3.10.26 \(\int \frac {(a+b x)^n (c+d x)^2}{x} \, dx\) [926]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {90, 67} \begin {gather*} \frac {d (2 b c-a d) (a+b x)^{n+1}}{b^2 (n+1)}+\frac {d^2 (a+b x)^{n+2}}{b^2 (n+2)}-\frac {c^2 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x)^2)/x,x]

[Out]

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

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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 77, normalized size = 0.88 \begin {gather*} -\frac {(a+b x)^{1+n} \left (a d (a d-b (2 c (2+n)+d (1+n) x))+b^2 c^2 (2+n) \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )\right )}{a b^2 (1+n) (2+n)} \end {gather*}

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

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{n} \left (d x +c \right )^{2}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(n>0)', see `assume?` for more
details)Is n

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (71) = 142\).
time = 3.69, size = 386, normalized size = 4.39 \begin {gather*} - \frac {b^{n} c^{2} 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^{2} \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 )} + 2 c d \left (\begin {cases} a^{n} x & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x\right )^{n + 1}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (a + b x \right )} & \text {otherwise} \end {cases}}{b} & \text {otherwise} \end {cases}\right ) + d^{2} \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^{2} 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^{2} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^n*(c + d*x)^2)/x,x)

[Out]

int(((a + b*x)^n*(c + d*x)^2)/x, x)

________________________________________________________________________________________