3.339 \(\int \frac{1}{x (d+e x) (a+c x^2)^{3/2}} \, dx\)
Optimal. Leaf size=147 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{e (a e+c d x)}{a d \sqrt{a+c x^2} \left (a e^2+c d^2\right )}+\frac{e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d \left (a e^2+c d^2\right )^{3/2}}+\frac{1}{a d \sqrt{a+c x^2}} \]
[Out]
1/(a*d*Sqrt[a + c*x^2]) - (e*(a*e + c*d*x))/(a*d*(c*d^2 + a*e^2)*Sqrt[a + c*x^2]) + (e^3*ArcTanh[(a*e - c*d*x)
/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d*(c*d^2 + a*e^2)^(3/2)) - ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]]/(a^(3/2)
*d)
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Rubi [A] time = 0.134966, antiderivative size = 147, normalized size of antiderivative = 1.,
number of steps used = 10, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used
= {961, 266, 51, 63, 208, 741, 12, 725, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{e (a e+c d x)}{a d \sqrt{a+c x^2} \left (a e^2+c d^2\right )}+\frac{e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d \left (a e^2+c d^2\right )^{3/2}}+\frac{1}{a d \sqrt{a+c x^2}} \]
Antiderivative was successfully verified.
[In]
Int[1/(x*(d + e*x)*(a + c*x^2)^(3/2)),x]
[Out]
1/(a*d*Sqrt[a + c*x^2]) - (e*(a*e + c*d*x))/(a*d*(c*d^2 + a*e^2)*Sqrt[a + c*x^2]) + (e^3*ArcTanh[(a*e - c*d*x)
/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d*(c*d^2 + a*e^2)^(3/2)) - ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]]/(a^(3/2)
*d)
Rule 961
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) && !(IGtQ[m, 0] || IGtQ[n, 0])
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 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
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 741
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 725
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx &=\int \left (\frac{1}{d x \left (a+c x^2\right )^{3/2}}-\frac{e}{d (d+e x) \left (a+c x^2\right )^{3/2}}\right ) \, dx\\ &=\frac{\int \frac{1}{x \left (a+c x^2\right )^{3/2}} \, dx}{d}-\frac{e \int \frac{1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx}{d}\\ &=-\frac{e (a e+c d x)}{a d \left (c d^2+a e^2\right ) \sqrt{a+c x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+c x)^{3/2}} \, dx,x,x^2\right )}{2 d}-\frac{e \int \frac{a e^2}{(d+e x) \sqrt{a+c x^2}} \, dx}{a d \left (c d^2+a e^2\right )}\\ &=\frac{1}{a d \sqrt{a+c x^2}}-\frac{e (a e+c d x)}{a d \left (c d^2+a e^2\right ) \sqrt{a+c x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{2 a d}-\frac{e^3 \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{d \left (c d^2+a e^2\right )}\\ &=\frac{1}{a d \sqrt{a+c x^2}}-\frac{e (a e+c d x)}{a d \left (c d^2+a e^2\right ) \sqrt{a+c x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{a c d}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{c d^2+a e^2-x^2} \, dx,x,\frac{a e-c d x}{\sqrt{a+c x^2}}\right )}{d \left (c d^2+a e^2\right )}\\ &=\frac{1}{a d \sqrt{a+c x^2}}-\frac{e (a e+c d x)}{a d \left (c d^2+a e^2\right ) \sqrt{a+c x^2}}+\frac{e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{d \left (c d^2+a e^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d}\\ \end{align*}
Mathematica [C] time = 0.191429, size = 132, normalized size = 0.9 \[ \frac{-\frac{e (a e+c d x)}{a \sqrt{a+c x^2} \left (a e^2+c d^2\right )}+\frac{e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}+\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x^2}{a}+1\right )}{a \sqrt{a+c x^2}}}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(x*(d + e*x)*(a + c*x^2)^(3/2)),x]
[Out]
(-((e*(a*e + c*d*x))/(a*(c*d^2 + a*e^2)*Sqrt[a + c*x^2])) + (e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sq
