3.5.35 \(\int \frac {(g x)^m (a+c x^2)^p}{(d+e x)^2} \, dx\) [435]

Optimal. Leaf size=238 \[ \frac {x (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,2;\frac {3+m}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 (1+m)}-\frac {2 e x^2 (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} F_1\left (\frac {2+m}{2};-p,2;\frac {4+m}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3 (2+m)}+\frac {e^2 x^3 (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} F_1\left (\frac {3+m}{2};-p,2;\frac {5+m}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^4 (3+m)} \]

[Out]

x*(g*x)^m*(c*x^2+a)^p*AppellF1(1/2+1/2*m,2,-p,3/2+1/2*m,e^2*x^2/d^2,-c*x^2/a)/d^2/(1+m)/((c*x^2/a+1)^p)-2*e*x^
2*(g*x)^m*(c*x^2+a)^p*AppellF1(1+1/2*m,2,-p,2+1/2*m,e^2*x^2/d^2,-c*x^2/a)/d^3/(2+m)/((c*x^2/a+1)^p)+e^2*x^3*(g
*x)^m*(c*x^2+a)^p*AppellF1(3/2+1/2*m,2,-p,5/2+1/2*m,e^2*x^2/d^2,-c*x^2/a)/d^4/(3+m)/((c*x^2/a+1)^p)

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Rubi [A]
time = 0.18, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {976, 525, 524} \begin {gather*} \frac {x (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+1}{2};-p,2;\frac {m+3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 (m+1)}+\frac {e^2 x^3 (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+3}{2};-p,2;\frac {m+5}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^4 (m+3)}-\frac {2 e x^2 (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+2}{2};-p,2;\frac {m+4}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((g*x)^m*(a + c*x^2)^p)/(d + e*x)^2,x]

[Out]

(x*(g*x)^m*(a + c*x^2)^p*AppellF1[(1 + m)/2, -p, 2, (3 + m)/2, -((c*x^2)/a), (e^2*x^2)/d^2])/(d^2*(1 + m)*(1 +
(c*x^2)/a)^p) - (2*e*x^2*(g*x)^m*(a + c*x^2)^p*AppellF1[(2 + m)/2, -p, 2, (4 + m)/2, -((c*x^2)/a), (e^2*x^2)/
d^2])/(d^3*(2 + m)*(1 + (c*x^2)/a)^p) + (e^2*x^3*(g*x)^m*(a + c*x^2)^p*AppellF1[(3 + m)/2, -p, 2, (5 + m)/2, -
((c*x^2)/a), (e^2*x^2)/d^2])/(d^4*(3 + m)*(1 + (c*x^2)/a)^p)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 976

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

Rubi steps

\begin {align*} \int \frac {(g x)^m \left (a+c x^2\right )^p}{(d+e x)^2} \, dx &=\left (x^{-m} (g x)^m\right ) \int \left (\frac {d^2 x^m \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^2}-\frac {2 d e x^{1+m} \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^2}+\frac {e^2 x^{2+m} \left (a+c x^2\right )^p}{\left (-d^2+e^2 x^2\right )^2}\right ) \, dx\\ &=\left (d^2 x^{-m} (g x)^m\right ) \int \frac {x^m \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx-\left (2 d e x^{-m} (g x)^m\right ) \int \frac {x^{1+m} \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx+\left (e^2 x^{-m} (g x)^m\right ) \int \frac {x^{2+m} \left (a+c x^2\right )^p}{\left (-d^2+e^2 x^2\right )^2} \, dx\\ &=\left (d^2 x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int \frac {x^m \left (1+\frac {c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx-\left (2 d e x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int \frac {x^{1+m} \left (1+\frac {c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx+\left (e^2 x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int \frac {x^{2+m} \left (1+\frac {c x^2}{a}\right )^p}{\left (-d^2+e^2 x^2\right )^2} \, dx\\ &=\frac {x (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,2;\frac {3+m}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 (1+m)}-\frac {2 e x^2 (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} F_1\left (\frac {2+m}{2};-p,2;\frac {4+m}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3 (2+m)}+\frac {e^2 x^3 (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} F_1\left (\frac {3+m}{2};-p,2;\frac {5+m}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^4 (3+m)}\\ \end {align*}

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Mathematica [F]
time = 0.10, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(g x)^m \left (a+c x^2\right )^p}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((g*x)^m*(a + c*x^2)^p)/(d + e*x)^2,x]

[Out]

Integrate[((g*x)^m*(a + c*x^2)^p)/(d + e*x)^2, x]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(c*x^2+a)^p/(e*x+d)^2,x)

[Out]

int((g*x)^m*(c*x^2+a)^p/(e*x+d)^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*(c*x^2+a)^p/(e*x+d)^2,x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^p*(g*x)^m/(x*e + d)^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*(c*x^2+a)^p/(e*x+d)^2,x, algorithm="fricas")

[Out]

integral((c*x^2 + a)^p*(g*x)^m/(x^2*e^2 + 2*d*x*e + d^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)**m*(c*x**2+a)**p/(e*x+d)**2,x)

[Out]

Timed out

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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*(c*x^2+a)^p/(e*x+d)^2,x, algorithm="giac")

[Out]

integrate((c*x^2 + a)^p*(g*x)^m/(x*e + d)^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 (c\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((g*x)^m*(a + c*x^2)^p)/(d + e*x)^2,x)

[Out]

int(((g*x)^m*(a + c*x^2)^p)/(d + e*x)^2, x)

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