3.3.51 \(\int \frac {1}{x^2 (d+e x) (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=194 \[ \frac {e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^2}-\frac {2 c x}{a^2 d \sqrt {a+c x^2}}+\frac {e^2 (a e+c d x)}{a d^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {e^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^{3/2}}-\frac {e}{a d^2 \sqrt {a+c x^2}}-\frac {1}{a d x \sqrt {a+c x^2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {961, 271, 191, 266, 51, 63, 208, 741, 12, 725, 206} \begin {gather*} \frac {e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^2}-\frac {2 c x}{a^2 d \sqrt {a+c x^2}}+\frac {e^2 (a e+c d x)}{a d^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {e^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^{3/2}}-\frac {e}{a d^2 \sqrt {a+c x^2}}-\frac {1}{a d x \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(d + e*x)*(a + c*x^2)^(3/2)),x]

[Out]

-(e/(a*d^2*Sqrt[a + c*x^2])) - 1/(a*d*x*Sqrt[a + c*x^2]) - (2*c*x)/(a^2*d*Sqrt[a + c*x^2]) + (e^2*(a*e + c*d*x
))/(a*d^2*(c*d^2 + a*e^2)*Sqrt[a + c*x^2]) - (e^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])]
)/(d^2*(c*d^2 + a*e^2)^(3/2)) + (e*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(a^(3/2)*d^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

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])

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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

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 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 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])

Rubi steps

\begin {align*} \int \frac {1}{x^2 (d+e x) \left (a+c x^2\right )^{3/2}} \, dx &=\int \left (\frac {1}{d x^2 \left (a+c x^2\right )^{3/2}}-\frac {e}{d^2 x \left (a+c x^2\right )^{3/2}}+\frac {e^2}{d^2 (d+e x) \left (a+c x^2\right )^{3/2}}\right ) \, dx\\ &=\frac {\int \frac {1}{x^2 \left (a+c x^2\right )^{3/2}} \, dx}{d}-\frac {e \int \frac {1}{x \left (a+c x^2\right )^{3/2}} \, dx}{d^2}+\frac {e^2 \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx}{d^2}\\ &=-\frac {1}{a d x \sqrt {a+c x^2}}+\frac {e^2 (a e+c d x)}{a d^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {(2 c) \int \frac {1}{\left (a+c x^2\right )^{3/2}} \, dx}{a d}-\frac {e \operatorname {Subst}\left (\int \frac {1}{x (a+c x)^{3/2}} \, dx,x,x^2\right )}{2 d^2}+\frac {e^2 \int \frac {a e^2}{(d+e x) \sqrt {a+c x^2}} \, dx}{a d^2 \left (c d^2+a e^2\right )}\\ &=-\frac {e}{a d^2 \sqrt {a+c x^2}}-\frac {1}{a d x \sqrt {a+c x^2}}-\frac {2 c x}{a^2 d \sqrt {a+c x^2}}+\frac {e^2 (a e+c d x)}{a d^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {e \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 a d^2}+\frac {e^4 \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^2 \left (c d^2+a e^2\right )}\\ &=-\frac {e}{a d^2 \sqrt {a+c x^2}}-\frac {1}{a d x \sqrt {a+c x^2}}-\frac {2 c x}{a^2 d \sqrt {a+c x^2}}+\frac {e^2 (a e+c d x)}{a d^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {e \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{a c d^2}-\frac {e^4 \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^2 \left (c d^2+a e^2\right )}\\ &=-\frac {e}{a d^2 \sqrt {a+c x^2}}-\frac {1}{a d x \sqrt {a+c x^2}}-\frac {2 c x}{a^2 d \sqrt {a+c x^2}}+\frac {e^2 (a e+c d x)}{a d^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {e^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )^{3/2}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^2}\\ \end {align*}

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Mathematica [C]  time = 0.40, size = 163, normalized size = 0.84 \begin {gather*} -\frac {\frac {d \left (a+2 c x^2\right )}{a^2 x \sqrt {a+c x^2}}-\frac {e^2 (a e+c d x)}{a \sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {e^4 \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 {e \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c x^2}{a}+1\right )}{a \sqrt {a+c x^2}}}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(d + e*x)*(a + c*x^2)^(3/2)),x]

[Out]

-((-((e^2*(a*e + c*d*x))/(a*(c*d^2 + a*e^2)*Sqrt[a + c*x^2])) + (d*(a + 2*c*x^2))/(a^2*x*Sqrt[a + c*x^2]) + (e
^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(c*d^2 + a*e^2)^(3/2) + (e*Hypergeometric2F1[
-1/2, 1, 1/2, 1 + (c*x^2)/a])/(a*Sqrt[a + c*x^2]))/d^2)

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IntegrateAlgebraic [A]  time = 0.73, size = 242, normalized size = 1.25 \begin {gather*} -\frac {2 e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^2}+\frac {-a^2 e^2-a c d^2-a c d e x-a c e^2 x^2-2 c^2 d^2 x^2}{a^2 d x \sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {2 e^4 \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(d + e*x)*(a + c*x^2)^(3/2)),x]

[Out]

(-(a*c*d^2) - a^2*e^2 - a*c*d*e*x - 2*c^2*d^2*x^2 - a*c*e^2*x^2)/(a^2*d*(c*d^2 + a*e^2)*x*Sqrt[a + c*x^2]) + (
2*e^4*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqrt[c]*d)/Sqrt[-(c*d^2) - a*e^2] + (Sqrt[c]*e*x)/Sqrt[-(c*d^2) - a*e^2]
- (e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/(d^2*(c*d^2 + a*e^2)^2) - (2*e*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sq
rt[a + c*x^2]/Sqrt[a]])/(a^(3/2)*d^2)

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fricas [B]  time = 0.72, size = 1556, normalized size = 8.02

