3.3.51 \(\int \frac {e^{3 \tanh ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx\) [251]

Optimal. Leaf size=192 \[ -\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {\sqrt {1-a^2 x^2}}{64 a c (c-a c x)^{5/2}}+\frac {3 \sqrt {1-a^2 x^2}}{256 a c^2 (c-a c x)^{3/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{256 \sqrt {2} a c^{7/2}} \]

[Out]

1/4*c^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(11/2)+3/512*arctanh(1/2*c^(1/2)*(-a^2*x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c)
^(1/2))/a/c^(7/2)*2^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(7/2)+1/64*(-a^2*x^2+1)^(1/2)/a/c/(-a*c*x+c)^(5/
2)+3/256*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*c*x+c)^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6262, 677, 687, 675, 214} \begin {gather*} \frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{256 \sqrt {2} a c^{7/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}+\frac {3 \sqrt {1-a^2 x^2}}{256 a c^2 (c-a c x)^{3/2}}-\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {\sqrt {1-a^2 x^2}}{64 a c (c-a c x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])/(c - a*c*x)^(7/2),x]

[Out]

-1/8*Sqrt[1 - a^2*x^2]/(a*(c - a*c*x)^(7/2)) + Sqrt[1 - a^2*x^2]/(64*a*c*(c - a*c*x)^(5/2)) + (3*Sqrt[1 - a^2*
x^2])/(256*a*c^2*(c - a*c*x)^(3/2)) + (c^2*(1 - a^2*x^2)^(3/2))/(4*a*(c - a*c*x)^(11/2)) + (3*ArcTanh[(Sqrt[c]
*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])])/(256*Sqrt[2]*a*c^(7/2))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 675

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

Rule 677

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

Rule 687

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

Rule 6262

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 \frac {e^{3 \tanh ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx &=c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{13/2}} \, dx\\ &=\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}-\frac {1}{8} (3 c) \int \frac {\sqrt {1-a^2 x^2}}{(c-a c x)^{9/2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}+\frac {\int \frac {1}{(c-a c x)^{5/2} \sqrt {1-a^2 x^2}} \, dx}{16 c}\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {\sqrt {1-a^2 x^2}}{64 a c (c-a c x)^{5/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}+\frac {3 \int \frac {1}{(c-a c x)^{3/2} \sqrt {1-a^2 x^2}} \, dx}{128 c^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {\sqrt {1-a^2 x^2}}{64 a c (c-a c x)^{5/2}}+\frac {3 \sqrt {1-a^2 x^2}}{256 a c^2 (c-a c x)^{3/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}+\frac {3 \int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}} \, dx}{512 c^3}\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {\sqrt {1-a^2 x^2}}{64 a c (c-a c x)^{5/2}}+\frac {3 \sqrt {1-a^2 x^2}}{256 a c^2 (c-a c x)^{3/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}-\frac {(3 a) \text {Subst}\left (\int \frac {1}{-2 a^2 c+a^2 c^2 x^2} \, dx,x,\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )}{256 c^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a (c-a c x)^{7/2}}+\frac {\sqrt {1-a^2 x^2}}{64 a c (c-a c x)^{5/2}}+\frac {3 \sqrt {1-a^2 x^2}}{256 a c^2 (c-a c x)^{3/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{4 a (c-a c x)^{11/2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{256 \sqrt {2} a c^{7/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 57, normalized size = 0.30 \begin {gather*} \frac {(1+a x)^{5/2} (c-a c x)^{3/2} \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {1}{2} (1+a x)\right )}{80 a c^5 (1-a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcTanh[a*x])/(c - a*c*x)^(7/2),x]

[Out]

((1 + a*x)^(5/2)*(c - a*c*x)^(3/2)*Hypergeometric2F1[5/2, 5, 7/2, (1 + a*x)/2])/(80*a*c^5*(1 - a*x)^(3/2))

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Maple [A]
time = 1.25, size = 258, normalized size = 1.34

