3.2070 \(\int \frac{1}{(d+e x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=269 \[ -\frac{15 c^2 d^2 \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{15 c^2 d^2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{7/2}}+\frac{5 c d}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[Out]

1/(2*(c*d^2 - a*e^2)*(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*c*d)/(4*(c*d^2 - a*e^2)
^2*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (15*c^2*d^2*Sqrt[d + e*x])/(4*(c*d^2 - a*e^2)^
3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (15*c^2*d^2*Sqrt[e]*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*(c*d^2 - a*e^2)^(7/2))

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Rubi [A]  time = 0.17749, antiderivative size = 269, 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 = {672, 666, 660, 205} \[ -\frac{15 c^2 d^2 \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{15 c^2 d^2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{7/2}}+\frac{5 c d}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

1/(2*(c*d^2 - a*e^2)*(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*c*d)/(4*(c*d^2 - a*e^2)
^2*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (15*c^2*d^2*Sqrt[d + e*x])/(4*(c*d^2 - a*e^2)^
3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (15*c^2*d^2*Sqrt[e]*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*(c*d^2 - a*e^2)^(7/2))

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 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), 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] && LtQ[p, -1] && LtQ[0, m, 1] && 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{1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{1}{2 \left (c d^2-a e^2\right ) (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{(5 c d) \int \frac{1}{\sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=\frac{1}{2 \left (c d^2-a e^2\right ) (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d}{4 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (15 c^2 d^2\right ) \int \frac{\sqrt{d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac{1}{2 \left (c d^2-a e^2\right ) (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d}{4 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{15 c^2 d^2 \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (15 c^2 d^2 e\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 \left (c d^2-a e^2\right )^3}\\ &=\frac{1}{2 \left (c d^2-a e^2\right ) (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d}{4 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{15 c^2 d^2 \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (15 c^2 d^2 e^2\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 \left (c d^2-a e^2\right )^3}\\ &=\frac{1}{2 \left (c d^2-a e^2\right ) (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d}{4 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{15 c^2 d^2 \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{15 c^2 d^2 \sqrt{e} \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 \left (c d^2-a e^2\right )^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0272629, size = 81, normalized size = 0.3 \[ -\frac{2 c^2 d^2 \sqrt{d+e x} \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{\left (c d^2-a e^2\right )^3 \sqrt{(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

(-2*c^2*d^2*Sqrt[d + e*x]*Hypergeometric2F1[-1/2, 3, 1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/((c*d^2 - a*e
^2)^3*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [A]  time = 0.248, size = 384, normalized size = 1.4 \begin{align*} -{\frac{1}{ \left ( 4\,cdx+4\,ae \right ) \left ( a{e}^{2}-c{d}^{2} \right ) ^{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 ) \sqrt{cdx+ae}{x}^{2}{c}^{2}{d}^{2}{e}^{3}+30\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}x{c}^{2}{d}^{3}{e}^{2}+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}{c}^{2}{d}^{4}e-15\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{2}{c}^{2}{d}^{2}{e}^{2}-5\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}xacd{e}^{3}-25\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}x{c}^{2}{d}^{3}e+2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{a}^{2}{e}^{4}-9\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}ac{d}^{2}{e}^{2}-8\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{c}^{2}{d}^{4} \right ) \left ( ex+d \right ) ^{-{\frac{5}{2}}}{\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(1/(e*x+d)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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))*(c*d*x+a
*e)^(1/2)*x^2*c^2*d^2*e^3+30*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*x*c^2*d^3*
e^2+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*c^2*d^4*e-15*((a*e^2-c*d^2)*e)^(
1/2)*x^2*c^2*d^2*e^2-5*((a*e^2-c*d^2)*e)^(1/2)*x*a*c*d*e^3-25*((a*e^2-c*d^2)*e)^(1/2)*x*c^2*d^3*e+2*((a*e^2-c*
d^2)*e)^(1/2)*a^2*e^4-9*((a*e^2-c*d^2)*e)^(1/2)*a*c*d^2*e^2-8*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^4)/(e*x+d)^(5/2)/(
c*d*x+a*e)/(a*e^2-c*d^2)^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{1}{{\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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)*(e*x + d)^(3/2)), x)

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Fricas [B]  time = 2.47221, size = 2273, normalized size = 8.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*(c^3*d^3*e^3*x^4 + a*c^2*d^5*e + (3*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^3 + 3*(c^3*d^5*e + a*c^2*d^3*e^3)*
x^2 + (c^3*d^6 + 3*a*c^2*d^4*e^2)*x)*sqrt(-e/(c*d^2 - a*e^2))*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)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-e/(c*d^2 - a*e^2)))/(e^2
*x^2 + 2*d*e*x + d^2)) - 2*(15*c^2*d^2*e^2*x^2 + 8*c^2*d^4 + 9*a*c*d^2*e^2 - 2*a^2*e^4 + 5*(5*c^2*d^3*e + 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))/(a*c^3*d^9*e - 3*a^2*c^2*d^7*e^3 + 3*a^3*
c*d^5*e^5 - a^4*d^3*e^7 + (c^4*d^7*e^3 - 3*a*c^3*d^5*e^5 + 3*a^2*c^2*d^3*e^7 - a^3*c*d*e^9)*x^4 + (3*c^4*d^8*e
^2 - 8*a*c^3*d^6*e^4 + 6*a^2*c^2*d^4*e^6 - a^4*e^10)*x^3 + 3*(c^4*d^9*e - 2*a*c^3*d^7*e^3 + 2*a^3*c*d^3*e^7 -
a^4*d*e^9)*x^2 + (c^4*d^10 - 6*a^2*c^2*d^6*e^4 + 8*a^3*c*d^4*e^6 - 3*a^4*d^2*e^8)*x), -1/4*(15*(c^3*d^3*e^3*x^
4 + a*c^2*d^5*e + (3*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^3 + 3*(c^3*d^5*e + a*c^2*d^3*e^3)*x^2 + (c^3*d^6 + 3*a*c^2
*d^4*e^2)*x)*sqrt(e/(c*d^2 - a*e^2))*arctan(-sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2 - a*e^2)*sqrt(
e*x + d)*sqrt(e/(c*d^2 - a*e^2))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) + (15*c^2*d^2*e^2*x^2 + 8*c^2*
d^4 + 9*a*c*d^2*e^2 - 2*a^2*e^4 + 5*(5*c^2*d^3*e + a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*s
qrt(e*x + d))/(a*c^3*d^9*e - 3*a^2*c^2*d^7*e^3 + 3*a^3*c*d^5*e^5 - a^4*d^3*e^7 + (c^4*d^7*e^3 - 3*a*c^3*d^5*e^
5 + 3*a^2*c^2*d^3*e^7 - a^3*c*d*e^9)*x^4 + (3*c^4*d^8*e^2 - 8*a*c^3*d^6*e^4 + 6*a^2*c^2*d^4*e^6 - a^4*e^10)*x^
3 + 3*(c^4*d^9*e - 2*a*c^3*d^7*e^3 + 2*a^3*c*d^3*e^7 - a^4*d*e^9)*x^2 + (c^4*d^10 - 6*a^2*c^2*d^6*e^4 + 8*a^3*
c*d^4*e^6 - 3*a^4*d^2*e^8)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac{3}{2}} \left (d + e x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(3/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral(1/(((d + e*x)*(a*e + c*d*x))**(3/2)*(d + e*x)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 2\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

[undef, undef, undef, undef, undef, undef, undef, 2]