3.2012 \(\int \frac{(d+e x)^{7/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=112 \[ -\frac{3 e \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}-\frac{(d+e x)^{3/2}}{c d (a e+c d x)}+\frac{3 e \sqrt{d+e x}}{c^2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0699287, antiderivative size = 112, 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, 50, 63, 208} \[ -\frac{3 e \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}-\frac{(d+e x)^{3/2}}{c d (a e+c d x)}+\frac{3 e \sqrt{d+e x}}{c^2 d^2} \]

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)^2,x]

[Out]

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

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)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac{(d+e x)^{3/2}}{(a e+c d x)^2} \, dx\\ &=-\frac{(d+e x)^{3/2}}{c d (a e+c d x)}+\frac{(3 e) \int \frac{\sqrt{d+e x}}{a e+c d x} \, dx}{2 c d}\\ &=\frac{3 e \sqrt{d+e x}}{c^2 d^2}-\frac{(d+e x)^{3/2}}{c d (a e+c d x)}+\frac{\left (3 e \left (c d^2-a e^2\right )\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{2 c^2 d^2}\\ &=\frac{3 e \sqrt{d+e x}}{c^2 d^2}-\frac{(d+e x)^{3/2}}{c d (a e+c d x)}+\frac{\left (3 \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 )}{c^2 d^2}\\ &=\frac{3 e \sqrt{d+e x}}{c^2 d^2}-\frac{(d+e x)^{3/2}}{c d (a e+c d x)}-\frac{3 e \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 )}{c^{5/2} d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0149621, size = 59, normalized size = 0.53 \[ \frac{2 e (d+e x)^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{5 \left (a e^2-c d^2\right )^2} \]

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)^2,x]

[Out]

(2*e*(d + e*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(5*(-(c*d^2) + a*e
^2)^2)

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Maple [A]  time = 0.203, size = 183, normalized size = 1.6 \begin{align*} 2\,{\frac{e\sqrt{ex+d}}{{c}^{2}{d}^{2}}}+{\frac{a{e}^{3}}{{c}^{2}{d}^{2} \left ( cdex+a{e}^{2} \right ) }\sqrt{ex+d}}-{\frac{e}{c \left ( cdex+a{e}^{2} \right ) }\sqrt{ex+d}}-3\,{\frac{a{e}^{3}}{{c}^{2}{d}^{2}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}\arctan \left ({\frac{\sqrt{ex+d}cd}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}} \right ) }+3\,{\frac{e}{c\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}\arctan \left ({\frac{\sqrt{ex+d}cd}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}} \right ) } \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)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96643, size = 576, normalized size = 5.14 \begin{align*} \left [\frac{3 \,{\left (c d e x + a e^{2}\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 (2 \, c d e x - c d^{2} + 3 \, a e^{2}\right )} \sqrt{e x + d}}{2 \,{\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )}}, -\frac{3 \,{\left (c d e x + a e^{2}\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 (2 \, c d e x - c d^{2} + 3 \, a e^{2}\right )} \sqrt{e x + d}}{c^{3} d^{3} x + a c^{2} d^{2} e}\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)^2,x, algorithm="fricas")

[Out]

[1/2*(3*(c*d*e*x + a*e^2)*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*sqr
t((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(2*c*d*e*x - c*d^2 + 3*a*e^2)*sqrt(e*x + d))/(c^3*d^3*x + a*c^2*d
^2*e), -(3*(c*d*e*x + a*e^2)*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)) - (2*c*d*e*x - c*d^2 + 3*a*e^2)*sqrt(e*x + d))/(c^3*d^3*x + a*c^2*d^2*e)]

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

[Out]

Timed out