3.354 \(\int \frac {(a+b x)^n (c+d x^2)}{x} \, dx\)

Optimal. Leaf size=77 \[ -\frac {a d (a+b x)^{n+1}}{b^2 (n+1)}+\frac {d (a+b x)^{n+2}}{b^2 (n+2)}-\frac {c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {952, 80, 65} \[ -\frac {a d (a+b x)^{n+1}}{b^2 (n+1)}+\frac {d (a+b x)^{n+2}}{b^2 (n+2)}-\frac {c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x^2))/x,x]

[Out]

-((a*d*(a + b*x)^(1 + n))/(b^2*(1 + n))) + (d*(a + b*x)^(2 + n))/(b^2*(2 + n)) - (c*(a + b*x)^(1 + n)*Hypergeo
metric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*(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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 952

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(c^p*(d
+ e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0]
&& NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 64, normalized size = 0.83 \[ -\frac {(a+b x)^{n+1} \left (b^2 c (n+2) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )+a d (a-b (n+1) x)\right )}{a b^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^n*(c + d*x^2))/x,x]

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x^{2} + c\right )} {\left (b x + a\right )}^{n}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x^2 + c)*(b*x + a)^n/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x^2 + c)*(b*x + a)^n/x, x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \,x^{2}+c \right ) \left (b x +a \right )^{n}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^n*(d*x^2+c)/x,x)

[Out]

int((b*x+a)^n*(d*x^2+c)/x,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x^2 + c)*(b*x + a)^n/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c + d*x^2)*(a + b*x)^n)/x,x)

[Out]

int(((c + d*x^2)*(a + b*x)^n)/x, x)

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sympy [B]  time = 7.22, size = 345, normalized size = 4.48 \[ - \frac {b^{n} c n \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{\Gamma \left (n + 2\right )} - \frac {b^{n} c \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{\Gamma \left (n + 2\right )} + d \left (\begin {cases} \frac {a^{n} x^{2}}{2} & \text {for}\: b = 0 \\\frac {a \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac {a}{a b^{2} + b^{3} x} + \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text {for}\: n = -2 \\- \frac {a \log {\left (\frac {a}{b} + x \right )}}{b^{2}} + \frac {x}{b} & \text {for}\: n = -1 \\- \frac {a^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac {a b n x \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac {b^{2} n x^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac {b^{2} x^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} & \text {otherwise} \end {cases}\right ) - \frac {b b^{n} c n x \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac {b b^{n} c x \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**n*(d*x**2+c)/x,x)

[Out]

-b**n*c*n*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) - b**n*c*(a/b + x)**n*lerchphi(
1 + b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) + d*Piecewise((a**n*x**2/2, Eq(b, 0)), (a*log(a/b + x)/(a*b**2
+ b**3*x) + a/(a*b**2 + b**3*x) + b*x*log(a/b + x)/(a*b**2 + b**3*x), Eq(n, -2)), (-a*log(a/b + x)/b**2 + x/b,
Eq(n, -1)), (-a**2*(a + b*x)**n/(b**2*n**2 + 3*b**2*n + 2*b**2) + a*b*n*x*(a + b*x)**n/(b**2*n**2 + 3*b**2*n
+ 2*b**2) + b**2*n*x**2*(a + b*x)**n/(b**2*n**2 + 3*b**2*n + 2*b**2) + b**2*x**2*(a + b*x)**n/(b**2*n**2 + 3*b
**2*n + 2*b**2), True)) - b*b**n*c*n*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)
) - b*b**n*c*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2))

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