3.3.39 \(\int \frac {(g x)^m}{(d+e x)^3 (d^2-e^2 x^2)^{7/2}} \, dx\) [239]

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

[Out]

4/11*(g*x)^(1+m)*(-e*x+d)/g/(-e^2*x^2+d^2)^(11/2)+1/11*(7-4*m)*(g*x)^(1+m)*hypergeom([11/2, 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^9/g/(1+m)/(-e^2*x^2+d^2)^(1/2)-1/11*e*(25-4*m)*(g*x)^(2+m)*hypergeom
([11/2, 1+1/2*m],[2+1/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/d^10/g^2/(2+m)/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 214, normalized size of antiderivative = 1.00, 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 = {866, 1820, 822, 372, 371} \begin {gather*} \frac {4 (d-e x) (g x)^{m+1}}{11 g \left (d^2-e^2 x^2\right )^{11/2}}-\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}} \end {gather*}

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 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 866

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 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)^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*}

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Mathematica [A]
time = 1.51, size = 200, normalized size = 0.93 \begin {gather*} \frac {x (g x)^m \sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {d^3 \, _2F_1\left (\frac {13}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{1+m}+e x \left (-\frac {3 d^2 \, _2F_1\left (\frac {13}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{2+m}+e x \left (\frac {3 d \, _2F_1\left (\frac {13}{2},\frac {3+m}{2};\frac {5+m}{2};\frac {e^2 x^2}{d^2}\right )}{3+m}-\frac {e x \, _2F_1\left (\frac {13}{2},\frac {4+m}{2};\frac {6+m}{2};\frac {e^2 x^2}{d^2}\right )}{4+m}\right )\right )\right )}{d^{12} \sqrt {d^2-e^2 x^2}} \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (g x \right )^{m}}{\left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m/(e*x+d)^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 = 0.00, 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)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(-x^2*e^2 + d^2)*(g*x)^m/(d^8*x^3*e^3 + 3*d^9*x^2*e^2 + 3*d^10*x*e + d^11 + (x^11*e^3 + 3*d*x^10*
e^2 + 3*d^2*x^9*e + d^3*x^8)*e^8 - 4*(d^2*x^9*e^3 + 3*d^3*x^8*e^2 + 3*d^4*x^7*e + d^5*x^6)*e^6 + 6*(d^4*x^7*e^
3 + 3*d^5*x^6*e^2 + 3*d^6*x^5*e + d^7*x^4)*e^4 - 4*(d^6*x^5*e^3 + 3*d^7*x^4*e^2 + 3*d^8*x^3*e + d^9*x^2)*e^2),
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 (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}} \left (d + e x\right )^{3}}\, dx \end {gather*}

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]

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

[Out]

integrate((g*x)^m/((-x^2*e^2 + d^2)^(7/2)*(x*e + d)^3), 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^2-e^2\,x^2\right )}^{7/2}\,{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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