 \begin{align*} -\frac{4096 \, \sqrt{2} c^{\frac{5}{2}}{\left | c \right |}}{315 \, a} + \frac{2 \,{\left (35 \,{\left (a c x + c\right )}^{\frac{9}{2}} - 360 \,{\left (a c x + c\right )}^{\frac{7}{2}} c + 1512 \,{\left (a c x + c\right )}^{\frac{5}{2}} c^{2} - 3360 \,{\left (a c x + c\right )}^{\frac{3}{2}} c^{3} + 5040 \, \sqrt{a c x + c} c^{4}\right )}{\left | c \right |}}{315 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4096/315*sqrt(2)*c^(5/2)*abs(c)/a + 2/315*(35*(a*c*x + c)^(9/2) - 360*(a*c*x + c)^(7/2)*c + 1512*(a*c*x + c)^
(5/2)*c^2 - 3360*(a*c*x + c)^(3/2)*c^3 + 5040*sqrt(a*c*x + c)*c^4)*abs(c)/(a*c^2)