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

Optimal. Leaf size=38 \[ -\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{5 c e (d+e x)^{5/2}} \]

[Out]

-2/5*(-c*e^2*x^2+c*d^2)^(5/2)/c/e/(e*x+d)^(5/2)

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Rubi [A]  time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {649} \[ -\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{5 c e (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c*d^2 - c*e^2*x^2)^(5/2))/(5*c*e*(d + e*x)^(5/2))

Rule 649

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))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rubi steps

\begin {align*} \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx &=-\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{5 c e (d+e x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 43, normalized size = 1.13 \[ -\frac {2 c (d-e x)^2 \sqrt {c \left (d^2-e^2 x^2\right )}}{5 e \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(d - e*x)^2*Sqrt[c*(d^2 - e^2*x^2)])/(5*e*Sqrt[d + e*x])

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fricas [A]  time = 0.92, size = 57, normalized size = 1.50 \[ -\frac {2 \, {\left (c e^{2} x^{2} - 2 \, c d e x + c d^{2}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{5 \, {\left (e^{2} x + d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(c*e^2*x^2 - 2*c*d*e*x + c*d^2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)/(e^2*x + d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 36, normalized size = 0.95 \[ -\frac {2 \left (-e x +d \right ) \left (-c \,e^{2} x^{2}+c \,d^{2}\right )^{\frac {3}{2}}}{5 \left (e x +d \right )^{\frac {3}{2}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(-e*x+d)*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(3/2)

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maxima [A]  time = 1.50, size = 39, normalized size = 1.03 \[ -\frac {2 \, {\left (c^{\frac {3}{2}} e^{2} x^{2} - 2 \, c^{\frac {3}{2}} d e x + c^{\frac {3}{2}} d^{2}\right )} \sqrt {-e x + d}}{5 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.55, size = 48, normalized size = 1.26 \[ -\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {2\,c\,d^2}{5\,e}-\frac {4\,c\,d\,x}{5}+\frac {2\,c\,e\,x^2}{5}\right )}{\sqrt {d+e\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c*d^2 - c*e^2*x^2)^(1/2)*((2*c*d^2)/(5*e) - (4*c*d*x)/5 + (2*c*e*x^2)/5))/(d + e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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