3.3.90 \(\int e^{\tanh ^{-1}(a x)} (c-a c x) \, dx\) [290]

Optimal. Leaf size=33 \[ \frac {1}{2} c x \sqrt {1-a^2 x^2}+\frac {c \text {ArcSin}(a x)}{2 a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6262, 201, 222} \begin {gather*} \frac {1}{2} c x \sqrt {1-a^2 x^2}+\frac {c \text {ArcSin}(a x)}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - a*c*x),x]

[Out]

(c*x*Sqrt[1 - a^2*x^2])/2 + (c*ArcSin[a*x])/(2*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 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) \, dx &=c \int \sqrt {1-a^2 x^2} \, dx\\ &=\frac {1}{2} c x \sqrt {1-a^2 x^2}+\frac {1}{2} c \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {1}{2} c x \sqrt {1-a^2 x^2}+\frac {c \sin ^{-1}(a x)}{2 a}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 30, normalized size = 0.91 \begin {gather*} \frac {c \left (a x \sqrt {1-a^2 x^2}+\text {ArcSin}(a x)\right )}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(a*x*Sqrt[1 - a^2*x^2] + ArcSin[a*x]))/(2*a)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(84\) vs. \(2(27)=54\).
time = 0.00, size = 85, normalized size = 2.58

method	result	size
risch	\(-\frac {x \left (a^{2} x^{2}-1\right ) c}{2 \sqrt {-a^{2} x^{2}+1}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c}{2 \sqrt {a^{2}}}\)	\(55\)
meijerg	\(\frac {c \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{2 \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c \arcsin \left (a x \right )}{a}\)	\(71\)
default	\(-c \left (a^{2} \left (-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}\right )\)	\(85\)

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),x,method=_RETURNVERBOSE)

[Out]

-c*(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)))-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 = 0.47, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} c x + \frac {c \arcsin \left (a x\right )}{2 \, 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),x, algorithm="maxima")

[Out]

1/2*sqrt(-a^2*x^2 + 1)*c*x + 1/2*c*arcsin(a*x)/a

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Fricas [A]
time = 0.35, size = 47, normalized size = 1.42 \begin {gather*} \frac {\sqrt {-a^{2} x^{2} + 1} a c x - 2 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{2 \, 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),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 2.04, size = 44, normalized size = 1.33 \begin {gather*} \begin {cases} \frac {- c \left (\begin {cases} - \frac {a x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {\operatorname {asin}{\left (a x \right )}}{2} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) + c \operatorname {asin}{\left (a x \right )}}{a} & \text {for}\: a \neq 0 \\c 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),x)

[Out]

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

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Giac [A]
time = 0.42, size = 30, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} c x + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, {\left | a \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),x, algorithm="giac")

[Out]

1/2*sqrt(-a^2*x^2 + 1)*c*x + 1/2*c*arcsin(a*x)*sgn(a)/abs(a)

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Mupad [B]
time = 0.00, size = 37, normalized size = 1.12 \begin {gather*} \frac {c\,x\,\sqrt {1-a^2\,x^2}}{2}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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