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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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 &=\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 = 0.04, 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])

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 80, normalized size = 0.47 \[ -\frac {2 \, {\left (35 \, a^{4} c^{3} x^{4} - 220 \, a^{3} c^{3} x^{3} + 642 \, a^{2} c^{3} x^{2} - 1276 \, a c^{3} x + 2867 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{315 \, {\left (a^{2} x - a\right )}} \]

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)

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 64, normalized size = 0.37 \[ \frac {2 \sqrt {-a^{2} x^{2}+1}\, \left (-a c x +c \right )^{\frac {7}{2}} \left (35 x^{4} a^{4}-220 x^{3} a^{3}+642 a^{2} x^{2}-1276 a x +2867\right )}{315 \left (a x -1\right )^{4} a} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 71, normalized size = 0.42 \[ \frac {2 \, {\left (35 \, a^{4} c^{\frac {7}{2}} x^{4} - 220 \, a^{3} c^{\frac {7}{2}} x^{3} + 642 \, a^{2} c^{\frac {7}{2}} x^{2} - 1276 \, a c^{\frac {7}{2}} x + 2867 \, c^{\frac {7}{2}}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{315 \, {\left (a^{2} x - a\right )}} \]

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)

________________________________________________________________________________________

mupad [B]  time = 1.02, size = 88, normalized size = 0.51 \[ -\frac {4096\,c^3\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{315\,a\,\left (a\,x-1\right )}-\frac {2\,c^3\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}\,\left (35\,a^3\,x^3-185\,a^2\,x^2+457\,a\,x-819\right )}{315\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (4096*c^3*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2))/(315*a*(a*x - 1)) - (2*c^3*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^
(1/2)*(457*a*x - 185*a^2*x^2 + 35*a^3*x^3 - 819))/(315*a)

________________________________________________________________________________________

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

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]

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

________________________________________________________________________________________