3.1139 \(\int (e x)^m (a+b x^2)^p (c+d x^2)^q \, dx\)

Optimal. Leaf size=101 \[ \frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} F_1\left (\frac {m+1}{2};-p,-q;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {511, 510} \[ \frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} F_1\left (\frac {m+1}{2};-p,-q;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 97, normalized size = 0.96 \[ \frac {x (e x)^m \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} F_1\left (\frac {m+1}{2};-p,-q;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} \left (e x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p*(d*x^2 + c)^q*(e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p*(d*x^2 + c)^q*(e*x)^m, x)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{q}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(b*x^2+a)^p*(d*x^2+c)^q,x)

[Out]

int((e*x)^m*(b*x^2+a)^p*(d*x^2+c)^q,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^p*(d*x^2 + c)^q*(e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q,x)

[Out]

int((e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, 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((e*x)**m*(b*x**2+a)**p*(d*x**2+c)**q,x)

[Out]

Timed out

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