3.1145 \(\int \frac {e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^5} \, dx\)

Optimal. Leaf size=115 \[ -\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {15}{8} a^4 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {3 a^3 c \sqrt {1-a^2 x^2}}{x} \]

[Out]

-15/8*a^4*c*arctanh((-a^2*x^2+1)^(1/2))-1/4*c*(-a^2*x^2+1)^(1/2)/x^4-a*c*(-a^2*x^2+1)^(1/2)/x^3-15/8*a^2*c*(-a
^2*x^2+1)^(1/2)/x^2-3*a^3*c*(-a^2*x^2+1)^(1/2)/x

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Rubi [A]  time = 0.22, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6148, 1807, 835, 807, 266, 63, 208} \[ -\frac {3 a^3 c \sqrt {1-a^2 x^2}}{x}-\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {15}{8} a^4 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2))/x^5,x]

[Out]

-(c*Sqrt[1 - a^2*x^2])/(4*x^4) - (a*c*Sqrt[1 - a^2*x^2])/x^3 - (15*a^2*c*Sqrt[1 - a^2*x^2])/(8*x^2) - (3*a^3*c
*Sqrt[1 - a^2*x^2])/x - (15*a^4*c*ArcTanh[Sqrt[1 - a^2*x^2]])/8

Rule 63

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 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 6148

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

Rubi steps

\begin {align*} \int \frac {e^{3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )}{x^5} \, dx &=c \int \frac {(1+a x)^3}{x^5 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {1}{4} c \int \frac {-12 a-15 a^2 x-4 a^3 x^2}{x^4 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}+\frac {1}{12} c \int \frac {45 a^2+36 a^3 x}{x^3 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {1}{24} c \int \frac {-72 a^3-45 a^4 x}{x^2 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {3 a^3 c \sqrt {1-a^2 x^2}}{x}+\frac {1}{8} \left (15 a^4 c\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {3 a^3 c \sqrt {1-a^2 x^2}}{x}+\frac {1}{16} \left (15 a^4 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {3 a^3 c \sqrt {1-a^2 x^2}}{x}-\frac {1}{8} \left (15 a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{4 x^4}-\frac {a c \sqrt {1-a^2 x^2}}{x^3}-\frac {15 a^2 c \sqrt {1-a^2 x^2}}{8 x^2}-\frac {3 a^3 c \sqrt {1-a^2 x^2}}{x}-\frac {15}{8} a^4 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 97, normalized size = 0.84 \[ \frac {1}{2} a c \left (-\frac {\sqrt {1-a^2 x^2} \left (6 a^2 x^2+3 a x+2\right )}{x^3}-2 a^3 \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-a^2 x^2\right )-3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2))/x^5,x]

[Out]

(a*c*(-((Sqrt[1 - a^2*x^2]*(2 + 3*a*x + 6*a^2*x^2))/x^3) - 3*a^3*ArcTanh[Sqrt[1 - a^2*x^2]] - 2*a^3*Sqrt[1 - a
^2*x^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 - a^2*x^2]))/2

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fricas [A]  time = 0.55, size = 75, normalized size = 0.65 \[ \frac {15 \, a^{4} c x^{4} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (24 \, a^{3} c x^{3} + 15 \, a^{2} c x^{2} + 8 \, a c x + 2 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^5,x, algorithm="fricas")

[Out]

1/8*(15*a^4*c*x^4*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (24*a^3*c*x^3 + 15*a^2*c*x^2 + 8*a*c*x + 2*c)*sqrt(-a^2*x^
2 + 1))/x^4

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giac [B]  time = 0.18, size = 280, normalized size = 2.43 \[ \frac {{\left (a^{5} c + \frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{3} c}{x} + \frac {32 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a c}{x^{2}} + \frac {104 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c}{a x^{3}}\right )} a^{8} x^{4}}{64 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} {\left | a \right |}} - \frac {15 \, a^{5} c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{8 \, {\left | a \right |}} - \frac {\frac {104 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{5} c {\left | a \right |}}{x} + \frac {32 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{3} c {\left | a \right |}}{x^{2}} + \frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} a c {\left | a \right |}}{x^{3}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} c {\left | a \right |}}{a x^{4}}}{64 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^5,x, algorithm="giac")

[Out]

1/64*(a^5*c + 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^3*c/x + 32*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a*c/x^2 + 104*(
sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c/(a*x^3))*a^8*x^4/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*abs(a)) - 15/8*a^5*c*lo
g(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 1/64*(104*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
*a^5*c*abs(a)/x + 32*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^3*c*abs(a)/x^2 + 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*
a*c*abs(a)/x^3 + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c*abs(a)/(a*x^4))/a^4

