### 3.2044 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{11/2}} \, dx$$

Optimal. Leaf size=250 $\frac{c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}+\frac{c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 e^{5/2} \left (c d^2-a e^2\right )^{3/2}}-\frac{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^2 (d+e x)^{5/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}$

[Out]

-(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e^2*(d + e*x)^(5/2)) + (c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*
e^2)*x + c*d*e*x^2])/(8*e^2*(c*d^2 - a*e^2)*(d + e*x)^(3/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(
3*e*(d + e*x)^(9/2)) + (c^3*d^3*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])])/(8*e^(5/2)*(c*d^2 - a*e^2)^(3/2))

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Rubi [A]  time = 0.184425, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {662, 672, 660, 205} $\frac{c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}+\frac{c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 e^{5/2} \left (c d^2-a e^2\right )^{3/2}}-\frac{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^2 (d+e x)^{5/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}$

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)^(11/2),x]

[Out]

-(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e^2*(d + e*x)^(5/2)) + (c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*
e^2)*x + c*d*e*x^2])/(8*e^2*(c*d^2 - a*e^2)*(d + e*x)^(3/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(
3*e*(d + e*x)^(9/2)) + (c^3*d^3*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])])/(8*e^(5/2)*(c*d^2 - a*e^2)^(3/2))

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 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + 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)^{11/2}} \, dx &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{(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)^{7/2}} \, dx}{2 e}\\ &=-\frac{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{5/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{\left (c^2 d^2\right ) \int \frac{1}{(d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 e^2}\\ &=-\frac{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{5/2}}+\frac{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{\left (c^3 d^3\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}{16 e^2 \left (c d^2-a e^2\right )}\\ &=-\frac{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{5/2}}+\frac{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{\left (c^3 d^3\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 )}{8 e \left (c d^2-a e^2\right )}\\ &=-\frac{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{5/2}}+\frac{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{c^3 d^3 \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 )}{8 e^{5/2} \left (c d^2-a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0527971, size = 83, normalized size = 0.33 $\frac{2 c^3 d^3 ((d+e x) (a e+c d x))^{5/2} \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^4}$

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)^(11/2),x]

[Out]

(2*c^3*d^3*((a*e + c*d*x)*(d + e*x))^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)
])/(5*(c*d^2 - a*e^2)^4*(d + e*x)^(5/2))

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Maple [B]  time = 0.26, size = 457, normalized size = 1.8 \begin{align*}{\frac{1}{ \left ( 24\,a{e}^{2}-24\,c{d}^{2} \right ){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 ){x}^{3}{c}^{3}{d}^{3}{e}^{3}+9\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{2}{c}^{3}{d}^{4}{e}^{2}+9\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) x{c}^{3}{d}^{5}e+3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{3}{d}^{6}-3\,{x}^{2}{c}^{2}{d}^{2}{e}^{2}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}-14\,xacd{e}^{3}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+8\,x{c}^{2}{d}^{3}e\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}-8\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{a}^{2}{e}^{4}+2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}ac{d}^{2}{e}^{2}+3\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{c}^{2}{d}^{4} \right ) \left ( ex+d \right ) ^{-{\frac{7}{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)^(11/2),x)

[Out]

1/24*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^3*c^3*d
^3*e^3+9*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^2*c^3*d^4*e^2+9*arctanh(e*(c*d*x+a*e)^(1/2)/((
a*e^2-c*d^2)*e)^(1/2))*x*c^3*d^5*e+3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c^3*d^6-3*x^2*c^2*d^
2*e^2*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)-14*x*a*c*d*e^3*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+8*x*c
^2*d^3*e*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)-8*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a^2*e^4+2*((a*e
^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d^2*e^2+3*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^4)/(e*x+d)^
(7/2)/(c*d*x+a*e)^(1/2)/(a*e^2-c*d^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{11}{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)^(11/2),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)^(11/2), x)

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Fricas [B]  time = 2.09629, size = 1968, normalized size = 7.87 \begin{align*} \text{result too large to display} \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)^(11/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(c^3*d^3*e^4*x^4 + 4*c^3*d^4*e^3*x^3 + 6*c^3*d^5*e^2*x^2 + 4*c^3*d^6*e*x + c^3*d^7)*sqrt(-c*d^2*e +
a*e^3)*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(
-c*d^2*e + a*e^3)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(3*c^3*d^6*e - a*c^2*d^4*e^3 - 10*a^2*c*d^2*e^
5 + 8*a^3*e^7 - 3*(c^3*d^4*e^3 - a*c^2*d^2*e^5)*x^2 + 2*(4*c^3*d^5*e^2 - 11*a*c^2*d^3*e^4 + 7*a^2*c*d*e^6)*x)*
sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c^2*d^8*e^3 - 2*a*c*d^6*e^5 + a^2*d^4*e^7 + (c^2*d
^4*e^7 - 2*a*c*d^2*e^9 + a^2*e^11)*x^4 + 4*(c^2*d^5*e^6 - 2*a*c*d^3*e^8 + a^2*d*e^10)*x^3 + 6*(c^2*d^6*e^5 - 2
*a*c*d^4*e^7 + a^2*d^2*e^9)*x^2 + 4*(c^2*d^7*e^4 - 2*a*c*d^5*e^6 + a^2*d^3*e^8)*x), -1/24*(3*(c^3*d^3*e^4*x^4
+ 4*c^3*d^4*e^3*x^3 + 6*c^3*d^5*e^2*x^2 + 4*c^3*d^6*e*x + c^3*d^7)*sqrt(c*d^2*e - a*e^3)*arctan(sqrt(c*d*e*x^2
+ a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d^2*e - a*e^3)*sqrt(e*x + d)/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x
)) + (3*c^3*d^6*e - a*c^2*d^4*e^3 - 10*a^2*c*d^2*e^5 + 8*a^3*e^7 - 3*(c^3*d^4*e^3 - a*c^2*d^2*e^5)*x^2 + 2*(4*
c^3*d^5*e^2 - 11*a*c^2*d^3*e^4 + 7*a^2*c*d*e^6)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/
(c^2*d^8*e^3 - 2*a*c*d^6*e^5 + a^2*d^4*e^7 + (c^2*d^4*e^7 - 2*a*c*d^2*e^9 + a^2*e^11)*x^4 + 4*(c^2*d^5*e^6 - 2
*a*c*d^3*e^8 + a^2*d*e^10)*x^3 + 6*(c^2*d^6*e^5 - 2*a*c*d^4*e^7 + a^2*d^2*e^9)*x^2 + 4*(c^2*d^7*e^4 - 2*a*c*d^
5*e^6 + a^2*d^3*e^8)*x)]

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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)**(11/2),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)^(11/2),x, algorithm="giac")

[Out]

Timed out