3.884 \(\int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx\)

Optimal. Leaf size=109 \[ -\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e} \]

[Out]

-1/4*arctanh(1/2*(-c*e^2*x^2+c*d^2)^(1/2)*2^(1/2)/c^(1/2)/d^(1/2)/(e*x+d)^(1/2))/d^(3/2)/e*2^(1/2)/c^(1/2)-1/2
*(-c*e^2*x^2+c*d^2)^(1/2)/c/d/e/(e*x+d)^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {673, 661, 208} \[ -\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 661

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

Rule 673

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 122, normalized size = 1.12 \[ \frac {-\sqrt {2} \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{\sqrt {2} \sqrt {d} \sqrt {d+e x}}\right )-2 \sqrt {d} (d-e x)}{4 d^{3/2} e \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 300, normalized size = 2.75 \[ \left [\frac {\sqrt {2} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{8 \, {\left (c d^{2} e^{3} x^{2} + 2 \, c d^{3} e^{2} x + c d^{4} e\right )}}, -\frac {\sqrt {2} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{4 \, {\left (c d^{2} e^{3} x^{2} + 2 \, c d^{3} e^{2} x + c d^{4} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(2)*(e^2*x^2 + 2*d*e*x + d^2)*sqrt(c*d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqrt(2)*sqrt(-c*e^
2*x^2 + c*d^2)*sqrt(c*d)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 4*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*
d)/(c*d^2*e^3*x^2 + 2*c*d^3*e^2*x + c*d^4*e), -1/4*(sqrt(2)*(e^2*x^2 + 2*d*e*x + d^2)*sqrt(-c*d)*arctan(sqrt(2
)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(-c*d)*sqrt(e*x + d)/(c*e^2*x^2 - c*d^2)) + 2*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x
+ d)*d)/(c*d^2*e^3*x^2 + 2*c*d^3*e^2*x + c*d^4*e)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c e^{2} x^{2} + c d^{2}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 133, normalized size = 1.22 \[ -\frac {\sqrt {-\left (e^{2} x^{2}-d^{2}\right ) c}\, \left (\sqrt {2}\, c e x \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+\sqrt {2}\, c d \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+2 \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\right )}{4 \left (e x +d \right )^{\frac {3}{2}} \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, c d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(e*x+d)^(3/2)*(-(e^2*x^2-d^2)*c)^(1/2)/c*(2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x*c
*e+c*d*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))+2*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2))/(-(e*x-d)
*c)^(1/2)/e/d/(c*d)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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