3.2020 \(\int \frac{(d+e x)^{11/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac{15 e^2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}}-\frac{5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac{15 e^2 \sqrt{d+e x}}{4 c^3 d^3} \]

[Out]

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

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Rubi [A]  time = 0.100075, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {626, 47, 50, 63, 208} \[ -\frac{15 e^2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}}-\frac{5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac{15 e^2 \sqrt{d+e x}}{4 c^3 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

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

Rule 626

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac{(d+e x)^{5/2}}{(a e+c d x)^3} \, dx\\ &=-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac{(5 e) \int \frac{(d+e x)^{3/2}}{(a e+c d x)^2} \, dx}{4 c d}\\ &=-\frac{5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac{\left (15 e^2\right ) \int \frac{\sqrt{d+e x}}{a e+c d x} \, dx}{8 c^2 d^2}\\ &=\frac{15 e^2 \sqrt{d+e x}}{4 c^3 d^3}-\frac{5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac{\left (15 e^2 \left (c d^2-a e^2\right )\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 c^3 d^3}\\ &=\frac{15 e^2 \sqrt{d+e x}}{4 c^3 d^3}-\frac{5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac{\left (15 e \left (c d^2-a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c d^2}{e}+a e+\frac{c d x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 c^3 d^3}\\ &=\frac{15 e^2 \sqrt{d+e x}}{4 c^3 d^3}-\frac{5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac{15 e^2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0181947, size = 61, normalized size = 0.4 \[ \frac{2 e^2 (d+e x)^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{7 \left (a e^2-c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

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

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Maple [B]  time = 0.202, size = 288, normalized size = 1.9 \begin{align*} 2\,{\frac{{e}^{2}\sqrt{ex+d}}{{c}^{3}{d}^{3}}}+{\frac{9\,{e}^{4}a}{4\,{c}^{2}{d}^{2} \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{9\,{e}^{2}}{4\,c \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{e}^{6}{a}^{2}}{4\,{c}^{3}{d}^{3} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{7\,{e}^{4}a}{2\,{c}^{2}d \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{7\,d{e}^{2}}{4\,c \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{15\,{e}^{4}a}{4\,{c}^{3}{d}^{3}}\arctan \left ({cd\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \right ){\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}}+{\frac{15\,{e}^{2}}{4\,{c}^{2}d}\arctan \left ({cd\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \right ){\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

2*e^2*(e*x+d)^(1/2)/c^3/d^3+9/4*e^4/c^2/d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*a-9/4*e^2/c/(c*d*e*x+a*e^2)^2*(e*x
+d)^(3/2)+7/4*e^6/c^3/d^3/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^2-7/2*e^4/c^2/d/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+
7/4*e^2/c*d/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)-15/4*e^4/c^3/d^3/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c*
d/((a*e^2-c*d^2)*c*d)^(1/2))*a+15/4*e^2/c^2/d/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c*d/((a*e^2-c*d^2
)*c*d)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00768, size = 892, normalized size = 5.87 \begin{align*} \left [\frac{15 \,{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt{\frac{c d^{2} - a e^{2}}{c d}} \log \left (\frac{c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt{e x + d} c d \sqrt{\frac{c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \,{\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} -{\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac{15 \,{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt{-\frac{c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac{\sqrt{e x + d} c d \sqrt{-\frac{c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) -{\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} -{\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2 - a*e^
2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(8*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 5*a*c
*d^2*e^2 + 15*a^2*e^4 - (9*c^2*d^3*e - 25*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^
3*d^3*e^2), -1/4*(15*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x
+ d)*c*d*sqrt(-(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (8*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 5*a*c*d^2*e^2 + 15*
a^2*e^4 - (9*c^2*d^3*e - 25*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*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((e*x+d)**(11/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,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((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out