### 3.862 $$\int (d+e x)^{5/2} \sqrt{c d^2-c e^2 x^2} \, dx$$

Optimal. Leaf size=160 $-\frac{256 d^3 \left (c d^2-c e^2 x^2\right )^{3/2}}{315 c e (d+e x)^{3/2}}-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e \sqrt{d+e x}}-\frac{8 d \sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{21 c e}-\frac{2 (d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}}{9 c e}$

[Out]

(-256*d^3*(c*d^2 - c*e^2*x^2)^(3/2))/(315*c*e*(d + e*x)^(3/2)) - (64*d^2*(c*d^2 - c*e^2*x^2)^(3/2))/(105*c*e*S
qrt[d + e*x]) - (8*d*Sqrt[d + e*x]*(c*d^2 - c*e^2*x^2)^(3/2))/(21*c*e) - (2*(d + e*x)^(3/2)*(c*d^2 - c*e^2*x^2
)^(3/2))/(9*c*e)

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Rubi [A]  time = 0.0729391, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.069, Rules used = {657, 649} $-\frac{256 d^3 \left (c d^2-c e^2 x^2\right )^{3/2}}{315 c e (d+e x)^{3/2}}-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e \sqrt{d+e x}}-\frac{8 d \sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{21 c e}-\frac{2 (d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}}{9 c e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-256*d^3*(c*d^2 - c*e^2*x^2)^(3/2))/(315*c*e*(d + e*x)^(3/2)) - (64*d^2*(c*d^2 - c*e^2*x^2)^(3/2))/(105*c*e*S
qrt[d + e*x]) - (8*d*Sqrt[d + e*x]*(c*d^2 - c*e^2*x^2)^(3/2))/(21*c*e) - (2*(d + e*x)^(3/2)*(c*d^2 - c*e^2*x^2
)^(3/2))/(9*c*e)

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 (d+e x)^{5/2} \sqrt{c d^2-c e^2 x^2} \, dx &=-\frac{2 (d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}}{9 c e}+\frac{1}{3} (4 d) \int (d+e x)^{3/2} \sqrt{c d^2-c e^2 x^2} \, dx\\ &=-\frac{8 d \sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{21 c e}-\frac{2 (d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}}{9 c e}+\frac{1}{21} \left (32 d^2\right ) \int \sqrt{d+e x} \sqrt{c d^2-c e^2 x^2} \, dx\\ &=-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e \sqrt{d+e x}}-\frac{8 d \sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{21 c e}-\frac{2 (d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}}{9 c e}+\frac{1}{105} \left (128 d^3\right ) \int \frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{d+e x}} \, dx\\ &=-\frac{256 d^3 \left (c d^2-c e^2 x^2\right )^{3/2}}{315 c e (d+e x)^{3/2}}-\frac{64 d^2 \left (c d^2-c e^2 x^2\right )^{3/2}}{105 c e \sqrt{d+e x}}-\frac{8 d \sqrt{d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}{21 c e}-\frac{2 (d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}}{9 c e}\\ \end{align*}

Mathematica [A]  time = 0.0629891, size = 75, normalized size = 0.47 $-\frac{2 \left (-156 d^2 e^2 x^2+2 d^3 e x+319 d^4-130 d e^3 x^3-35 e^4 x^4\right ) \sqrt{c \left (d^2-e^2 x^2\right )}}{315 e \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*Sqrt[c*(d^2 - e^2*x^2)]*(319*d^4 + 2*d^3*e*x - 156*d^2*e^2*x^2 - 130*d*e^3*x^3 - 35*e^4*x^4))/(315*e*Sqrt[
d + e*x])

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Maple [A]  time = 0.044, size = 66, normalized size = 0.4 \begin{align*} -{\frac{ \left ( -2\,ex+2\,d \right ) \left ( 35\,{e}^{3}{x}^{3}+165\,d{e}^{2}{x}^{2}+321\,{d}^{2}xe+319\,{d}^{3} \right ) }{315\,e}\sqrt{-c{e}^{2}{x}^{2}+c{d}^{2}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(-e*x+d)*(35*e^3*x^3+165*d*e^2*x^2+321*d^2*e*x+319*d^3)*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(1/2)

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Maxima [A]  time = 1.10864, size = 111, normalized size = 0.69 \begin{align*} \frac{2 \,{\left (35 \, \sqrt{c} e^{4} x^{4} + 130 \, \sqrt{c} d e^{3} x^{3} + 156 \, \sqrt{c} d^{2} e^{2} x^{2} - 2 \, \sqrt{c} d^{3} e x - 319 \, \sqrt{c} 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)^(5/2)*(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*sqrt(c)*e^4*x^4 + 130*sqrt(c)*d*e^3*x^3 + 156*sqrt(c)*d^2*e^2*x^2 - 2*sqrt(c)*d^3*e*x - 319*sqrt(c)*
d^4)*(e*x + d)*sqrt(-e*x + d)/(e^2*x + d*e)

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Fricas [A]  time = 2.08586, size = 174, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (35 \, e^{4} x^{4} + 130 \, d e^{3} x^{3} + 156 \, d^{2} e^{2} x^{2} - 2 \, d^{3} e x - 319 \, 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)^(5/2)*(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*e^4*x^4 + 130*d*e^3*x^3 + 156*d^2*e^2*x^2 - 2*d^3*e*x - 319*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 \sqrt{- c \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{\frac{5}{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)*(-c*e**2*x**2+c*d**2)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-c e^{2} x^{2} + c d^{2}}{\left (e x + d\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")

[Out]

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