### 3.2019 $$\int \frac{(d+e x)^{13/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx$$

Optimal. Leaf size=186 $\frac{35 e^2 \sqrt{d+e x} \left (c d^2-a e^2\right )}{4 c^4 d^4}-\frac{35 e^2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{9/2} d^{9/2}}-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}$

[Out]

(35*e^2*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(4*c^4*d^4) + (35*e^2*(d + e*x)^(3/2))/(12*c^3*d^3) - (7*e*(d + e*x)^(5
/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(7/2)/(2*c*d*(a*e + c*d*x)^2) - (35*e^2*(c*d^2 - a*e^2)^(3/2)*ArcTa
nh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*c^(9/2)*d^(9/2))

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Rubi [A]  time = 0.137784, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 7, 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} $\frac{35 e^2 \sqrt{d+e x} \left (c d^2-a e^2\right )}{4 c^4 d^4}-\frac{35 e^2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{9/2} d^{9/2}}-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(35*e^2*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(4*c^4*d^4) + (35*e^2*(d + e*x)^(3/2))/(12*c^3*d^3) - (7*e*(d + e*x)^(5
/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(7/2)/(2*c*d*(a*e + c*d*x)^2) - (35*e^2*(c*d^2 - a*e^2)^(3/2)*ArcTa
nh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*c^(9/2)*d^(9/2))

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)^{13/2}}{\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)^{7/2}}{(a e+c d x)^3} \, dx\\ &=-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{(7 e) \int \frac{(d+e x)^{5/2}}{(a e+c d x)^2} \, dx}{4 c d}\\ &=-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{\left (35 e^2\right ) \int \frac{(d+e x)^{3/2}}{a e+c d x} \, dx}{8 c^2 d^2}\\ &=\frac{35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{\left (35 e^2 \left (c d^2-a e^2\right )\right ) \int \frac{\sqrt{d+e x}}{a e+c d x} \, dx}{8 c^3 d^3}\\ &=\frac{35 e^2 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{4 c^4 d^4}+\frac{35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{\left (35 e^2 \left (c d^2-a e^2\right )^2\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 c^4 d^4}\\ &=\frac{35 e^2 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{4 c^4 d^4}+\frac{35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac{\left (35 e \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 )}{4 c^4 d^4}\\ &=\frac{35 e^2 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{4 c^4 d^4}+\frac{35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac{7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{7/2}}{2 c d (a e+c d x)^2}-\frac{35 e^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 )}{4 c^{9/2} d^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0199576, size = 61, normalized size = 0.33 $\frac{2 e^2 (d+e x)^{9/2} \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{9 \left (a e^2-c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(2*e^2*(d + e*x)^(9/2)*Hypergeometric2F1[3, 9/2, 11/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(9*(-(c*d^2) +
a*e^2)^3)

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Maple [B]  time = 0.204, size = 449, normalized size = 2.4 \begin{align*}{\frac{2\,{e}^{2}}{3\,{c}^{3}{d}^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-6\,{\frac{{e}^{4}a\sqrt{ex+d}}{{c}^{4}{d}^{4}}}+6\,{\frac{{e}^{2}\sqrt{ex+d}}{{c}^{3}{d}^{2}}}-{\frac{13\,{e}^{6}{a}^{2}}{4\,{c}^{3}{d}^{3} \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{13\,{e}^{4}a}{2\,{c}^{2}d \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{13\,d{e}^{2}}{4\,c \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{11\,{e}^{8}{a}^{3}}{4\,{c}^{4}{d}^{4} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{33\,{e}^{6}{a}^{2}}{4\,{c}^{3}{d}^{2} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{33\,{e}^{4}a}{4\,{c}^{2} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{11\,{d}^{2}{e}^{2}}{4\,c \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{35\,{e}^{6}{a}^{2}}{4\,{c}^{4}{d}^{4}}\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{35\,{e}^{4}a}{2\,{c}^{3}{d}^{2}}\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{35\,{e}^{2}}{4\,{c}^{2}}\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)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

2/3*e^2*(e*x+d)^(3/2)/c^3/d^3-6*e^4/c^4/d^4*a*(e*x+d)^(1/2)+6*e^2/c^3/d^2*(e*x+d)^(1/2)-13/4*e^6/c^3/d^3/(c*d*
e*x+a*e^2)^2*(e*x+d)^(3/2)*a^2+13/2*e^4/c^2/d/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*a-13/4*e^2/c*d/(c*d*e*x+a*e^2)^2
*(e*x+d)^(3/2)-11/4*e^8/c^4/d^4/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^3+33/4*e^6/c^3/d^2/(c*d*e*x+a*e^2)^2*(e*x+d)
^(1/2)*a^2-33/4*e^4/c^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+11/4*e^2/c*d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)+35/4*
e^6/c^4/d^4/((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^2-35/2*e^4/c^3/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+35/4*e^2/c^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)^(13/2)/(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 = 2.03103, size = 1289, normalized size = 6.93 \begin{align*} \left [\frac{105 \,{\left (a^{2} c d^{2} e^{4} - a^{3} e^{6} +{\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\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^{3} d^{3} e^{3} x^{3} - 6 \, c^{3} d^{6} - 21 \, a c^{2} d^{4} e^{2} + 140 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 8 \,{\left (10 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} -{\left (39 \, c^{3} d^{5} e - 238 \, a c^{2} d^{3} e^{3} + 175 \, a^{2} c d e^{5}\right )} x\right )} \sqrt{e x + d}}{24 \,{\left (c^{6} d^{6} x^{2} + 2 \, a c^{5} d^{5} e x + a^{2} c^{4} d^{4} e^{2}\right )}}, -\frac{105 \,{\left (a^{2} c d^{2} e^{4} - a^{3} e^{6} +{\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\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^{3} d^{3} e^{3} x^{3} - 6 \, c^{3} d^{6} - 21 \, a c^{2} d^{4} e^{2} + 140 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 8 \,{\left (10 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} -{\left (39 \, c^{3} d^{5} e - 238 \, a c^{2} d^{3} e^{3} + 175 \, a^{2} c d e^{5}\right )} x\right )} \sqrt{e x + d}}{12 \,{\left (c^{6} d^{6} x^{2} + 2 \, a c^{5} d^{5} e x + a^{2} c^{4} d^{4} e^{2}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/24*(105*(a^2*c*d^2*e^4 - a^3*e^6 + (c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 + 2*(a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)*s
qrt((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^3*d^3*e^3*x^3 - 6*c^3*d^6 - 21*a*c^2*d^4*e^2 + 140*a^2*c*d^2*e^4 - 105*a^3*e^6 + 8*(10*
c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 - (39*c^3*d^5*e - 238*a*c^2*d^3*e^3 + 175*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c
^6*d^6*x^2 + 2*a*c^5*d^5*e*x + a^2*c^4*d^4*e^2), -1/12*(105*(a^2*c*d^2*e^4 - a^3*e^6 + (c^3*d^4*e^2 - a*c^2*d^
2*e^4)*x^2 + 2*(a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)*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^3*d^3*e^3*x^3 - 6*c^3*d^6 - 21*a*c^2*d^4*e^2 + 140*a^2*c*d^2*e^4
- 105*a^3*e^6 + 8*(10*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 - (39*c^3*d^5*e - 238*a*c^2*d^3*e^3 + 175*a^2*c*d*e^
5)*x)*sqrt(e*x + d))/(c^6*d^6*x^2 + 2*a*c^5*d^5*e*x + a^2*c^4*d^4*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)**(13/2)/(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)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out