### 3.2042 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{7/2}} \, dx$$

Optimal. Leaf size=175 $-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}+\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^2 \sqrt{d+e x}}-\frac{3 c d \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{e^{5/2}}$

[Out]

(3*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e^2*Sqrt[d + e*x]) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x
^2)^(3/2)/(e*(d + e*x)^(5/2)) - (3*c*d*Sqrt[c*d^2 - a*e^2]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*
d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/e^(5/2)

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Rubi [A]  time = 0.11926, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {662, 664, 660, 205} $-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}+\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^2 \sqrt{d+e x}}-\frac{3 c d \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{e^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

(3*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e^2*Sqrt[d + e*x]) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x
^2)^(3/2)/(e*(d + e*x)^(5/2)) - (3*c*d*Sqrt[c*d^2 - a*e^2]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*
d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/e^(5/2)

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}+\frac{(3 c d) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{3/2}} \, dx}{2 e}\\ &=\frac{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^2 \sqrt{d+e x}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\frac{\left (3 c d \left (c d^2-a e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 e^2}\\ &=\frac{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^2 \sqrt{d+e x}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\frac{\left (3 c d \left (c d^2-a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}}\right )}{e}\\ &=\frac{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^2 \sqrt{d+e x}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\frac{3 c d \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{c d^2-a e^2} \sqrt{d+e x}}\right )}{e^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0514579, size = 79, normalized size = 0.45 $\frac{2 c d ((d+e x) (a e+c d x))^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

(2*c*d*((a*e + c*d*x)*(d + e*x))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(
5*(c*d^2 - a*e^2)^2*(d + e*x)^(5/2))

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Maple [B]  time = 0.247, size = 314, normalized size = 1.8 \begin{align*}{\frac{1}{{e}^{2}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( -3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) xacd{e}^{3}+3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) x{c}^{2}{d}^{3}e-3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) ac{d}^{2}{e}^{2}+3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{2}{d}^{4}+2\,xcde\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}-\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}a{e}^{2}+3\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}c{d}^{2} \right ) \left ( ex+d \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cdx+ae}}}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(7/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}}{{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)/(e*x + d)^(7/2), x)

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Fricas [A]  time = 2.01484, size = 867, normalized size = 4.95 \begin{align*} \left [\frac{3 \,{\left (c d e^{2} x^{2} + 2 \, c d^{2} e x + c d^{3}\right )} \sqrt{-\frac{c d^{2} - a e^{2}}{e}} \log \left (-\frac{c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} e \sqrt{-\frac{c d^{2} - a e^{2}}{e}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + 3 \, c d^{2} - a e^{2}\right )} \sqrt{e x + d}}{2 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}}, \frac{3 \,{\left (c d e^{2} x^{2} + 2 \, c d^{2} e x + c d^{3}\right )} \sqrt{\frac{c d^{2} - a e^{2}}{e}} \arctan \left (\frac{\sqrt{e x + d} \sqrt{\frac{c d^{2} - a e^{2}}{e}}}{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}\right ) + \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + 3 \, c d^{2} - a e^{2}\right )} \sqrt{e x + d}}{e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/(e*x+d)**(7/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

Timed out