3.6.21 \(\int e^{2 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^{3/2} \, dx\) [521]

Optimal. Leaf size=71 \[ \frac {c \sqrt {c-\frac {c}{a x}}}{a}-\left (c-\frac {c}{a x}\right )^{3/2} x-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6268, 25, 528, 382, 79, 52, 65, 214} \begin {gather*} -\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}+\frac {c \sqrt {c-\frac {c}{a x}}}{a}-x \left (c-\frac {c}{a x}\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c - c/(a*x)])/a - (c - c/(a*x))^(3/2)*x - (c^(3/2)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[u*((
a + b*x^n)^(m + p)/x^(n*p)), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[(a + b/x^n)^p*((c +
d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6268

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Int[u*(c + d/x)^p*((1 + a*x)^(n/
2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c, d, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rubi steps

\begin {align*} \int e^{2 \tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^{3/2} \, dx &=\int \frac {\left (c-\frac {c}{a x}\right )^{3/2} (1+a x)}{1-a x} \, dx\\ &=-\frac {c \int \frac {\sqrt {c-\frac {c}{a x}} (1+a x)}{x} \, dx}{a}\\ &=-\frac {c \int \left (a+\frac {1}{x}\right ) \sqrt {c-\frac {c}{a x}} \, dx}{a}\\ &=\frac {c \text {Subst}\left (\int \frac {(a+x) \sqrt {c-\frac {c x}{a}}}{x^2} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\left (c-\frac {c}{a x}\right )^{3/2} x+\frac {c \text {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}}}{x} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=\frac {c \sqrt {c-\frac {c}{a x}}}{a}-\left (c-\frac {c}{a x}\right )^{3/2} x+\frac {c^2 \text {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=\frac {c \sqrt {c-\frac {c}{a x}}}{a}-\left (c-\frac {c}{a x}\right )^{3/2} x-c \text {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )\\ &=\frac {c \sqrt {c-\frac {c}{a x}}}{a}-\left (c-\frac {c}{a x}\right )^{3/2} x-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 57, normalized size = 0.80 \begin {gather*} \frac {c \sqrt {c-\frac {c}{a x}} (2-a x)-c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c - c/(a*x)]*(2 - a*x) - c^(3/2)*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/a

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Maple [A]
time = 0.79, size = 104, normalized size = 1.46

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c \left (2 a^{\frac {3}{2}} \sqrt {a \,x^{2}-x}\, x^{2}-4 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}-\ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \,x^{2}\right )}{2 x \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}}}\)	\(104\)
risch	\(-\frac {\left (a^{2} x^{2}-3 a x +2\right ) c \sqrt {\frac {c \left (a x -1\right )}{a x}}}{a \left (a x -1\right )}-\frac {\ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c x a}\right ) c \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {a c x \left (a x -1\right )}}{2 \sqrt {a^{2} c}\, \left (a x -1\right )}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.40, size = 136, normalized size = 1.92 \begin {gather*} \left [\frac {c^{\frac {3}{2}} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) - 2 \, {\left (a c x - 2 \, c\right )} \sqrt {\frac {a c x - c}{a x}}}{2 \, a}, \frac {\sqrt {-c} c \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - {\left (a c x - 2 \, c\right )} \sqrt {\frac {a c x - c}{a x}}}{a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(c^(3/2)*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) - 2*(a*c*x - 2*c)*sqrt((a*c*x - c)/(a*
x)))/a, (sqrt(-c)*c*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - (a*c*x - 2*c)*sqrt((a*c*x - c)/(a*x)))/a]

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Sympy [C] Result contains complex when optimal does not.
time = 17.03, size = 165, normalized size = 2.32 \begin {gather*} - c \left (\begin {cases} - \frac {\sqrt {c} \operatorname {acosh}{\left (\sqrt {a} \sqrt {x} \right )}}{a} + \frac {\sqrt {c} \sqrt {x} \sqrt {a x - 1}}{\sqrt {a}} & \text {for}\: \left |{a x}\right | > 1 \\- \frac {i \sqrt {a} \sqrt {c} x^{\frac {3}{2}}}{\sqrt {- a x + 1}} + \frac {i \sqrt {c} \operatorname {asin}{\left (\sqrt {a} \sqrt {x} \right )}}{a} + \frac {i \sqrt {c} \sqrt {x}}{\sqrt {a} \sqrt {- a x + 1}} & \text {otherwise} \end {cases}\right ) + \frac {2 c^{2} \operatorname {atan}{\left (\frac {\sqrt {c - \frac {c}{a x}}}{\sqrt {- c}} \right )}}{a \sqrt {- c}} + \frac {2 c \sqrt {c - \frac {c}{a x}}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*Piecewise((-sqrt(c)*acosh(sqrt(a)*sqrt(x))/a + sqrt(c)*sqrt(x)*sqrt(a*x - 1)/sqrt(a), Abs(a*x) > 1), (-I*sq
rt(a)*sqrt(c)*x**(3/2)/sqrt(-a*x + 1) + I*sqrt(c)*asin(sqrt(a)*sqrt(x))/a + I*sqrt(c)*sqrt(x)/(sqrt(a)*sqrt(-a
*x + 1)), True)) + 2*c**2*atan(sqrt(c - c/(a*x))/sqrt(-c))/(a*sqrt(-c)) + 2*c*sqrt(c - c/(a*x))/a

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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