### 3.2054 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{13/2}} \, dx$$

Optimal. Leaf size=236 $-\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3 (d+e x)^{3/2}}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt{c d^2-a e^2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}$

[Out]

(-5*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*e^3*(d + e*x)^(3/2)) - (5*c*d*(a*d*e + (c*d^2 + a*
e^2)*x + c*d*e*x^2)^(3/2))/(12*e^2*(d + e*x)^(7/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(3*e*(d +
e*x)^(11/2)) + (5*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]*Sq
rt[d + e*x])])/(8*e^(7/2)*Sqrt[c*d^2 - a*e^2])

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Rubi [A]  time = 0.151789, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {662, 660, 205} $-\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3 (d+e x)^{3/2}}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt{c d^2-a e^2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(13/2),x]

[Out]

(-5*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*e^3*(d + e*x)^(3/2)) - (5*c*d*(a*d*e + (c*d^2 + a*
e^2)*x + c*d*e*x^2)^(3/2))/(12*e^2*(d + e*x)^(7/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(3*e*(d +
e*x)^(11/2)) + (5*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]*Sq
rt[d + e*x])])/(8*e^(7/2)*Sqrt[c*d^2 - a*e^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 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 )^{5/2}}{(d+e x)^{13/2}} \, dx &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{(5 c d) \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)^{9/2}} \, dx}{6 e}\\ &=-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{\left (5 c^2 d^2\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{5/2}} \, dx}{8 e^2}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{\left (5 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^3}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{\left (5 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^2}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{5 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^{7/2} \sqrt{c d^2-a e^2}}\\ \end{align*}

Mathematica [A]  time = 0.258554, size = 220, normalized size = 0.93 $\frac{15 c^3 d^3 (d+e x)^3 \sqrt{a e+c d x} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d^2-a e^2}}\right )-\sqrt{e} \sqrt{c d^2-a e^2} \left (2 a^2 c d e^3 (5 d+17 e x)+8 a^3 e^5+a c^2 d^2 e \left (15 d^2+50 d e x+59 e^2 x^2\right )+c^3 d^3 x \left (15 d^2+40 d e x+33 e^2 x^2\right )\right )}{24 e^{7/2} (d+e x)^{5/2} \sqrt{c d^2-a e^2} \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(13/2),x]

[Out]

(-(Sqrt[e]*Sqrt[c*d^2 - a*e^2]*(8*a^3*e^5 + 2*a^2*c*d*e^3*(5*d + 17*e*x) + c^3*d^3*x*(15*d^2 + 40*d*e*x + 33*e
^2*x^2) + a*c^2*d^2*e*(15*d^2 + 50*d*e*x + 59*e^2*x^2))) + 15*c^3*d^3*Sqrt[a*e + c*d*x]*(d + e*x)^3*ArcTan[(Sq
rt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]])/(24*e^(7/2)*Sqrt[c*d^2 - a*e^2]*(d + e*x)^(5/2)*Sqrt[(a*e + c*d
*x)*(d + e*x)])

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Maple [B]  time = 0.247, size = 443, normalized size = 1.9 \begin{align*} -{\frac{1}{24\,{e}^{3}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 15\,{\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}+45\,{\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}+45\,{\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+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{3}{d}^{6}+33\,{x}^{2}{c}^{2}{d}^{2}{e}^{2}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+26\,xacd{e}^{3}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+40\,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}+10\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}ac{d}^{2}{e}^{2}+15\,\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)^(5/2)/(e*x+d)^(13/2),x)

[Out]

-1/24*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(15*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+45*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+45*arctanh(e*(c*d*x+a*e)^(1/2
)/((a*e^2-c*d^2)*e)^(1/2))*x*c^3*d^5*e+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c^3*d^6+33*x^2*
c^2*d^2*e^2*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+26*x*a*c*d*e^3*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)
+40*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+
10*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d^2*e^2+15*((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)/e^3/((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{5}{2}}}{{\left (e x + d\right )}^{\frac{13}{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)^(5/2)/(e*x+d)^(13/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/(e*x + d)^(13/2), x)

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Fricas [B]  time = 1.99794, size = 1709, normalized size = 7.24 \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)^(5/2)/(e*x+d)^(13/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(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*(15*c^3*d^6*e - 5*a*c^2*d^4*e^3 - 2*a^2*c*d^2
*e^5 - 8*a^3*e^7 + 33*(c^3*d^4*e^3 - a*c^2*d^2*e^5)*x^2 + 2*(20*c^3*d^5*e^2 - 7*a*c^2*d^3*e^4 - 13*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*d^6*e^4 - a*d^4*e^6 + (c*d^2*e^8 - a*e^10)*
x^4 + 4*(c*d^3*e^7 - a*d*e^9)*x^3 + 6*(c*d^4*e^6 - a*d^2*e^8)*x^2 + 4*(c*d^5*e^5 - a*d^3*e^7)*x), -1/24*(15*(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)) + (15*c^3*d^6*e - 5*a*c^2*d^4*e^3 - 2*a^2*c*d^2*e^5 - 8*a^3*e^7 + 33*(c^3*d^4*e^3 - a*c^2*d^
2*e^5)*x^2 + 2*(20*c^3*d^5*e^2 - 7*a*c^2*d^3*e^4 - 13*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*d^6*e^4 - a*d^4*e^6 + (c*d^2*e^8 - a*e^10)*x^4 + 4*(c*d^3*e^7 - a*d*e^9)*x^3 + 6*(c*d^4*
e^6 - a*d^2*e^8)*x^2 + 4*(c*d^5*e^5 - a*d^3*e^7)*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)**(5/2)/(e*x+d)**(13/2),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)^(5/2)/(e*x+d)^(13/2),x, algorithm="giac")

[Out]

Timed out