3.3.34 \(\int \frac {(g x)^m (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [234]

Optimal. Leaf size=216 \[ \frac {2 (g x)^{1+m} (d+e x)}{5 d g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {(3-2 m) (g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 g (1+m) \sqrt {d^2-e^2 x^2}}+\frac {2 e (3-m) (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {5}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^5 g^2 (2+m) \sqrt {d^2-e^2 x^2}} \]

[Out]

2/5*(g*x)^(1+m)*(e*x+d)/d/g/(-e^2*x^2+d^2)^(5/2)+1/5*(3-2*m)*(g*x)^(1+m)*hypergeom([5/2, 1/2+1/2*m],[3/2+1/2*m
],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/d^4/g/(1+m)/(-e^2*x^2+d^2)^(1/2)+2/5*e*(3-m)*(g*x)^(2+m)*hypergeom([5/2,
1+1/2*m],[2+1/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/d^5/g^2/(2+m)/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1820, 822, 372, 371} \begin {gather*} \frac {2 (d+e x) (g x)^{m+1}}{5 d g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 e (3-m) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac {5}{2},\frac {m+2}{2};\frac {m+4}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^5 g^2 (m+2) \sqrt {d^2-e^2 x^2}}+\frac {(3-2 m) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac {5}{2},\frac {m+1}{2};\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 g (m+1) \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(g*x)^(1 + m)*(d + e*x))/(5*d*g*(d^2 - e^2*x^2)^(5/2)) + ((3 - 2*m)*(g*x)^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*H
ypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(5*d^4*g*(1 + m)*Sqrt[d^2 - e^2*x^2]) + (2*e*(3 -
m)*(g*x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[5/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(5*d^5*g
^2*(2 + m)*Sqrt[d^2 - e^2*x^2])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 822

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 1820

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)), I
nt[(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]] /;
FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(g x)^m (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 (g x)^{1+m} (d+e x)}{5 d g \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(g x)^m \left (-d^2 (3-2 m)-2 d e (3-m) x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 (g x)^{1+m} (d+e x)}{5 d g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {(2 e (3-m)) \int \frac {(g x)^{1+m}}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d g}-\frac {1}{5} (-3+2 m) \int \frac {(g x)^m}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac {2 (g x)^{1+m} (d+e x)}{5 d g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\left (2 e (3-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 )^{5/2}} \, dx}{5 d^5 g \sqrt {d^2-e^2 x^2}}-\frac {\left ((-3+2 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 )^{5/2}} \, dx}{5 d^4 \sqrt {d^2-e^2 x^2}}\\ &=\frac {2 (g x)^{1+m} (d+e x)}{5 d g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {(3-2 m) (g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 g (1+m) \sqrt {d^2-e^2 x^2}}+\frac {2 e (3-m) (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {5}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^5 g^2 (2+m) \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.88, size = 174, normalized size = 0.81 \begin {gather*} \frac {x (g x)^m \sqrt {1-\frac {e^2 x^2}{d^2}} \left (d^2 \left (6+5 m+m^2\right ) \, _2F_1\left (\frac {7}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )+e (1+m) x \left (2 d (3+m) \, _2F_1\left (\frac {7}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )+e (2+m) x \, _2F_1\left (\frac {7}{2},\frac {3+m}{2};\frac {5+m}{2};\frac {e^2 x^2}{d^2}\right )\right )\right )}{d^6 (1+m) (2+m) (3+m) \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*Sqrt[1 - (e^2*x^2)/d^2]*(d^2*(6 + 5*m + m^2)*Hypergeometric2F1[7/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)
/d^2] + e*(1 + m)*x*(2*d*(3 + m)*Hypergeometric2F1[7/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2] + e*(2 + m)*x*Hyp
ergeometric2F1[7/2, (3 + m)/2, (5 + m)/2, (e^2*x^2)/d^2])))/(d^6*(1 + m)*(2 + m)*(3 + m)*Sqrt[d^2 - e^2*x^2])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (g x \right )^{m} \left (e x +d \right )^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^2*e^2 + 2*d*x*e + d^2)*sqrt(-x^2*e^2 + d^2)*(g*x)^m/(x^8*e^8 - 4*d^2*x^6*e^6 + 6*d^4*x^4*e^4 - 4*d
^6*x^2*e^2 + d^8), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (g x\right )^{m} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((g*x)**m*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (g\,x\right )}^m\,{\left (d+e\,x\right )}^2}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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