rt[a + c*x^2])])/(c*d^2 + a*e^2)^(3/2) + Hypergeometric2F1[-1/2, 1, 1/2, 1 + (c*x^2)/a]/(a*Sqrt[a + c*x^2]))/d
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Maple [B] time = 0.236, size = 318, normalized size = 2.2 \begin{align*}{\frac{1}{ad}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{1}{d}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{{e}^{2}}{d \left ( a{e}^{2}+c{d}^{2} \right ) }{\frac{1}{\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{cex}{ \left ( a{e}^{2}+c{d}^{2} \right ) a}{\frac{1}{\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}+{\frac{{e}^{2}}{d \left ( a{e}^{2}+c{d}^{2} \right ) }\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x/(e*x+d)/(c*x^2+a)^(3/2),x)
[Out]
1/a/d/(c*x^2+a)^(1/2)-1/d/a^(3/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)-1/d/(a*e^2+c*d^2)*e^2/((d/e+x)^2*c-2*c
*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)-e/(a*e^2+c*d^2)/a/((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*
x*c+1/d/(a*e^2+c*d^2)*e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e
^2)^(1/2)*((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + a\right )}^{\frac{3}{2}}{\left (e x + d\right )} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((c*x^2 + a)^(3/2)*(e*x + d)*x), x)
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Fricas [B] time = 4.00096, size = 2695, normalized size = 18.33 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="fricas")
[Out]
[1/2*((a^2*c*e^3*x^2 + a^3*e^3)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*
e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + (a*c^2*d^4 + 2*a^
2*c*d^2*e^2 + a^3*e^4 + (c^3*d^4 + 2*a*c^2*d^2*e^2 + a^2*c*e^4)*x^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*s
qrt(a) + 2*a)/x^2) + 2*(a*c^2*d^4 + a^2*c*d^2*e^2 - (a*c^2*d^3*e + a^2*c*d*e^3)*x)*sqrt(c*x^2 + a))/(a^3*c^2*d
^5 + 2*a^4*c*d^3*e^2 + a^5*d*e^4 + (a^2*c^3*d^5 + 2*a^3*c^2*d^3*e^2 + a^4*c*d*e^4)*x^2), 1/2*(2*(a^2*c*e^3*x^2
+ a^3*e^3)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2
+ (c^2*d^2 + a*c*e^2)*x^2)) + (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4 + (c^3*d^4 + 2*a*c^2*d^2*e^2 + a^2*c*e^4)
*x^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(a*c^2*d^4 + a^2*c*d^2*e^2 - (a*c^2*d^3*
e + a^2*c*d*e^3)*x)*sqrt(c*x^2 + a))/(a^3*c^2*d^5 + 2*a^4*c*d^3*e^2 + a^5*d*e^4 + (a^2*c^3*d^5 + 2*a^3*c^2*d^3
*e^2 + a^4*c*d*e^4)*x^2), 1/2*(2*(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4 + (c^3*d^4 + 2*a*c^2*d^2*e^2 + a^2*c*e
^4)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (a^2*c*e^3*x^2 + a^3*e^3)*sqrt(c*d^2 + a*e^2)*log((2*a*c*
d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))
/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(a*c^2*d^4 + a^2*c*d^2*e^2 - (a*c^2*d^3*e + a^2*c*d*e^3)*x)*sqrt(c*x^2 + a))/(
a^3*c^2*d^5 + 2*a^4*c*d^3*e^2 + a^5*d*e^4 + (a^2*c^3*d^5 + 2*a^3*c^2*d^3*e^2 + a^4*c*d*e^4)*x^2), ((a^2*c*e^3*
x^2 + a^3*e^3)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e
^2 + (c^2*d^2 + a*c*e^2)*x^2)) + (a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4 + (c^3*d^4 + 2*a*c^2*d^2*e^2 + a^2*c*e
^4)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (a*c^2*d^4 + a^2*c*d^2*e^2 - (a*c^2*d^3*e + a^2*c*d*e^3)*
x)*sqrt(c*x^2 + a))/(a^3*c^2*d^5 + 2*a^4*c*d^3*e^2 + a^5*d*e^4 + (a^2*c^3*d^5 + 2*a^3*c^2*d^3*e^2 + a^4*c*d*e^
4)*x^2)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (a + c x^{2}\right )^{\frac{3}{2}} \left (d + e x\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(e*x+d)/(c*x**2+a)**(3/2),x)
[Out]
Integral(1/(x*(a + c*x**2)**(3/2)*(d + e*x)), x)
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError