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

Optimal. Leaf size=163 \[ \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac {9}{2},\frac {m+1}{2};\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )}{d^7 g (m+1) \sqrt {d^2-e^2 x^2}}-\frac {e \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac {9}{2},\frac {m+2}{2};\frac {m+4}{2};\frac {e^2 x^2}{d^2}\right )}{d^8 g^2 (m+2) \sqrt {d^2-e^2 x^2}} \]

[Out]

(g*x)^(1+m)*hypergeom([9/2, 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^7/g/(1+m)/(-e^2*x^2+d^
2)^(1/2)-e*(g*x)^(2+m)*hypergeom([9/2, 1+1/2*m],[2+1/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/d^8/g^2/(2+m)/(-e
^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {892, 82, 126, 365, 364} \[ \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac {9}{2},\frac {m+1}{2};\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )}{d^7 g (m+1) \sqrt {d^2-e^2 x^2}}-\frac {e \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac {9}{2},\frac {m+2}{2};\frac {m+4}{2};\frac {e^2 x^2}{d^2}\right )}{d^8 g^2 (m+2) \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((g*x)^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[9/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(d^7*g*(1
+ m)*Sqrt[d^2 - e^2*x^2]) - (e*(g*x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[9/2, (2 + m)/2, (4 + m)
/2, (e^2*x^2)/d^2])/(d^8*g^2*(2 + m)*Sqrt[d^2 - e^2*x^2])

Rule 82

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

Rule 126

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[((a + b*x)^Fra
cPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[(a*c + b*d*x^2)^m*(f*x)^p, x], x] /; FreeQ[{a
, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 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])

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 892

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

Rubi steps

\begin {align*} \int \frac {(g x)^m}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {\left (\sqrt {d-e x} \sqrt {d+e x}\right ) \int \frac {(g x)^m}{(d-e x)^{7/2} (d+e x)^{9/2}} \, dx}{\sqrt {d^2-e^2 x^2}}\\ &=\frac {\left (d \sqrt {d-e x} \sqrt {d+e x}\right ) \int \frac {(g x)^m}{(d-e x)^{9/2} (d+e x)^{9/2}} \, dx}{\sqrt {d^2-e^2 x^2}}-\frac {\left (e \sqrt {d-e x} \sqrt {d+e x}\right ) \int \frac {(g x)^{1+m}}{(d-e x)^{9/2} (d+e x)^{9/2}} \, dx}{g \sqrt {d^2-e^2 x^2}}\\ &=d \int \frac {(g x)^m}{\left (d^2-e^2 x^2\right )^{9/2}} \, dx-\frac {e \int \frac {(g x)^{1+m}}{\left (d^2-e^2 x^2\right )^{9/2}} \, dx}{g}\\ &=\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {(g x)^m}{\left (1-\frac {e^2 x^2}{d^2}\right )^{9/2}} \, dx}{d^7 \sqrt {d^2-e^2 x^2}}-\frac {\left (e \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 )^{9/2}} \, dx}{d^8 g \sqrt {d^2-e^2 x^2}}\\ &=\frac {(g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {9}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{d^7 g (1+m) \sqrt {d^2-e^2 x^2}}-\frac {e (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {9}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{d^8 g^2 (2+m) \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 122, normalized size = 0.75 \[ \frac {x \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^m \left (d (m+2) \, _2F_1\left (\frac {9}{2},\frac {m+1}{2};\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )-e (m+1) x \, _2F_1\left (\frac {9}{2},\frac {m}{2}+1;\frac {m}{2}+2;\frac {e^2 x^2}{d^2}\right )\right )}{d^8 (m+1) (m+2) \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*Sqrt[1 - (e^2*x^2)/d^2]*(-(e*(1 + m)*x*Hypergeometric2F1[9/2, 1 + m/2, 2 + m/2, (e^2*x^2)/d^2]) + d
*(2 + m)*Hypergeometric2F1[9/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2]))/(d^8*(1 + m)*(2 + m)*Sqrt[d^2 - e^2*x^2
])

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-e^{2} x^{2} + d^{2}} \left (g x\right )^{m}}{e^{9} x^{9} + d e^{8} x^{8} - 4 \, d^{2} e^{7} x^{7} - 4 \, d^{3} e^{6} x^{6} + 6 \, d^{4} e^{5} x^{5} + 6 \, d^{5} e^{4} x^{4} - 4 \, d^{6} e^{3} x^{3} - 4 \, d^{7} e^{2} x^{2} + d^{8} e x + d^{9}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m/(e*x+d)/(-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)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x \right )^{m}}{\left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m/(e*x+d)/(-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)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (g\,x\right )}^m}{{\left (d^2-e^2\,x^2\right )}^{7/2}\,\left (d+e\,x\right )} \,d x \]

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)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)**m/(e*x+d)/(-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)), x)

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