3.3.59 \(\int \frac {1}{x (d+e x)^2 \sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=179 \[ \frac {e^2 \sqrt {a+c x^2}}{d (d+e x) \left (a e^2+c d^2\right )}+\frac {e \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \sqrt {a e^2+c d^2}}+\frac {c e \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {961, 266, 63, 208, 731, 725, 206} \begin {gather*} \frac {e^2 \sqrt {a+c x^2}}{d (d+e x) \left (a e^2+c d^2\right )}+\frac {e \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \sqrt {a e^2+c d^2}}+\frac {c e \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 {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 731

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

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

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Mathematica [A]  time = 0.22, size = 178, normalized size = 0.99 \begin {gather*} \frac {\frac {d e^2 \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}+\frac {e \left (a e^2+2 c d^2\right ) \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {e \left (a e^2+2 c d^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}-\frac {\log \left (\sqrt {a} \sqrt {a+c x^2}+a\right )}{\sqrt {a}}+\frac {\log (x)}{\sqrt {a}}}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.80, size = 178, normalized size = 0.99 \begin {gather*} \frac {e^2 \sqrt {a+c x^2}}{d (d+e x) \left (a e^2+c d^2\right )}-\frac {2 \sqrt {-a e^2-c d^2} \left (a e^3+2 c d^2 e\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^2}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 1261, normalized size = 7.04

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((2*a*c*d^3*e + a^2*d*e^3 + (2*a*c*d^2*e^2 + a^2*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^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(a)*log(-(c*x
^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(a*c*d^3*e^2 + a^2*d*e^4)*sqrt(c*x^2 + a))/(a*c^2*d^7 + 2*a^2*c
*d^5*e^2 + a^3*d^3*e^4 + (a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x), 1/2*(2*(2*a*c*d^3*e + a^2*d*e^3 + (
2*a*c*d^2*e^2 + a^2*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^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^
3 + a^2*e^5)*x)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(a*c*d^3*e^2 + a^2*d*e^4)*sqrt
(c*x^2 + a))/(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a^3*d^3*e^4 + (a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x), 1/
2*(2*(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-a)*arctan(sqrt(-a)/
sqrt(c*x^2 + a)) + (2*a*c*d^3*e + a^2*d*e^3 + (2*a*c*d^2*e^2 + a^2*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*d^3*e^2 + a^2*d*e^4)*sqrt(c*x^2 + a))/(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a^3*d^3*
e^4 + (a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x), ((2*a*c*d^3*e + a^2*d*e^3 + (2*a*c*d^2*e^2 + a^2*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^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-a)*
arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (a*c*d^3*e^2 + a^2*d*e^4)*sqrt(c*x^2 + a))/(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a
^3*d^3*e^4 + (a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x)]

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giac [A]  time = 0.57, size = 126, normalized size = 0.70 \begin {gather*} {\left (\frac {\sqrt {c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}} d^{2} e^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{c d^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{2} + a d^{3} e^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{2}} - \frac {\sqrt {c} e^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{c d^{3} + a d e^{2}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2)*d^2*e^2*sgn(1/(x*e + d))/(c*d^5*sgn(1/(x*e
+ d))^2 + a*d^3*e^2*sgn(1/(x*e + d))^2) - sqrt(c)*e^2*sgn(1/(x*e + d))/(c*d^3 + a*d*e^2))*e^(-1)

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maple [B]  time = 0.01, size = 364, normalized size = 2.03 \begin {gather*} \frac {c \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}}}}+\frac {\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}}}\, e}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) d}+\frac {\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 )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{2}}-\frac {\ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}\, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^2/a^(1/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+1/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))+1
/d*e/(a*e^2+c*d^2)/(x+d/e)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)+c/(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}{\sqrt {c x^{2} + a} {\left (e x + d\right )}^{2} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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