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{4} c \,x^{4}-12 \sqrt {2}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{3} c \,x^{3}-6 a^{3} x^{3} \sqrt {\left (a x +1\right ) c}\, \sqrt {c}+18 \sqrt {2}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{2} c \,x^{2}+26 a^{2} x^{2} \sqrt {c}\, \sqrt {\left (a x +1\right ) c}-12 \sqrt {2}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c a x +158 \sqrt {c}\, \sqrt {\left (a x +1\right ) c}\, a x +3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +78 \sqrt {\left (a x +1\right ) c}\, \sqrt {c}\right )}{512 c^{\frac {9}{2}} \left (a x -1\right )^{5} \sqrt {\left (a x +1\right ) c}\, a}\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/512*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(9/2)*(3*2^(1/2)*arctanh(1/2*((a*x+1)*c)^(1/2)*2^(1/2)/c^(1/2))
*a^4*c*x^4-12*2^(1/2)*arctanh(1/2*((a*x+1)*c)^(1/2)*2^(1/2)/c^(1/2))*a^3*c*x^3-6*a^3*x^3*((a*x+1)*c)^(1/2)*c^(
1/2)+18*2^(1/2)*arctanh(1/2*((a*x+1)*c)^(1/2)*2^(1/2)/c^(1/2))*a^2*c*x^2+26*a^2*x^2*c^(1/2)*((a*x+1)*c)^(1/2)-
12*2^(1/2)*arctanh(1/2*((a*x+1)*c)^(1/2)*2^(1/2)/c^(1/2))*c*a*x+158*c^(1/2)*((a*x+1)*c)^(1/2)*a*x+3*2^(1/2)*ar
ctanh(1/2*((a*x+1)*c)^(1/2)*2^(1/2)/c^(1/2))*c+78*((a*x+1)*c)^(1/2)*c^(1/2))/(a*x-1)^5/((a*x+1)*c)^(1/2)/a

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)^3/((-a^2*x^2 + 1)^(3/2)*(-a*c*x + c)^(7/2)), x)

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Fricas [A]
time = 0.37, size = 420, normalized size = 2.19 \begin {gather*} \left [\frac {3 \, \sqrt {2} {\left (a^{5} x^{5} - 5 \, a^{4} x^{4} + 10 \, a^{3} x^{3} - 10 \, a^{2} x^{2} + 5 \, a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, {\left (3 \, a^{3} x^{3} - 13 \, a^{2} x^{2} - 79 \, a x - 39\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{1024 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{5} x^{5} - 5 \, a^{4} x^{4} + 10 \, a^{3} x^{3} - 10 \, a^{2} x^{2} + 5 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + 2 \, {\left (3 \, a^{3} x^{3} - 13 \, a^{2} x^{2} - 79 \, a x - 39\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{512 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(7/2),x, algorithm="fricas")

[Out]

[1/1024*(3*sqrt(2)*(a^5*x^5 - 5*a^4*x^4 + 10*a^3*x^3 - 10*a^2*x^2 + 5*a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c
*x - 2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a^2*x^2 - 2*a*x + 1)) + 4*(3*a^3*x^3 - 13*a
^2*x^2 - 79*a*x - 39)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^6*c^4*x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*
a^3*c^4*x^2 + 5*a^2*c^4*x - a*c^4), 1/512*(3*sqrt(2)*(a^5*x^5 - 5*a^4*x^4 + 10*a^3*x^3 - 10*a^2*x^2 + 5*a*x -
1)*sqrt(-c)*arctan(sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + 2*(3*a^3*x^3 - 13*a
^2*x^2 - 79*a*x - 39)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^6*c^4*x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*
a^3*c^4*x^2 + 5*a^2*c^4*x - a*c^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x + 1\right )^{3}}{\left (- c \left (a x - 1\right )\right )^{\frac {7}{2}} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**(7/2),x)

[Out]

Integral((a*x + 1)**3/((-c*(a*x - 1))**(7/2)*(-(a*x - 1)*(a*x + 1))**(3/2)), x)

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Giac [A]
time = 0.45, size = 105, normalized size = 0.55 \begin {gather*} -\frac {\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2}} + \frac {2 \, {\left (3 \, {\left (a c x + c\right )}^{\frac {7}{2}} - 22 \, {\left (a c x + c\right )}^{\frac {5}{2}} c - 44 \, {\left (a c x + c\right )}^{\frac {3}{2}} c^{2} + 24 \, \sqrt {a c x + c} c^{3}\right )}}{{\left (a c x - c\right )}^{4} c^{2}}}{512 \, a {\left | c \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(7/2),x, algorithm="giac")

[Out]

-1/512*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c^2) + 2*(3*(a*c*x + c)^(7/2) - 22*(a
*c*x + c)^(5/2)*c - 44*(a*c*x + c)^(3/2)*c^2 + 24*sqrt(a*c*x + c)*c^3)/((a*c*x - c)^4*c^2))/(a*abs(c))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}\,{\left (c-a\,c\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(7/2)),x)

[Out]

int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(7/2)), x)

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