### 3.871 $$\int \sqrt{d+e x} (c d^2-c e^2 x^2)^{3/2} \, dx$$

Optimal. Leaf size=119 $-\frac{2 \left (c d^2-c e^2 x^2\right )^{5/2}}{9 c e \sqrt{d+e x}}-\frac{16 d \left (c d^2-c e^2 x^2\right )^{5/2}}{63 c e (d+e x)^{3/2}}-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{315 c e (d+e x)^{5/2}}$

[Out]

(-64*d^2*(c*d^2 - c*e^2*x^2)^(5/2))/(315*c*e*(d + e*x)^(5/2)) - (16*d*(c*d^2 - c*e^2*x^2)^(5/2))/(63*c*e*(d +
e*x)^(3/2)) - (2*(c*d^2 - c*e^2*x^2)^(5/2))/(9*c*e*Sqrt[d + e*x])

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Rubi [A]  time = 0.0498016, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.069, Rules used = {657, 649} $-\frac{2 \left (c d^2-c e^2 x^2\right )^{5/2}}{9 c e \sqrt{d+e x}}-\frac{16 d \left (c d^2-c e^2 x^2\right )^{5/2}}{63 c e (d+e x)^{3/2}}-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{315 c e (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-64*d^2*(c*d^2 - c*e^2*x^2)^(5/2))/(315*c*e*(d + e*x)^(5/2)) - (16*d*(c*d^2 - c*e^2*x^2)^(5/2))/(63*c*e*(d +
e*x)^(3/2)) - (2*(c*d^2 - c*e^2*x^2)^(5/2))/(9*c*e*Sqrt[d + e*x])

Rule 657

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

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 \sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2} \, dx &=-\frac{2 \left (c d^2-c e^2 x^2\right )^{5/2}}{9 c e \sqrt{d+e x}}+\frac{1}{9} (8 d) \int \frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt{d+e x}} \, dx\\ &=-\frac{16 d \left (c d^2-c e^2 x^2\right )^{5/2}}{63 c e (d+e x)^{3/2}}-\frac{2 \left (c d^2-c e^2 x^2\right )^{5/2}}{9 c e \sqrt{d+e x}}+\frac{1}{63} \left (32 d^2\right ) \int \frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx\\ &=-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{5/2}}{315 c e (d+e x)^{5/2}}-\frac{16 d \left (c d^2-c e^2 x^2\right )^{5/2}}{63 c e (d+e x)^{3/2}}-\frac{2 \left (c d^2-c e^2 x^2\right )^{5/2}}{9 c e \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.0475711, size = 62, normalized size = 0.52 $-\frac{2 c (d-e x)^2 \left (107 d^2+110 d e x+35 e^2 x^2\right ) \sqrt{c \left (d^2-e^2 x^2\right )}}{315 e \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*c*(d - e*x)^2*Sqrt[c*(d^2 - e^2*x^2)]*(107*d^2 + 110*d*e*x + 35*e^2*x^2))/(315*e*Sqrt[d + e*x])

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Maple [A]  time = 0.046, size = 55, normalized size = 0.5 \begin{align*} -{\frac{ \left ( -2\,ex+2\,d \right ) \left ( 35\,{e}^{2}{x}^{2}+110\,dex+107\,{d}^{2} \right ) }{315\,e} \left ( -c{e}^{2}{x}^{2}+c{d}^{2} \right ) ^{{\frac{3}{2}}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(-e*x+d)*(35*e^2*x^2+110*d*e*x+107*d^2)*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(3/2)

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Maxima [A]  time = 1.21692, size = 111, normalized size = 0.93 \begin{align*} -\frac{2 \,{\left (35 \, c^{\frac{3}{2}} e^{4} x^{4} + 40 \, c^{\frac{3}{2}} d e^{3} x^{3} - 78 \, c^{\frac{3}{2}} d^{2} e^{2} x^{2} - 104 \, c^{\frac{3}{2}} d^{3} e x + 107 \, c^{\frac{3}{2}} d^{4}\right )}{\left (e x + d\right )} \sqrt{-e x + d}}{315 \,{\left (e^{2} x + d e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*c^(3/2)*e^4*x^4 + 40*c^(3/2)*d*e^3*x^3 - 78*c^(3/2)*d^2*e^2*x^2 - 104*c^(3/2)*d^3*e*x + 107*c^(3/2)
*d^4)*(e*x + d)*sqrt(-e*x + d)/(e^2*x + d*e)

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Fricas [A]  time = 2.25751, size = 189, normalized size = 1.59 \begin{align*} -\frac{2 \,{\left (35 \, c e^{4} x^{4} + 40 \, c d e^{3} x^{3} - 78 \, c d^{2} e^{2} x^{2} - 104 \, c d^{3} e x + 107 \, c d^{4}\right )} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}}{315 \,{\left (e^{2} x + d e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*c*e^4*x^4 + 40*c*d*e^3*x^3 - 78*c*d^2*e^2*x^2 - 104*c*d^3*e*x + 107*c*d^4)*sqrt(-c*e^2*x^2 + c*d^2)
*sqrt(e*x + d)/(e^2*x + d*e)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac{3}{2}} \sqrt{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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