3.9.89 \(\int \frac {(d+e x)^{3/2}}{(c d^2-c e^2 x^2)^{3/2}} \, dx\) [889]

Optimal. Leaf size=36 \[ \frac {2 \sqrt {d+e x}}{c e \sqrt {c d^2-c e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, 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 = {663} \begin {gather*} \frac {2 \sqrt {d+e x}}{c e \sqrt {c d^2-c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 663

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

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Mathematica [A]
time = 0.25, size = 35, normalized size = 0.97 \begin {gather*} \frac {2 \sqrt {d+e x}}{c e \sqrt {c \left (d^2-e^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.56, size = 40, normalized size = 1.11

method	result	size
gosper	\(\frac {2 \left (-e x +d \right ) \left (e x +d \right )^{\frac {3}{2}}}{e \left (-x^{2} c \,e^{2}+c \,d^{2}\right )^{\frac {3}{2}}}\)	\(36\)
default	\(\frac {2 \sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}}{\sqrt {e x +d}\, c^{2} \left (-e x +d \right ) e}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 16, normalized size = 0.44 \begin {gather*} \frac {2 \, e^{\left (-1\right )}}{\sqrt {-x e + d} c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^(-1)/(sqrt(-x*e + d)*c^(3/2))

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Fricas [A]
time = 2.10, size = 48, normalized size = 1.33 \begin {gather*} -\frac {2 \, \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d}}{c^{2} x^{2} e^{3} - c^{2} d^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.28, size = 39, normalized size = 1.08 \begin {gather*} -\frac {\sqrt {2} e^{\left (-1\right )}}{\sqrt {c d} c} + \frac {2 \, e^{\left (-1\right )}}{\sqrt {-{\left (x e + d\right )} c + 2 \, c d} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.65, size = 50, normalized size = 1.39 \begin {gather*} \frac {2\,\sqrt {c\,d^2-c\,e^2\,x^2}\,\sqrt {d+e\,x}}{e\,\left (c^2\,d^2-c^2\,e^2\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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