3.2494 \(\int \frac{1}{(d+e x)^2 (a+b x+c x^2)^{7/3}} \, dx\)

Optimal. Leaf size=189 \[ -\frac{3 \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{7/3} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{7/3} F_1\left (\frac{17}{3};\frac{7}{3},\frac{7}{3};\frac{20}{3};\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{272\ 2^{2/3} e (d+e x) \left (a+b x+c x^2\right )^{7/3}} \]

[Out]

(-3*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(7/3)*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x
)))^(7/3)*AppellF1[17/3, 7/3, 7/3, 20/3, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)), (2*d - ((b + Sqr
t[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(272*2^(2/3)*e*(d + e*x)*(a + b*x + c*x^2)^(7/3))

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Rubi [A]  time = 0.0913116, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {758, 133} \[ -\frac{3 \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{7/3} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{7/3} F_1\left (\frac{17}{3};\frac{7}{3},\frac{7}{3};\frac{20}{3};\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{272\ 2^{2/3} e (d+e x) \left (a+b x+c x^2\right )^{7/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*(a + b*x + c*x^2)^(7/3)),x]

[Out]

(-3*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(7/3)*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x
)))^(7/3)*AppellF1[17/3, 7/3, 7/3, 20/3, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)), (2*d - ((b + Sqr
t[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(272*2^(2/3)*e*(d + e*x)*(a + b*x + c*x^2)^(7/3))

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, -Dist[((1/(d + e*x))^(2*p)*(a + b*x + c*x^2)^p)/(e*((e*(b - q + 2*c*x))/(2*c*(d + e*x)))^p*((e*(b + q +
2*c*x))/(2*c*(d + e*x)))^p), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - (e*(b - q))/(2*c))*x, x]^p*Simp[1 - (d
- (e*(b + q))/(2*c))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p] && ILtQ[m, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^2 \left (a+b x+c x^2\right )^{7/3}} \, dx &=-\frac{\left (\left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{7/3} \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{7/3}\right ) \operatorname{Subst}\left (\int \frac{x^{14/3}}{\left (1-\frac{1}{2} \left (2 d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{c}\right ) x\right )^{7/3} \left (1-\frac{1}{2} \left (2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}\right ) x\right )^{7/3}} \, dx,x,\frac{1}{d+e x}\right )}{16\ 2^{2/3} e \left (\frac{1}{d+e x}\right )^{14/3} \left (a+b x+c x^2\right )^{7/3}}\\ &=-\frac{3 \left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{7/3} \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{7/3} F_1\left (\frac{17}{3};\frac{7}{3},\frac{7}{3};\frac{20}{3};\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{272\ 2^{2/3} e (d+e x) \left (a+b x+c x^2\right )^{7/3}}\\ \end{align*}

Mathematica [A]  time = 1.0946, size = 190, normalized size = 1.01 \[ -\frac{3 e^3 \sqrt [3]{\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}} \sqrt [3]{\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}} F_1\left (\frac{17}{3};\frac{7}{3},\frac{7}{3};\frac{20}{3};\frac{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 c d-b e+\sqrt{b^2-4 a c} e}{2 c d+2 c e x}\right )}{17\ 2^{2/3} c^2 (d+e x)^5 \sqrt [3]{a+x (b+c x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((d + e*x)^2*(a + b*x + c*x^2)^(7/3)),x]

[Out]

(-3*e^3*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(1/3)*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d +
e*x)))^(1/3)*AppellF1[17/3, 7/3, 7/3, 20/3, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)), (2*c*d - b*e
+ Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x)])/(17*2^(2/3)*c^2*(d + e*x)^5*(a + x*(b + c*x))^(1/3))

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Maple [F]  time = 1.274, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( ex+d \right ) ^{2}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{7}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(c*x^2+b*x+a)^(7/3),x)

[Out]

int(1/(e*x+d)^2/(c*x^2+b*x+a)^(7/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{7}{3}}{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(7/3),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x + a)^(7/3)*(e*x + d)^2), x)

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Fricas [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(1/(e*x+d)^2/(c*x^2+b*x+a)^(7/3),x, algorithm="fricas")

[Out]

Timed out

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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(1/(e*x+d)**2/(c*x**2+b*x+a)**(7/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{7}{3}}{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(7/3),x, algorithm="giac")

[Out]

integrate(1/((c*x^2 + b*x + a)^(7/3)*(e*x + d)^2), x)