3.2002 \(\int \frac{(d+e x)^{5/2}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\)

Optimal. Leaf size=114 \[ \frac{2 \sqrt{d+e x} \left (c d^2-a e^2\right )}{c^2 d^2}-\frac{2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}+\frac{2 (d+e x)^{3/2}}{3 c d} \]

[Out]

(2*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(c^2*d^2) + (2*(d + e*x)^(3/2))/(3*c*d) - (2*(c*d^2 - a*e^2)^(3/2)*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 = 0.0712063, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {626, 50, 63, 208} \[ \frac{2 \sqrt{d+e x} \left (c d^2-a e^2\right )}{c^2 d^2}-\frac{2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}+\frac{2 (d+e x)^{3/2}}{3 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(c^2*d^2) + (2*(d + e*x)^(3/2))/(3*c*d) - (2*(c*d^2 - a*e^2)^(3/2)*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 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)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\int \frac{(d+e x)^{3/2}}{a e+c d x} \, dx\\ &=\frac{2 (d+e x)^{3/2}}{3 c d}+\frac{\left (c d^2-a e^2\right ) \int \frac{\sqrt{d+e x}}{a e+c d x} \, dx}{c d}\\ &=\frac{2 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{c^2 d^2}+\frac{2 (d+e x)^{3/2}}{3 c d}+\frac{\left (c d^2-a e^2\right )^2 \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{c^2 d^2}\\ &=\frac{2 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{c^2 d^2}+\frac{2 (d+e x)^{3/2}}{3 c d}+\frac{\left (2 \left (c d^2-a e^2\right )^2\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 e}\\ &=\frac{2 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{c^2 d^2}+\frac{2 (d+e x)^{3/2}}{3 c d}-\frac{2 \left (c d^2-a e^2\right )^{3/2} \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 [A]  time = 0.0776083, size = 102, normalized size = 0.89 \[ \frac{2 \sqrt{d+e x} \left (c d (4 d+e x)-3 a e^2\right )}{3 c^2 d^2}-\frac{2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(-3*a*e^2 + c*d*(4*d + e*x)))/(3*c^2*d^2) - (2*(c*d^2 - a*e^2)^(3/2)*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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Maple [B]  time = 0.192, size = 211, normalized size = 1.9 \begin{align*}{\frac{2}{3\,cd} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-2\,{\frac{a{e}^{2}\sqrt{ex+d}}{{c}^{2}{d}^{2}}}+2\,{\frac{\sqrt{ex+d}}{c}}+2\,{\frac{{a}^{2}{e}^{4}}{{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 ) }-4\,{\frac{a{e}^{2}}{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 ) }+2\,{\frac{{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x)

[Out]

2/3*(e*x+d)^(3/2)/c/d-2/c^2/d^2*a*e^2*(e*x+d)^(1/2)+2/c*(e*x+d)^(1/2)+2/c^2/d^2/((a*e^2-c*d^2)*c*d)^(1/2)*arct
an((e*x+d)^(1/2)*c*d/((a*e^2-c*d^2)*c*d)^(1/2))*a^2*e^4-4/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))*a*e^2+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))

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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)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93921, size = 533, normalized size = 4.68 \begin{align*} \left [\frac{3 \,{\left (c d^{2} - 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 (c d e x + 4 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt{e x + d}}{3 \, c^{2} d^{2}}, -\frac{2 \,{\left (3 \,{\left (c d^{2} - 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 (c d e x + 4 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt{e x + d}\right )}}{3 \, c^{2} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out