3.239 \(\int \frac{(g x)^m}{(d+e x)^3 (d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac{e (25-4 m) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac{11}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{11 d^{10} g^2 (m+2) \sqrt{d^2-e^2 x^2}}+\frac{(7-4 m) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac{11}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{11 d^9 g (m+1) \sqrt{d^2-e^2 x^2}}+\frac{4 (d-e x) (g x)^{m+1}}{11 g \left (d^2-e^2 x^2\right )^{11/2}} \]

[Out]

(4*(g*x)^(1 + m)*(d - e*x))/(11*g*(d^2 - e^2*x^2)^(11/2)) + ((7 - 4*m)*(g*x)^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*H
ypergeometric2F1[11/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(11*d^9*g*(1 + m)*Sqrt[d^2 - e^2*x^2]) - (e*(25 -
4*m)*(g*x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[11/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(11*
d^10*g^2*(2 + m)*Sqrt[d^2 - e^2*x^2])

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Rubi [A]  time = 0.229159, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {852, 1806, 808, 365, 364} \[ -\frac{e (25-4 m) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac{11}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{11 d^{10} g^2 (m+2) \sqrt{d^2-e^2 x^2}}+\frac{(7-4 m) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac{11}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{11 d^9 g (m+1) \sqrt{d^2-e^2 x^2}}+\frac{4 (d-e x) (g x)^{m+1}}{11 g \left (d^2-e^2 x^2\right )^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(g*x)^m/((d + e*x)^3*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(4*(g*x)^(1 + m)*(d - e*x))/(11*g*(d^2 - e^2*x^2)^(11/2)) + ((7 - 4*m)*(g*x)^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*H
ypergeometric2F1[11/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(11*d^9*g*(1 + m)*Sqrt[d^2 - e^2*x^2]) - (e*(25 -
4*m)*(g*x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[11/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(11*
d^10*g^2*(2 + m)*Sqrt[d^2 - e^2*x^2])

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1806

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, -Simp[((c*x)^(m + 1)*(f + g*x)*(a + b*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int
[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; F
reeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(g x)^m}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\int \frac{(g x)^m (d-e x)^3}{\left (d^2-e^2 x^2\right )^{13/2}} \, dx\\ &=\frac{4 (g x)^{1+m} (d-e x)}{11 g \left (d^2-e^2 x^2\right )^{11/2}}-\frac{\int \frac{(g x)^m \left (-d^3 (7-4 m)+d^2 e (25-4 m) x\right )}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx}{11 d^2}\\ &=\frac{4 (g x)^{1+m} (d-e x)}{11 g \left (d^2-e^2 x^2\right )^{11/2}}+\frac{1}{11} (d (7-4 m)) \int \frac{(g x)^m}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx-\frac{(e (25-4 m)) \int \frac{(g x)^{1+m}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx}{11 g}\\ &=\frac{4 (g x)^{1+m} (d-e x)}{11 g \left (d^2-e^2 x^2\right )^{11/2}}+\frac{\left ((7-4 m) \sqrt{1-\frac{e^2 x^2}{d^2}}\right ) \int \frac{(g x)^m}{\left (1-\frac{e^2 x^2}{d^2}\right )^{11/2}} \, dx}{11 d^9 \sqrt{d^2-e^2 x^2}}-\frac{\left (e (25-4 m) \sqrt{1-\frac{e^2 x^2}{d^2}}\right ) \int \frac{(g x)^{1+m}}{\left (1-\frac{e^2 x^2}{d^2}\right )^{11/2}} \, dx}{11 d^{10} g \sqrt{d^2-e^2 x^2}}\\ &=\frac{4 (g x)^{1+m} (d-e x)}{11 g \left (d^2-e^2 x^2\right )^{11/2}}+\frac{(7-4 m) (g x)^{1+m} \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (\frac{11}{2},\frac{1+m}{2};\frac{3+m}{2};\frac{e^2 x^2}{d^2}\right )}{11 d^9 g (1+m) \sqrt{d^2-e^2 x^2}}-\frac{e (25-4 m) (g x)^{2+m} \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (\frac{11}{2},\frac{2+m}{2};\frac{4+m}{2};\frac{e^2 x^2}{d^2}\right )}{11 d^{10} g^2 (2+m) \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.224017, size = 200, normalized size = 0.93 \[ \frac{x \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^m \left (\frac{d^3 \, _2F_1\left (\frac{13}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{m+1}+e x \left (e x \left (\frac{3 d \, _2F_1\left (\frac{13}{2},\frac{m+3}{2};\frac{m+5}{2};\frac{e^2 x^2}{d^2}\right )}{m+3}-\frac{e x \, _2F_1\left (\frac{13}{2},\frac{m+4}{2};\frac{m+6}{2};\frac{e^2 x^2}{d^2}\right )}{m+4}\right )-\frac{3 d^2 \, _2F_1\left (\frac{13}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{m+2}\right )\right )}{d^{12} \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*x)^m/((d + e*x)^3*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(x*(g*x)^m*Sqrt[1 - (e^2*x^2)/d^2]*((d^3*Hypergeometric2F1[13/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(1 + m)
+ e*x*((-3*d^2*Hypergeometric2F1[13/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(2 + m) + e*x*((3*d*Hypergeometr
ic2F1[13/2, (3 + m)/2, (5 + m)/2, (e^2*x^2)/d^2])/(3 + m) - (e*x*Hypergeometric2F1[13/2, (4 + m)/2, (6 + m)/2,
(e^2*x^2)/d^2])/(4 + m)))))/(d^12*Sqrt[d^2 - e^2*x^2])

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Maple [F]  time = 0.645, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m}}{ \left ( ex+d \right ) ^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

int((g*x)^m/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

integrate((g*x)^m/((-e^2*x^2 + d^2)^(7/2)*(e*x + d)^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-e^{2} x^{2} + d^{2}} \left (g x\right )^{m}}{e^{11} x^{11} + 3 \, d e^{10} x^{10} - d^{2} e^{9} x^{9} - 11 \, d^{3} e^{8} x^{8} - 6 \, d^{4} e^{7} x^{7} + 14 \, d^{5} e^{6} x^{6} + 14 \, d^{6} e^{5} x^{5} - 6 \, d^{7} e^{4} x^{4} - 11 \, d^{8} e^{3} x^{3} - d^{9} e^{2} x^{2} + 3 \, d^{10} e x + d^{11}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(-e^2*x^2 + d^2)*(g*x)^m/(e^11*x^11 + 3*d*e^10*x^10 - d^2*e^9*x^9 - 11*d^3*e^8*x^8 - 6*d^4*e^7*x^
7 + 14*d^5*e^6*x^6 + 14*d^6*e^5*x^5 - 6*d^7*e^4*x^4 - 11*d^8*e^3*x^3 - d^9*e^2*x^2 + 3*d^10*e*x + d^11), 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((g*x)**m/(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((g*x)^m/((-e^2*x^2 + d^2)^(7/2)*(e*x + d)^3), x)