3.6.9 \(\int \frac {(d+e x) (a+c x^2)^p}{x^3} \, dx\) [509]

Optimal. Leaf size=92 \[ -\frac {e \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {c x^2}{a}\right )}{x}+\frac {c d \left (a+c x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;1+\frac {c x^2}{a}\right )}{2 a^2 (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {778, 272, 67, 372, 371} \begin {gather*} \frac {c d \left (a+c x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac {c x^2}{a}+1\right )}{2 a^2 (p+1)}-\frac {e \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {c x^2}{a}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(a + c*x^2)^p)/x^3,x]

[Out]

-((e*(a + c*x^2)^p*Hypergeometric2F1[-1/2, -p, 1/2, -((c*x^2)/a)])/(x*(1 + (c*x^2)/a)^p)) + (c*d*(a + c*x^2)^(
1 + p)*Hypergeometric2F1[2, 1 + p, 2 + p, 1 + (c*x^2)/a])/(2*a^2*(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 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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

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

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Mathematica [A]
time = 0.11, size = 89, normalized size = 0.97 \begin {gather*} \frac {1}{2} \left (a+c x^2\right )^p \left (-\frac {2 e \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {c x^2}{a}\right )}{x}+\frac {c d \left (a+c x^2\right ) \, _2F_1\left (2,1+p;2+p;1+\frac {c x^2}{a}\right )}{a^2 (1+p)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(a + c*x^2)^p)/x^3,x]

[Out]

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

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right ) \left (c \,x^{2}+a \right )^{p}}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(c*x^2+a)^p/x^3,x)

[Out]

int((e*x+d)*(c*x^2+a)^p/x^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x^2+a)^p/x^3,x, algorithm="maxima")

[Out]

integrate((x*e + d)*(c*x^2 + a)^p/x^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x^2+a)^p/x^3,x, algorithm="fricas")

[Out]

integral((x*e + d)*(c*x^2 + a)^p/x^3, x)

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Sympy [C] Result contains complex when optimal does not.
time = 6.71, size = 71, normalized size = 0.77 \begin {gather*} - \frac {a^{p} e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{x} - \frac {c^{p} d x^{2 p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{c x^{2}}} \right )}}{2 x^{2} \Gamma \left (2 - p\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x**2+a)**p/x**3,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x^2+a)^p/x^3,x, algorithm="giac")

[Out]

integrate((x*e + d)*(c*x^2 + a)^p/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^p\,\left (d+e\,x\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^p*(d + e*x))/x^3,x)

[Out]

int(((a + c*x^2)^p*(d + e*x))/x^3, x)

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