3.3.32 \(\int \frac {(g x)^m (d^2-e^2 x^2)^{5/2}}{(d+e x)^3} \, dx\) [232]

Optimal. Leaf size=250 \[ -\frac {3 d (g x)^{1+m} \sqrt {d^2-e^2 x^2}}{g (2+m)}+\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}+\frac {d^3 (5+4 m) (g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{g (1+m) (2+m) \sqrt {d^2-e^2 x^2}}-\frac {d^2 e (11+4 m) (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{g^2 (2+m) (3+m) \sqrt {d^2-e^2 x^2}} \]

[Out]

-3*d*(g*x)^(1+m)*(-e^2*x^2+d^2)^(1/2)/g/(2+m)+e*(g*x)^(2+m)*(-e^2*x^2+d^2)^(1/2)/g^2/(3+m)+d^3*(5+4*m)*(g*x)^(
1+m)*hypergeom([1/2, 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)/g/(1+m)/(2+m)/(-e^2*x^2+d^2)^(1
/2)-d^2*e*(11+4*m)*(g*x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/g^2/(2+m)
/(3+m)/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 250, 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 = {866, 1823, 822, 372, 371} \begin {gather*} -\frac {d^2 e (4 m+11) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\frac {e^2 x^2}{d^2}\right )}{g^2 (m+2) (m+3) \sqrt {d^2-e^2 x^2}}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}-\frac {3 d \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}+\frac {d^3 (4 m+5) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\frac {e^2 x^2}{d^2}\right )}{g (m+1) (m+2) \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*(g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2])/(g*(2 + m)) + (e*(g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2])/(g^2*(3 + m)) + (
d^3*(5 + 4*m)*(g*x)^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2
])/(g*(1 + m)*(2 + m)*Sqrt[d^2 - e^2*x^2]) - (d^2*e*(11 + 4*m)*(g*x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeom
etric2F1[1/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(g^2*(2 + m)*(3 + 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 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int \frac {(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx &=\int \frac {(g x)^m (d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}-\frac {\int \frac {(g x)^m \left (-d^3 e^2 (3+m)+d^2 e^3 (11+4 m) x-3 d e^4 (3+m) x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2 (3+m)}\\ &=-\frac {3 d (g x)^{1+m} \sqrt {d^2-e^2 x^2}}{g (2+m)}+\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}+\frac {\int \frac {(g x)^m \left (d^3 e^4 (3+m) (5+4 m)-d^2 e^5 (2+m) (11+4 m) x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4 (2+m) (3+m)}\\ &=-\frac {3 d (g x)^{1+m} \sqrt {d^2-e^2 x^2}}{g (2+m)}+\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}+\frac {\left (d^3 (5+4 m)\right ) \int \frac {(g x)^m}{\sqrt {d^2-e^2 x^2}} \, dx}{2+m}-\frac {\left (d^2 e (11+4 m)\right ) \int \frac {(g x)^{1+m}}{\sqrt {d^2-e^2 x^2}} \, dx}{g (3+m)}\\ &=-\frac {3 d (g x)^{1+m} \sqrt {d^2-e^2 x^2}}{g (2+m)}+\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}+\frac {\left (d^3 (5+4 m) \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {(g x)^m}{\sqrt {1-\frac {e^2 x^2}{d^2}}} \, dx}{(2+m) \sqrt {d^2-e^2 x^2}}-\frac {\left (d^2 e (11+4 m) \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {(g x)^{1+m}}{\sqrt {1-\frac {e^2 x^2}{d^2}}} \, dx}{g (3+m) \sqrt {d^2-e^2 x^2}}\\ &=-\frac {3 d (g x)^{1+m} \sqrt {d^2-e^2 x^2}}{g (2+m)}+\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}+\frac {d^3 (5+4 m) (g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )}{g (1+m) (2+m) \sqrt {d^2-e^2 x^2}}-\frac {d^2 e (11+4 m) (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )}{g^2 (2+m) (3+m) \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.72, size = 245, normalized size = 0.98 \begin {gather*} \frac {x (g x)^m \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}} \left (d^3 \left (24+26 m+9 m^2+m^3\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {e^2 x^2}{d^2}\right )-e (1+m) x \left (3 d^2 \left (12+7 m+m^2\right ) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\frac {e^2 x^2}{d^2}\right )+e (2+m) x \left (-3 d (4+m) \, _2F_1\left (\frac {1}{2},\frac {3+m}{2};\frac {5+m}{2};\frac {e^2 x^2}{d^2}\right )+e (3+m) x \, _2F_1\left (\frac {1}{2},\frac {4+m}{2};\frac {6+m}{2};\frac {e^2 x^2}{d^2}\right )\right )\right )\right )}{(1+m) (2+m) (3+m) (4+m) (d-e x) (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2]*(d^3*(24 + 26*m + 9*m^2 + m^3)*Hypergeometric2F1[1/2, (
1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2] - e*(1 + m)*x*(3*d^2*(12 + 7*m + m^2)*Hypergeometric2F1[1/2, (2 + m)/2, (4
+ m)/2, (e^2*x^2)/d^2] + e*(2 + m)*x*(-3*d*(4 + m)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, (e^2*x^2)/d^2
] + e*(3 + m)*x*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, (e^2*x^2)/d^2]))))/((1 + m)*(2 + m)*(3 + m)*(4 +
m)*(d - e*x)*(d + e*x))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (g x \right )^{m} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{\left (e x +d \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,x)

[Out]

int((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,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^2*x^2+d^2)^(5/2)/(e*x+d)^3,x, algorithm="maxima")

[Out]

integrate((-x^2*e^2 + d^2)^(5/2)*(g*x)^m/(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^2*x^2+d^2)^(5/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral((x^4*e^4 - 2*d^2*x^2*e^2 + d^4)*sqrt(-x^2*e^2 + d^2)*(g*x)^m/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3
), 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 {5}{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**2*x**2+d**2)**(5/2)/(e*x+d)**3,x)

[Out]

Integral((g*x)**m*(-(-d + e*x)*(d + e*x))**(5/2)/(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^2*x^2+d^2)^(5/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate((-x^2*e^2 + d^2)^(5/2)*(g*x)^m/(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 (d^2-e^2\,x^2\right )}^{5/2}\,{\left (g\,x\right )}^m}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(5/2)*(g*x)^m)/(d + e*x)^3,x)

[Out]

int(((d^2 - e^2*x^2)^(5/2)*(g*x)^m)/(d + e*x)^3, x)

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