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

Optimal. Leaf size=91 \[ -\frac {e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}-\frac {c d \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {731, 725, 206} \begin {gather*} -\frac {e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}-\frac {c d \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 115, normalized size = 1.26 \begin {gather*} -\frac {e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}-\frac {c d \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 {c d \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.50, size = 151, normalized size = 1.66 \begin {gather*} \frac {2 c d \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 )}{\left (a e^2+c d^2\right )^2}-\frac {e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 381, normalized size = 4.19 \begin {gather*} \left [\frac {{\left (c d e x + c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, -\frac {{\left (c d e x + c d^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 210, normalized size = 2.31 \begin {gather*} -\frac {c d \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}}}\, e}-\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}}}}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(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)-1/e*c*d/(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 [A]  time = 1.51, size = 93, normalized size = 1.02 \begin {gather*} -\frac {\sqrt {c x^{2} + a}}{c d^{2} x + a e^{2} x + \frac {c d^{3}}{e} + a d e} + \frac {c d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(c*x^2 + a)/(c*d^2*x + a*e^2*x + c*d^3/e + a*d*e) + c*d*arcsinh(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqr
t(a*c)*abs(e*x + d)))/((a + c*d^2/e^2)^(3/2)*e^3)

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

[Out]

int(1/((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}{\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/(e*x+d)**2/(c*x**2+a)**(1/2),x)

[Out]

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

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