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

Optimal. Leaf size=157 \[ \frac {x (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d (m+1)}-\frac {e x^2 (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+2}{2};-p,1;\frac {m+4}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {959, 511, 510} \[ \frac {x (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d (m+1)}-\frac {e x^2 (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {m+2}{2};-p,1;\frac {m+4}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rubi steps

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

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Mathematica [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {(g x)^m \left (a+c x^2\right )^p}{d+e x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x \right )^{m} \left (c \,x^{2}+a \right )^{p}}{e x +d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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