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maple [B]  time = 0.05, size = 231, normalized size = 2.01 \[ -c \left (\frac {x \,a^{5}}{\sqrt {-a^{2} x^{2}+1}}+3 a^{4} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )+2 a^{3} \left (-\frac {1}{x \sqrt {-a^{2} x^{2}+1}}+\frac {2 a^{2} x}{\sqrt {-a^{2} x^{2}+1}}\right )-3 a \left (-\frac {1}{3 x^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {4 a^{2} \left (-\frac {1}{x \sqrt {-a^{2} x^{2}+1}}+\frac {2 a^{2} x}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}\right )-\frac {13 a^{2} \left (-\frac {1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 a^{2} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}\right )}{4}+\frac {1}{4 x^{4} \sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^5,x)

[Out]

-c*(x*a^5/(-a^2*x^2+1)^(1/2)+3*a^4*(1/(-a^2*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2)))+2*a^3*(-1/x/(-a^2*x^2+
1)^(1/2)+2*a^2*x/(-a^2*x^2+1)^(1/2))-3*a*(-1/3/x^3/(-a^2*x^2+1)^(1/2)+4/3*a^2*(-1/x/(-a^2*x^2+1)^(1/2)+2*a^2*x
/(-a^2*x^2+1)^(1/2)))-13/4*a^2*(-1/2/x^2/(-a^2*x^2+1)^(1/2)+3/2*a^2*(1/(-a^2*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+
1)^(1/2))))+1/4/x^4/(-a^2*x^2+1)^(1/2))

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maxima [A]  time = 0.33, size = 149, normalized size = 1.30 \[ \frac {3 \, a^{5} c x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {15}{8} \, a^{4} c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {15 \, a^{4} c}{8 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {2 \, a^{3} c}{\sqrt {-a^{2} x^{2} + 1} x} - \frac {13 \, a^{2} c}{8 \, \sqrt {-a^{2} x^{2} + 1} x^{2}} - \frac {a c}{\sqrt {-a^{2} x^{2} + 1} x^{3}} - \frac {c}{4 \, \sqrt {-a^{2} x^{2} + 1} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^5,x, algorithm="maxima")

[Out]

3*a^5*c*x/sqrt(-a^2*x^2 + 1) - 15/8*a^4*c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 15/8*a^4*c/sqrt(-a^2*x
^2 + 1) - 2*a^3*c/(sqrt(-a^2*x^2 + 1)*x) - 13/8*a^2*c/(sqrt(-a^2*x^2 + 1)*x^2) - a*c/(sqrt(-a^2*x^2 + 1)*x^3)
- 1/4*c/(sqrt(-a^2*x^2 + 1)*x^4)

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mupad [B]  time = 0.88, size = 103, normalized size = 0.90 \[ -\frac {c\,\sqrt {1-a^2\,x^2}}{4\,x^4}-\frac {15\,a^2\,c\,\sqrt {1-a^2\,x^2}}{8\,x^2}-\frac {3\,a^3\,c\,\sqrt {1-a^2\,x^2}}{x}-\frac {a\,c\,\sqrt {1-a^2\,x^2}}{x^3}+\frac {a^4\,c\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,15{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a^2*c*x^2)*(a*x + 1)^3)/(x^5*(1 - a^2*x^2)^(3/2)),x)

[Out]

(a^4*c*atan((1 - a^2*x^2)^(1/2)*1i)*15i)/8 - (c*(1 - a^2*x^2)^(1/2))/(4*x^4) - (15*a^2*c*(1 - a^2*x^2)^(1/2))/
(8*x^2) - (3*a^3*c*(1 - a^2*x^2)^(1/2))/x - (a*c*(1 - a^2*x^2)^(1/2))/x^3

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sympy [C]  time = 12.00, size = 411, normalized size = 3.57 \[ a^{3} c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) + 3 a^{2} c \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) + 3 a c \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {3 a^{4} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {a}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {3 i a^{4} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} - \frac {3 i a^{3}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i a}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(-a**2*c*x**2+c)/x**5,x)

[Out]

a**3*c*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) + 3*a**2*c*P
iecewise((-a**2*acosh(1/(a*x))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*
x))/2 - I*a/(2*x*sqrt(1 - 1/(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True)) + 3*a*c*Piecewise((-2
*I*a**2*sqrt(a**2*x**2 - 1)/(3*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a**2*sqrt(-a**2*x
**2 + 1)/(3*x) - sqrt(-a**2*x**2 + 1)/(3*x**3), True)) + c*Piecewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*s
qrt(-1 + 1/(a**2*x**2))) - a/(8*x**3*sqrt(-1 + 1/(a**2*x**2))) - 1/(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(
a**2*x**2) > 1), (3*I*a**4*asin(1/(a*x))/8 - 3*I*a**3/(8*x*sqrt(1 - 1/(a**2*x**2))) + I*a/(8*x**3*sqrt(1 - 1/(
a**2*x**2))) + I/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True))

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