3.387 \(\int \frac{(d+e x) (a+b x^2)^p}{x} \, dx\)

Optimal. Leaf size=88 \[ e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-\frac{d \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)} \]

[Out]

(e*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p - (d*(a + b*x^2)^(1 + p)*H
ypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a*(1 + p))

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Rubi [A]  time = 0.0528735, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {764, 266, 65, 246, 245} \[ e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-\frac{d \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(a + b*x^2)^p)/x,x]

[Out]

(e*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p - (d*(a + b*x^2)^(1 + p)*H
ypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a*(1 + p))

Rule 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0551568, size = 88, normalized size = 1. \[ e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-\frac{d \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p - (d*(a + b*x^2)^(1 + p)*H
ypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a*(1 + p))

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Maple [F]  time = 0.399, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) \left ( b{x}^{2}+a \right ) ^{p}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)*(b*x^2 + a)^p/x, x)

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Sympy [C]  time = 15.9876, size = 65, normalized size = 0.74 \begin{align*} a^{p} e x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} - \frac{b^{p} d x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ 1 - p \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*e*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) - b**p*d*x**(2*p)*gamma(-p)*hyper((-p, -p), (1 - p
,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(1 - p))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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