### 3.2053 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{11/2}} \, dx$$

Optimal. Leaf size=236 $\frac{15 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^3 \sqrt{d+e x}}-\frac{15 c^2 d^2 \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 e^{7/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}$

[Out]

(15*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e^3*Sqrt[d + e*x]) - (5*c*d*(a*d*e + (c*d^2 + a*e^
2)*x + c*d*e*x^2)^(3/2))/(4*e^2*(d + e*x)^(5/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(2*e*(d + e*x
)^(9/2)) - (15*c^2*d^2*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])])/(4*e^(7/2))

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Rubi [A]  time = 0.154452, antiderivative size = 236, 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, 664, 660, 205} $\frac{15 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^3 \sqrt{d+e x}}-\frac{15 c^2 d^2 \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 e^{7/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}$

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)^(11/2),x]

[Out]

(15*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e^3*Sqrt[d + e*x]) - (5*c*d*(a*d*e + (c*d^2 + a*e^
2)*x + c*d*e*x^2)^(3/2))/(4*e^2*(d + e*x)^(5/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(2*e*(d + e*x
)^(9/2)) - (15*c^2*d^2*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])])/(4*e^(7/2))

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 )^{5/2}}{(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 )^{5/2}}{2 e (d+e x)^{9/2}}+\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)^{7/2}} \, dx}{4 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}}{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 )^{5/2}}{2 e (d+e x)^{9/2}}+\frac{\left (15 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)^{3/2}} \, dx}{8 e^2}\\ &=\frac{15 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt{d+e x}}-\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}}{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 )^{5/2}}{2 e (d+e x)^{9/2}}-\frac{\left (15 c^2 d^2 \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}{8 e^3}\\ &=\frac{15 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt{d+e x}}-\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}}{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 )^{5/2}}{2 e (d+e x)^{9/2}}-\frac{\left (15 c^2 d^2 \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 )}{4 e^2}\\ &=\frac{15 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt{d+e x}}-\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}}{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 )^{5/2}}{2 e (d+e x)^{9/2}}-\frac{15 c^2 d^2 \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 )}{4 e^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0730468, size = 83, normalized size = 0.35 $\frac{2 c^2 d^2 ((d+e x) (a e+c d x))^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{7 (d+e x)^{7/2} \left (c d^2-a e^2\right )^3}$

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)^(11/2),x]

[Out]

(2*c^2*d^2*((a*e + c*d*x)*(d + e*x))^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)
])/(7*(c*d^2 - a*e^2)^3*(d + e*x)^(7/2))

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Maple [B]  time = 0.286, size = 527, normalized size = 2.2 \begin{align*} -{\frac{1}{4\,{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}^{2}a{c}^{2}{d}^{2}{e}^{4}-15\,{\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}+30\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) xa{c}^{2}{d}^{3}{e}^{3}-30\,{\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 ) a{c}^{2}{d}^{4}{e}^{2}-15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{3}{d}^{6}-8\,{x}^{2}{c}^{2}{d}^{2}{e}^{2}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+9\,xacd{e}^{3}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}-25\,x{c}^{2}{d}^{3}e\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{a}^{2}{e}^{4}+5\,\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{5}{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)^(11/2),x)

[Out]

-1/4*(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^2*a*c^
2*d^2*e^4-15*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+30*arctanh(e*(c*d*x+a*e)^(1/
2)/((a*e^2-c*d^2)*e)^(1/2))*x*a*c^2*d^3*e^3-30*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))*a*c^2*d^4*e^2-15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2
-c*d^2)*e)^(1/2))*c^3*d^6-8*x^2*c^2*d^2*e^2*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+9*x*a*c*d*e^3*(c*d*x+a*e
)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)-25*x*c^2*d^3*e*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+2*((a*e^2-c*d^2)*e)^(
1/2)*(c*d*x+a*e)^(1/2)*a^2*e^4+5*((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)^(5/2)/(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{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)^(5/2)/(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)^(5/2)/(e*x + d)^(11/2), x)

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Fricas [A]  time = 2.49685, size = 1166, normalized size = 4.94 \begin{align*} \left [\frac{15 \,{\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\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 \,{\left (8 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} +{\left (25 \, c^{2} d^{3} e - 9 \, a c d e^{3}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{8 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}, \frac{15 \,{\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\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 ) +{\left (8 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} +{\left (25 \, c^{2} d^{3} e - 9 \, a c d e^{3}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{4 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}\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)^(5/2)/(e*x+d)^(11/2),x, algorithm="fricas")

[Out]

[1/8*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*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*(8*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 5*a*c*d^2*e^2 - 2*a^2*e^4 +
(25*c^2*d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^6*x^3 + 3*d*e^5
*x^2 + 3*d^2*e^4*x + d^3*e^3), 1/4*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*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)) + (
8*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 5*a*c*d^2*e^2 - 2*a^2*e^4 + (25*c^2*d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 +
a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)]

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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)**(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)^(5/2)/(e*x+d)^(11/2),x, algorithm="giac")

[Out]

Timed out