### 3.736 $$\int \frac{(a+c x^2)^p}{d+e x} \, dx$$

Optimal. Leaf size=125 $\frac{x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d}-\frac{e \left (a+c x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )}$

[Out]

(x*(a + c*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((c*x^2)/a), (e^2*x^2)/d^2])/(d*(1 + (c*x^2)/a)^p) - (e*(a + c*x^2
)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(2*(c*d^2 + a*e^2)*(1 + p))

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Rubi [A]  time = 0.118191, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.294, Rules used = {757, 430, 429, 444, 68} $\frac{x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d}-\frac{e \left (a+c x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + c*x^2)^p/(d + e*x),x]

[Out]

(x*(a + c*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((c*x^2)/a), (e^2*x^2)/d^2])/(d*(1 + (c*x^2)/a)^p) - (e*(a + c*x^2
)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(2*(c*d^2 + a*e^2)*(1 + p))

Rule 757

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

Rule 430

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

Rule 429

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

Rule 444

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{\left (a+c x^2\right )^p}{d+e x} \, dx &=\int \left (\frac{d \left (a+c x^2\right )^p}{d^2-e^2 x^2}+\frac{e x \left (a+c x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx\\ &=d \int \frac{\left (a+c x^2\right )^p}{d^2-e^2 x^2} \, dx+e \int \frac{x \left (a+c x^2\right )^p}{-d^2+e^2 x^2} \, dx\\ &=\frac{1}{2} e \operatorname{Subst}\left (\int \frac{(a+c x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )+\left (d \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{\left (1+\frac{c x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx\\ &=\frac{x \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d}-\frac{e \left (a+c x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac{e^2 \left (a+c x^2\right )}{c d^2+a e^2}\right )}{2 \left (c d^2+a e^2\right ) (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0890637, size = 131, normalized size = 1.05 $\frac{\left (a+c x^2\right )^p \left (\frac{e \left (x-\sqrt{-\frac{a}{c}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{c}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac{d-\sqrt{-\frac{a}{c}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{c}} e}{d+e x}\right )}{2 e p}$

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + c*x^2)^p/(d + e*x),x]

[Out]

((a + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/c)]*e)/(d + e*x), (d + Sqrt[-(a/c)]*e)/(d + e*x)]
)/(2*e*p*((e*(-Sqrt[-(a/c)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/c)] + x))/(d + e*x))^p)

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Maple [F]  time = 0.564, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+a \right ) ^{p}}{ex+d}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + a\right )}^{p}}{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^p/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + a\right )}^{p}}{e x + d}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + a)^p/(e*x + d), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + c x^{2}\right )^{p}}{d + e x}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + c*x**2)**p/(d + e*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + a\right )}^{p}}{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^p/(e*x + d), x)