### 3.2022 $$\int \frac{(d+e x)^{7/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx$$

Optimal. Leaf size=144 $\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}}-\frac{e \sqrt{d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{\sqrt{d+e x}}{2 c d (a e+c d x)^2}$

[Out]

-Sqrt[d + e*x]/(2*c*d*(a*e + c*d*x)^2) - (e*Sqrt[d + e*x])/(4*c*d*(c*d^2 - a*e^2)*(a*e + c*d*x)) + (e^2*ArcTan
h[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*c^(3/2)*d^(3/2)*(c*d^2 - a*e^2)^(3/2))

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Rubi [A]  time = 0.0882176, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.135, Rules used = {626, 47, 51, 63, 208} $\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}}-\frac{e \sqrt{d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{\sqrt{d+e x}}{2 c d (a e+c d x)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-Sqrt[d + e*x]/(2*c*d*(a*e + c*d*x)^2) - (e*Sqrt[d + e*x])/(4*c*d*(c*d^2 - a*e^2)*(a*e + c*d*x)) + (e^2*ArcTan
h[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*c^(3/2)*d^(3/2)*(c*d^2 - a*e^2)^(3/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac{\sqrt{d+e x}}{(a e+c d x)^3} \, dx\\ &=-\frac{\sqrt{d+e x}}{2 c d (a e+c d x)^2}+\frac{e \int \frac{1}{(a e+c d x)^2 \sqrt{d+e x}} \, dx}{4 c d}\\ &=-\frac{\sqrt{d+e x}}{2 c d (a e+c d x)^2}-\frac{e \sqrt{d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{e^2 \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 c d \left (c d^2-a e^2\right )}\\ &=-\frac{\sqrt{d+e x}}{2 c d (a e+c d x)^2}-\frac{e \sqrt{d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{e \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 d \left (c d^2-a e^2\right )}\\ &=-\frac{\sqrt{d+e x}}{2 c d (a e+c d x)^2}-\frac{e \sqrt{d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}+\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.016056, size = 61, normalized size = 0.42 $\frac{2 e^2 (d+e x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{3 \left (a e^2-c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.201, size = 142, normalized size = 1. \begin{align*}{\frac{{e}^{2}}{4\, \left ( cdex+a{e}^{2} \right ) ^{2} \left ( a{e}^{2}-c{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{2}}{4\, \left ( cdex+a{e}^{2} \right ) ^{2}dc}\sqrt{ex+d}}+{\frac{{e}^{2}}{ \left ( 4\,a{e}^{2}-4\,c{d}^{2} \right ) cd}\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)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

1/4*e^2/(c*d*e*x+a*e^2)^2/(a*e^2-c*d^2)*(e*x+d)^(3/2)-1/4*e^2/(c*d*e*x+a*e^2)^2/d/c*(e*x+d)^(1/2)+1/4*e^2/(a*e
^2-c*d^2)/c/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)^(7/2)/(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 [B]  time = 1.9764, size = 1129, normalized size = 7.84 \begin{align*} \left [\frac{{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt{c^{2} d^{3} - a c d e^{2}} \log \left (\frac{c d e x + 2 \, c d^{2} - a e^{2} + 2 \, \sqrt{c^{2} d^{3} - a c d e^{2}} \sqrt{e x + d}}{c d x + a e}\right ) - 2 \,{\left (2 \, c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4} +{\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} c^{4} d^{6} e^{2} - 2 \, a^{3} c^{3} d^{4} e^{4} + a^{4} c^{2} d^{2} e^{6} +{\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} x^{2} + 2 \,{\left (a c^{5} d^{7} e - 2 \, a^{2} c^{4} d^{5} e^{3} + a^{3} c^{3} d^{3} e^{5}\right )} x\right )}}, -\frac{{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt{-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac{\sqrt{-c^{2} d^{3} + a c d e^{2}} \sqrt{e x + d}}{c d e x + c d^{2}}\right ) +{\left (2 \, c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4} +{\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} c^{4} d^{6} e^{2} - 2 \, a^{3} c^{3} d^{4} e^{4} + a^{4} c^{2} d^{2} e^{6} +{\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} x^{2} + 2 \,{\left (a c^{5} d^{7} e - 2 \, a^{2} c^{4} d^{5} e^{3} + a^{3} c^{3} d^{3} e^{5}\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/8*((c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt(c^2*d^3 - a*c*d*e^2)*log((c*d*e*x + 2*c*d^2 - a*e^2 + 2
*sqrt(c^2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(c*d*x + a*e)) - 2*(2*c^3*d^5 - 3*a*c^2*d^3*e^2 + a^2*c*d*e^4 + (c^3
*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/(a^2*c^4*d^6*e^2 - 2*a^3*c^3*d^4*e^4 + a^4*c^2*d^2*e^6 + (c^6*d^8 -
2*a*c^5*d^6*e^2 + a^2*c^4*d^4*e^4)*x^2 + 2*(a*c^5*d^7*e - 2*a^2*c^4*d^5*e^3 + a^3*c^3*d^3*e^5)*x), -1/4*((c^2*
d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2)*arctan(sqrt(-c^2*d^3 + a*c*d*e^2)*sqrt(e*x +
d)/(c*d*e*x + c*d^2)) + (2*c^3*d^5 - 3*a*c^2*d^3*e^2 + a^2*c*d*e^4 + (c^3*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x
+ d))/(a^2*c^4*d^6*e^2 - 2*a^3*c^3*d^4*e^4 + a^4*c^2*d^2*e^6 + (c^6*d^8 - 2*a*c^5*d^6*e^2 + a^2*c^4*d^4*e^4)*x
^2 + 2*(a*c^5*d^7*e - 2*a^2*c^4*d^5*e^3 + a^3*c^3*d^3*e^5)*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((e*x+d)**(7/2)/(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)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out