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^2*c*e^4*x^3 + a^3*e^4*x)*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)) + ((c^3*d^4*e + 2
*a*c^2*d^2*e^3 + a^2*c*e^5)*x^3 + (a*c^2*d^4*e + 2*a^2*c*d^2*e^3 + a^3*e^5)*x)*sqrt(a)*log(-(c*x^2 + 2*sqrt(c*
x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(a*c^2*d^5 + 2*a^2*c*d^3*e^2 + a^3*d*e^4 + (2*c^3*d^5 + 3*a*c^2*d^3*e^2 + a^2
*c*d*e^4)*x^2 + (a*c^2*d^4*e + a^2*c*d^2*e^3)*x)*sqrt(c*x^2 + a))/((a^2*c^3*d^6 + 2*a^3*c^2*d^4*e^2 + a^4*c*d^
2*e^4)*x^3 + (a^3*c^2*d^6 + 2*a^4*c*d^4*e^2 + a^5*d^2*e^4)*x), -1/2*(2*(a^2*c*e^4*x^3 + a^3*e^4*x)*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)) - ((c^3*d^4*e + 2*a*c^2*d^2*e^3 + a^2*c*e^5)*x^3 + (a*c^2*d^4*e + 2*a^2*c*d^2*e^3 + a^3*e^5)*x)*sqrt(a)*l
og(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(a*c^2*d^5 + 2*a^2*c*d^3*e^2 + a^3*d*e^4 + (2*c^3*d^5 +
3*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x^2 + (a*c^2*d^4*e + a^2*c*d^2*e^3)*x)*sqrt(c*x^2 + a))/((a^2*c^3*d^6 + 2*a^3*
c^2*d^4*e^2 + a^4*c*d^2*e^4)*x^3 + (a^3*c^2*d^6 + 2*a^4*c*d^4*e^2 + a^5*d^2*e^4)*x), -1/2*(2*((c^3*d^4*e + 2*a
*c^2*d^2*e^3 + a^2*c*e^5)*x^3 + (a*c^2*d^4*e + 2*a^2*c*d^2*e^3 + a^3*e^5)*x)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x
^2 + a)) - (a^2*c*e^4*x^3 + a^3*e^4*x)*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
^5 + 2*a^2*c*d^3*e^2 + a^3*d*e^4 + (2*c^3*d^5 + 3*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x^2 + (a*c^2*d^4*e + a^2*c*d^2*
e^3)*x)*sqrt(c*x^2 + a))/((a^2*c^3*d^6 + 2*a^3*c^2*d^4*e^2 + a^4*c*d^2*e^4)*x^3 + (a^3*c^2*d^6 + 2*a^4*c*d^4*e
^2 + a^5*d^2*e^4)*x), -((a^2*c*e^4*x^3 + a^3*e^4*x)*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)) + ((c^3*d^4*e + 2*a*c^2*d^2*e^3 + a^2*c*e^
5)*x^3 + (a*c^2*d^4*e + 2*a^2*c*d^2*e^3 + a^3*e^5)*x)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (a*c^2*d^5 +
2*a^2*c*d^3*e^2 + a^3*d*e^4 + (2*c^3*d^5 + 3*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x^2 + (a*c^2*d^4*e + a^2*c*d^2*e^3)
*x)*sqrt(c*x^2 + a))/((a^2*c^3*d^6 + 2*a^3*c^2*d^4*e^2 + a^4*c*d^2*e^4)*x^3 + (a^3*c^2*d^6 + 2*a^4*c*d^4*e^2 +
a^5*d^2*e^4)*x)]

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giac [A]  time = 0.25, size = 266, normalized size = 1.37 \begin {gather*} -\frac {\frac {{\left (a c^{3} d^{3} + a^{2} c^{2} d e^{2}\right )} x}{a^{3} c^{2} d^{4} + 2 \, a^{4} c d^{2} e^{2} + a^{5} e^{4}} + \frac {a^{2} c^{2} d^{2} e + a^{3} c e^{3}}{a^{3} c^{2} d^{4} + 2 \, a^{4} c d^{2} e^{2} + a^{5} e^{4}}}{\sqrt {c x^{2} + a}} - \frac {2 \, \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{4}}{{\left (c d^{4} + a d^{2} e^{2}\right )} \sqrt {-c d^{2} - a e^{2}}} - \frac {2 \, \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right ) e}{\sqrt {-a} a d^{2}} + \frac {2 \, \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )} a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*c^3*d^3 + a^2*c^2*d*e^2)*x/(a^3*c^2*d^4 + 2*a^4*c*d^2*e^2 + a^5*e^4) + (a^2*c^2*d^2*e + a^3*c*e^3)/(a^3*c
^2*d^4 + 2*a^4*c*d^2*e^2 + a^5*e^4))/sqrt(c*x^2 + a) - 2*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/
sqrt(-c*d^2 - a*e^2))*e^4/((c*d^4 + a*d^2*e^2)*sqrt(-c*d^2 - a*e^2)) - 2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))
/sqrt(-a))*e/(sqrt(-a)*a*d^2) + 2*sqrt(c)/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)*a*d)

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maple [B]  time = 0.01, size = 363, normalized size = 1.87 \begin {gather*} \frac {c \,e^{2} x}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a d}-\frac {e^{3} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{2}}+\frac {e^{3}}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{2}}-\frac {2 c x}{\sqrt {c \,x^{2}+a}\, a^{2} d}+\frac {e \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{a^{\frac {3}{2}} d^{2}}-\frac {e}{\sqrt {c \,x^{2}+a}\, a \,d^{2}}-\frac {1}{\sqrt {c \,x^{2}+a}\, a d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (e x + d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*(a + c*x**2)**(3/2)*(d + e*x)), x)

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