\(\int \frac {(d+e x) (a+b x^2)^p}{x} \, dx\) [387]

Optimal result

Integrand size = 18, antiderivative size = 88 \[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=e x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )-\frac {d \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {778, 272, 67, 252, 251} \[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )-\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a (p+1)} \]

[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 67

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

Rule 251

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])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 272

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 778

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]

Rubi steps \begin{align*} \text {integral}& = 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 \text {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] (verified)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=e x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )-\frac {d \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a (1+p)} \]

[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))

Maple [F]

\[\int \frac {\left (e x +d \right ) \left (b \,x^{2}+a \right )^{p}}{x}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=\int { \frac {{\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p}}{x} \,d x } \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.70 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=a^{p} e x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} - \frac {b^{p} d x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} \]

[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))

Maxima [F]

\[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=\int { \frac {{\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p}}{x} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=\int { \frac {{\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p}}{x} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x) \left (a+b x^2\right )^p}{x} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,\left (d+e\,x\right )}{x} \,d x \]

[In]

int(((a + b*x^2)^p*(d + e*x))/x,x)

[Out]

int(((a + b*x^2)^p*(d + e*x))/x, x)