3.2.59 \(\int e^{\tanh ^{-1}(a x)} (c-a c x)^3 \, dx\) [159]

Optimal. Leaf size=91 \[ \frac {5}{8} c^3 x \sqrt {1-a^2 x^2}+\frac {5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac {5 c^3 \text {ArcSin}(a x)}{8 a} \]

[Out]

5/12*c^3*(-a^2*x^2+1)^(3/2)/a+1/4*c^3*(-a*x+1)*(-a^2*x^2+1)^(3/2)/a+5/8*c^3*arcsin(a*x)/a+5/8*c^3*x*(-a^2*x^2+
1)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6262, 685, 655, 201, 222} \begin {gather*} \frac {c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac {5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {5}{8} c^3 x \sqrt {1-a^2 x^2}+\frac {5 c^3 \text {ArcSin}(a x)}{8 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - a*c*x)^3,x]

[Out]

(5*c^3*x*Sqrt[1 - a^2*x^2])/8 + (5*c^3*(1 - a^2*x^2)^(3/2))/(12*a) + (c^3*(1 - a*x)*(1 - a^2*x^2)^(3/2))/(4*a)
+ (5*c^3*ArcSin[a*x])/(8*a)

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 655

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

Rule 685

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*((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, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*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 e^{\tanh ^{-1}(a x)} (c-a c x)^3 \, dx &=c \int (c-a c x)^2 \sqrt {1-a^2 x^2} \, dx\\ &=\frac {c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac {1}{4} \left (5 c^2\right ) \int (c-a c x) \sqrt {1-a^2 x^2} \, dx\\ &=\frac {5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac {1}{4} \left (5 c^3\right ) \int \sqrt {1-a^2 x^2} \, dx\\ &=\frac {5}{8} c^3 x \sqrt {1-a^2 x^2}+\frac {5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac {1}{8} \left (5 c^3\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5}{8} c^3 x \sqrt {1-a^2 x^2}+\frac {5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac {5 c^3 \sin ^{-1}(a x)}{8 a}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 67, normalized size = 0.74 \begin {gather*} \frac {c^3 \left (\sqrt {1-a^2 x^2} \left (16+9 a x-16 a^2 x^2+6 a^3 x^3\right )-30 \text {ArcSin}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{24 a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*(Sqrt[1 - a^2*x^2]*(16 + 9*a*x - 16*a^2*x^2 + 6*a^3*x^3) - 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(24*a)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(172\) vs. \(2(77)=154\).
time = 0.73, size = 173, normalized size = 1.90

method	result	size
risch	\(-\frac {\left (6 a^{3} x^{3}-16 a^{2} x^{2}+9 a x +16\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{24 a \sqrt {-a^{2} x^{2}+1}}+\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{8 \sqrt {a^{2}}}\)	\(83\)
meijerg	\(-\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{2 \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{3} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{a \sqrt {\pi }}+\frac {c^{3} \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{a \sqrt {\pi }}+\frac {c^{3} \arcsin \left (a x \right )}{a}\)	\(162\)
default	\(-c^{3} \left (a^{4} \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )-2 a^{3} \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )-\frac {2 \sqrt {-a^{2} x^{2}+1}}{a}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}\right )\)	\(173\)

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)^3,x,method=_RETURNVERBOSE)

[Out]

-c^3*(a^4*(-1/4*x^3*(-a^2*x^2+1)^(1/2)/a^2+3/4/a^2*(-1/2*x*(-a^2*x^2+1)^(1/2)/a^2+1/2/a^2/(a^2)^(1/2)*arctan((
a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))))-2*a^3*(-1/3*x^2*(-a^2*x^2+1)^(1/2)/a^2-2/3*(-a^2*x^2+1)^(1/2)/a^4)-2*(-a^2*
x^2+1)^(1/2)/a-1/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2)))

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Maxima [A]
time = 0.46, size = 95, normalized size = 1.04 \begin {gather*} \frac {1}{4} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{3} - \frac {2}{3} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{2} + \frac {3}{8} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x + \frac {5 \, c^{3} \arcsin \left (a x\right )}{8 \, a} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{3 \, a} \end {gather*}

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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 82, normalized size = 0.90 \begin {gather*} -\frac {30 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (6 \, a^{3} c^{3} x^{3} - 16 \, a^{2} c^{3} x^{2} + 9 \, a c^{3} x + 16 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{24 \, a} \end {gather*}

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)^3,x, algorithm="fricas")

[Out]

-1/24*(30*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (6*a^3*c^3*x^3 - 16*a^2*c^3*x^2 + 9*a*c^3*x + 16*c^3)*s
qrt(-a^2*x^2 + 1))/a

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Sympy [A]
time = 4.27, size = 134, normalized size = 1.47 \begin {gather*} \begin {cases} \frac {2 c^{3} \sqrt {- a^{2} x^{2} + 1} + 2 c^{3} \left (\begin {cases} \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} - \sqrt {- a^{2} x^{2} + 1} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) - c^{3} \left (\begin {cases} \frac {a x \left (- 2 a^{2} x^{2} + 1\right ) \sqrt {- a^{2} x^{2} + 1}}{8} - \frac {a x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {3 \operatorname {asin}{\left (a x \right )}}{8} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) + c^{3} \operatorname {asin}{\left (a x \right )}}{a} & \text {for}\: a \neq 0 \\c^{3} x & \text {otherwise} \end {cases} \end {gather*}

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)**3,x)

[Out]

Piecewise(((2*c**3*sqrt(-a**2*x**2 + 1) + 2*c**3*Piecewise(((-a**2*x**2 + 1)**(3/2)/3 - sqrt(-a**2*x**2 + 1),
(a*x > -1) & (a*x < 1))) - c**3*Piecewise((a*x*(-2*a**2*x**2 + 1)*sqrt(-a**2*x**2 + 1)/8 - a*x*sqrt(-a**2*x**2
+ 1)/2 + 3*asin(a*x)/8, (a*x > -1) & (a*x < 1))) + c**3*asin(a*x))/a, Ne(a, 0)), (c**3*x, True))

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Giac [A]
time = 0.42, size = 66, normalized size = 0.73 \begin {gather*} \frac {5 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, {\left | a \right |}} + \frac {1}{24} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {16 \, c^{3}}{a} + {\left (9 \, c^{3} + 2 \, {\left (3 \, a^{2} c^{3} x - 8 \, a c^{3}\right )} x\right )} x\right )} \end {gather*}

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)^3,x, algorithm="giac")

[Out]

5/8*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1/24*sqrt(-a^2*x^2 + 1)*(16*c^3/a + (9*c^3 + 2*(3*a^2*c^3*x - 8*a*c^3)*x)*
x)

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Mupad [B]
time = 0.04, size = 105, normalized size = 1.15 \begin {gather*} \frac {3\,c^3\,x\,\sqrt {1-a^2\,x^2}}{8}+\frac {5\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,\sqrt {-a^2}}+\frac {2\,c^3\,\sqrt {1-a^2\,x^2}}{3\,a}-\frac {2\,a\,c^3\,x^2\,\sqrt {1-a^2\,x^2}}{3}+\frac {a^2\,c^3\,x^3\,\sqrt {1-a^2\,x^2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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