3.10.33 \(\int \frac {(a+b x)^n (c+d x)^3}{x} \, dx\) [933]

Optimal. Leaf size=131 \[ \frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) (a+b x)^{1+n}}{b^3 (1+n)}+\frac {d^2 (3 b c-2 a d) (a+b x)^{2+n}}{b^3 (2+n)}+\frac {d^3 (a+b x)^{3+n}}{b^3 (3+n)}-\frac {c^3 (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )}{a (1+n)} \]

[Out]

d*(a^2*d^2-3*a*b*c*d+3*b^2*c^2)*(b*x+a)^(1+n)/b^3/(1+n)+d^2*(-2*a*d+3*b*c)*(b*x+a)^(2+n)/b^3/(2+n)+d^3*(b*x+a)
^(3+n)/b^3/(3+n)-c^3*(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 = 0.04, antiderivative size = 131, 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 \left (a^2 d^2-3 a b c d+3 b^2 c^2\right ) (a+b x)^{n+1}}{b^3 (n+1)}+\frac {d^2 (3 b c-2 a d) (a+b x)^{n+2}}{b^3 (n+2)}+\frac {d^3 (a+b x)^{n+3}}{b^3 (n+3)}-\frac {c^3 (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)^3)/x,x]

[Out]

(d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*(a + b*x)^(1 + n))/(b^3*(1 + n)) + (d^2*(3*b*c - 2*a*d)*(a + b*x)^(2 + n)
)/(b^3*(2 + n)) + (d^3*(a + b*x)^(3 + n))/(b^3*(3 + n)) - (c^3*(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)^3}{x} \, dx &=\int \left (\frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) (a+b x)^n}{b^2}+\frac {c^3 (a+b x)^n}{x}+\frac {d^2 (3 b c-2 a d) (a+b x)^{1+n}}{b^2}+\frac {d^3 (a+b x)^{2+n}}{b^2}\right ) \, dx\\ &=\frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) (a+b x)^{1+n}}{b^3 (1+n)}+\frac {d^2 (3 b c-2 a d) (a+b x)^{2+n}}{b^3 (2+n)}+\frac {d^3 (a+b x)^{3+n}}{b^3 (3+n)}+c^3 \int \frac {(a+b x)^n}{x} \, dx\\ &=\frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) (a+b x)^{1+n}}{b^3 (1+n)}+\frac {d^2 (3 b c-2 a d) (a+b x)^{2+n}}{b^3 (2+n)}+\frac {d^3 (a+b x)^{3+n}}{b^3 (3+n)}-\frac {c^3 (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 = 0.14, size = 117, normalized size = 0.89 \begin {gather*} (a+b x)^{1+n} \left (\frac {d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right )}{b^3 (1+n)}+\frac {d^2 (3 b c-2 a d) (a+b x)}{b^3 (2+n)}+\frac {d^3 (a+b x)^2}{b^3 (3+n)}-\frac {c^3 \, _2F_1\left (1,1+n;2+n;\frac {a+b x}{a}\right )}{a+a n}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{n} \left (d x +c \right )^{3}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^3/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

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

[Out]

integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(b*x + a)^n/x, x)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 600 vs. \(2 (116) = 232\).
time = 4.67, size = 993, normalized size = 7.58 \begin {gather*} - \frac {b^{n} c^{3} 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^{3} \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 )} + 3 c^{2} 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 ) + 3 c 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 ) + d^{3} \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^{3} 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^{3} 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)**3/x,x)

[Out]

-b**n*c**3*n*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) - b**n*c**3*(a/b + x)**n*ler
chphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) + 3*c**2*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)) + 3*c*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)) + d**3*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**3*n*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*ga
mma(n + 1)/(a*gamma(n + 2)) - b*b**n*c**3*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n
+ 2))

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

[Out]

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

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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 )}^3}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^n*(c + d*x)^3)/x,x)

[Out]

int(((a + b*x)^n*(c + d*x)^3)/x, x)

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