3.177 \(\int \frac {(a+b x)^n (c+d x^3)}{x} \, dx\)

Optimal. Leaf size=99 \[ \frac {a^2 d (a+b x)^{n+1}}{b^3 (n+1)}-\frac {2 a d (a+b x)^{n+2}}{b^3 (n+2)}+\frac {d (a+b x)^{n+3}}{b^3 (n+3)}-\frac {c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a (n+1)} \]

[Out]

a^2*d*(b*x+a)^(1+n)/b^3/(1+n)-2*a*d*(b*x+a)^(2+n)/b^3/(2+n)+d*(b*x+a)^(3+n)/b^3/(3+n)-c*(b*x+a)^(1+n)*hypergeo
m([1, 1+n],[2+n],1+b*x/a)/a/(1+n)

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Rubi [A]  time = 0.06, antiderivative size = 99, 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 = {1620, 65} \[ \frac {a^2 d (a+b x)^{n+1}}{b^3 (n+1)}-\frac {2 a d (a+b x)^{n+2}}{b^3 (n+2)}+\frac {d (a+b x)^{n+3}}{b^3 (n+3)}-\frac {c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x^3))/x,x]

[Out]

(a^2*d*(a + b*x)^(1 + n))/(b^3*(1 + n)) - (2*a*d*(a + b*x)^(2 + n))/(b^3*(2 + n)) + (d*(a + b*x)^(3 + n))/(b^3
*(3 + n)) - (c*(a + b*x)^(1 + n)*Hypergeometric2F1[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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {align*} \int \frac {(a+b x)^n \left (c+d x^3\right )}{x} \, dx &=\int \left (\frac {a^2 d (a+b x)^n}{b^2}+\frac {c (a+b x)^n}{x}-\frac {2 a d (a+b x)^{1+n}}{b^2}+\frac {d (a+b x)^{2+n}}{b^2}\right ) \, dx\\ &=\frac {a^2 d (a+b x)^{1+n}}{b^3 (1+n)}-\frac {2 a d (a+b x)^{2+n}}{b^3 (2+n)}+\frac {d (a+b x)^{3+n}}{b^3 (3+n)}+c \int \frac {(a+b x)^n}{x} \, dx\\ &=\frac {a^2 d (a+b x)^{1+n}}{b^3 (1+n)}-\frac {2 a d (a+b x)^{2+n}}{b^3 (2+n)}+\frac {d (a+b x)^{3+n}}{b^3 (3+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 = 0.06, size = 94, normalized size = 0.95 \[ \frac {(a+b x)^{n+1} \left (a d \left (2 a^2-2 a b (n+1) x+b^2 \left (n^2+3 n+2\right ) x^2\right )-b^3 c \left (n^2+5 n+6\right ) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )\right )}{a b^3 (n+1) (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^n*(c + d*x^3))/x,x]

[Out]

((a + b*x)^(1 + n)*(a*d*(2*a^2 - 2*a*b*(1 + n)*x + b^2*(2 + 3*n + n^2)*x^2) - b^3*c*(6 + 5*n + n^2)*Hypergeome
tric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a]))/(a*b^3*(1 + n)*(2 + n)*(3 + n))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x^{3} + c\right )} {\left (b x + a\right )}^{n}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n*(d*x^3+c)/x,x, algorithm="fricas")

[Out]

integral((d*x^3 + c)*(b*x + a)^n/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )} {\left (b x + a\right )}^{n}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n*(d*x^3+c)/x,x, algorithm="giac")

[Out]

integrate((d*x^3 + c)*(b*x + a)^n/x, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \,x^{3}+c \right ) \left (b x +a \right )^{n}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^n*(d*x^3+c)/x,x)

[Out]

int((b*x+a)^n*(d*x^3+c)/x,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )} {\left (b x + a\right )}^{n}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n*(d*x^3+c)/x,x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)*(b*x + a)^n/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (d\,x^3+c\right )\,{\left (a+b\,x\right )}^n}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c + d*x^3)*(a + b*x)^n)/x,x)

[Out]

int(((c + d*x^3)*(a + b*x)^n)/x, x)

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sympy [B]  time = 6.43, size = 741, normalized size = 7.48 \[ - \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^{3}}{3} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {3 a^{2}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {4 a b x \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {4 a b x}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {2 b^{2} x^{2} \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} & \text {for}\: n = -3 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + x \right )}}{a b^{3} + b^{4} x} - \frac {2 a^{2}}{a b^{3} + b^{4} x} - \frac {2 a b x \log {\left (\frac {a}{b} + x \right )}}{a b^{3} + b^{4} x} + \frac {b^{2} x^{2}}{a b^{3} + b^{4} x} & \text {for}\: n = -2 \\\frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{b^{3}} - \frac {a x}{b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: n = -1 \\\frac {2 a^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} - \frac {2 a^{2} b n x \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {a b^{2} n^{2} x^{2} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {a b^{2} n x^{2} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {b^{3} n^{2} x^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {3 b^{3} n x^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} + \frac {2 b^{3} x^{3} \left (a + b x\right )^{n}}{b^{3} n^{3} + 6 b^{3} n^{2} + 11 b^{3} n + 6 b^{3}} & \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**3+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**3/3, Eq(b, 0)), (2*a**2*log(a/b + x)/(2*
a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 3*a**2/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x*log(a/b + x)
/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 2*b**2*x**2*log
(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2), Eq(n, -3)), (-2*a**2*log(a/b + x)/(a*b**3 + b**4*x) - 2*a*
*2/(a*b**3 + b**4*x) - 2*a*b*x*log(a/b + x)/(a*b**3 + b**4*x) + b**2*x**2/(a*b**3 + b**4*x), Eq(n, -2)), (a**2
*log(a/b + x)/b**3 - a*x/b**2 + x**2/(2*b), Eq(n, -1)), (2*a**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**
3*n + 6*b**3) - 2*a**2*b*n*x*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + a*b**2*n**2*x**2*(a
+ b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + a*b**2*n*x**2*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**
2 + 11*b**3*n + 6*b**3) + b**3*n**2*x**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + 3*b**3*
n*x**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + 2*b**3*x**3*(a + b*x)**n/(b**3*n**3 + 6*b
**3*n**2 + 11*b**3*n + 6*b**3), 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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