3.10.34 \(\int \frac {(a+b x)^n (c+d x)^3}{x^2} \, dx\) [934]

Optimal. Leaf size=143 \[ \frac {d (a+b x)^{1+n} (c+d x)^2}{b (2+n) x}-\frac {(a+b x)^{1+n} \left (b c^2 (1+n) (a d+b c (2+n))+a d^2 (a d-b c (4+n)) x\right )}{a b^2 (1+n) (2+n) x}-\frac {c^2 (3 a d+b c n) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )}{a^2 (1+n)} \]

[Out]

d*(b*x+a)^(1+n)*(d*x+c)^2/b/(2+n)/x-(b*x+a)^(1+n)*(b*c^2*(1+n)*(a*d+b*c*(2+n))+a*d^2*(a*d-b*c*(4+n))*x)/a/b^2/
(1+n)/(2+n)/x-c^2*(b*c*n+3*a*d)*(b*x+a)^(1+n)*hypergeom([1, 1+n],[2+n],1+b*x/a)/a^2/(1+n)

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Rubi [A]
time = 0.09, antiderivative size = 143, 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 = {102, 151, 67} \begin {gather*} -\frac {c^2 (a+b x)^{n+1} (3 a d+b c n) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 (n+1)}-\frac {(a+b x)^{n+1} \left (b c^2 (n+1) (a d+b c (n+2))+a d^2 x (a d-b c (n+4))\right )}{a b^2 (n+1) (n+2) x}+\frac {d (c+d x)^2 (a+b x)^{n+1}}{b (n+2) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x)^3)/x^2,x]

[Out]

(d*(a + b*x)^(1 + n)*(c + d*x)^2)/(b*(2 + n)*x) - ((a + b*x)^(1 + n)*(b*c^2*(1 + n)*(a*d + b*c*(2 + n)) + a*d^
2*(a*d - b*c*(4 + n))*x))/(a*b^2*(1 + n)*(2 + n)*x) - (c^2*(3*a*d + b*c*n)*(a + b*x)^(1 + n)*Hypergeometric2F1
[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a^2*(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 102

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^n (c+d x)^3}{x^2} \, dx &=\frac {d (a+b x)^{1+n} (c+d x)^2}{b (2+n) x}+\frac {\int \frac {(a+b x)^n (c+d x) (c (a d+b c (2+n))-d (a d-b c (4+n)) x)}{x^2} \, dx}{b (2+n)}\\ &=\frac {d (a+b x)^{1+n} (c+d x)^2}{b (2+n) x}-\frac {(a+b x)^{1+n} \left (b c^2 (1+n) (a d+b c (2+n))+a d^2 (a d-b c (4+n)) x\right )}{a b^2 (1+n) (2+n) x}+\frac {\left (c^2 (3 a d+b c n)\right ) \int \frac {(a+b x)^n}{x} \, dx}{a}\\ &=\frac {d (a+b x)^{1+n} (c+d x)^2}{b (2+n) x}-\frac {(a+b x)^{1+n} \left (b c^2 (1+n) (a d+b c (2+n))+a d^2 (a d-b c (4+n)) x\right )}{a b^2 (1+n) (2+n) x}-\frac {c^2 (3 a d+b c n) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )}{a^2 (1+n)}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 114, normalized size = 0.80 \begin {gather*} -\frac {(a+b x)^{1+n} \left (a \left (b^2 c^3 \left (2+3 n+n^2\right )+a^2 d^3 x-a b d^2 x (3 c (2+n)+d (1+n) x)\right )+b^2 c^2 (2+n) (3 a d+b c n) x \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )\right )}{a^2 b^2 (1+n) (2+n) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{n} \left (d x +c \right )^{3}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 3.50, size = 770, normalized size = 5.38 \begin {gather*} \frac {b^{n} c^{3} n^{2} \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{x \Gamma \left (n + 2\right )} + \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 )}{x \Gamma \left (n + 2\right )} - \frac {b^{n} c^{3} n \left (\frac {a}{b} + x\right )^{n} \Gamma \left (n + 1\right )}{x \Gamma \left (n + 2\right )} - \frac {b^{n} c^{3} \left (\frac {a}{b} + x\right )^{n} \Gamma \left (n + 1\right )}{x \Gamma \left (n + 2\right )} - \frac {3 b^{n} c^{2} d 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 {3 b^{n} c^{2} d \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 d^{2} \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^{3} \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^{3} n^{2} \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} n \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} n \left (\frac {a}{b} + x\right )^{n} \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac {b b^{n} c^{3} \left (\frac {a}{b} + x\right )^{n} \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac {3 b b^{n} c^{2} d 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 {3 b b^{n} c^{2} d 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^{2} b^{n} c^{3} n^{2} \left (\frac {a}{b} + x\right )^{2} \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a^{2} x \Gamma \left (n + 2\right )} - \frac {b^{2} b^{n} c^{3} n \left (\frac {a}{b} + x\right )^{2} \left (\frac {a}{b} + x\right )^{n} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a^{2} x \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**2,x)

[Out]

b**n*c**3*n**2*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(x*gamma(n + 2)) + b**n*c**3*n*(a/b + x
)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(x*gamma(n + 2)) - b**n*c**3*n*(a/b + x)**n*gamma(n + 1)/(x*ga
mma(n + 2)) - b**n*c**3*(a/b + x)**n*gamma(n + 1)/(x*gamma(n + 2)) - 3*b**n*c**2*d*n*(a/b + x)**n*lerchphi(1 +
b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) - 3*b**n*c**2*d*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n
+ 1)/gamma(n + 2) + 3*c*d**2*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**3*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**3*n**2*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)) + b*
b**n*c**3*n*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)) - b*b**n*c**3*n*(a/b + x)
**n*gamma(n + 1)/(a*gamma(n + 2)) - b*b**n*c**3*(a/b + x)**n*gamma(n + 1)/(a*gamma(n + 2)) - 3*b*b**n*c**2*d*n
*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)) - 3*b*b**n*c**2*d*x*(a/b + x)**n*l
erchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2)) - b**2*b**n*c**3*n**2*(a/b + x)**2*(a/b + x)**n*ler
chphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a**2*x*gamma(n + 2)) - b**2*b**n*c**3*n*(a/b + x)**2*(a/b + x)**n*ler
chphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a**2*x*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^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^3*(b*x + a)^n/x^2, 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^2} \,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^2,x)

[Out]

int(((a + b*x)^n*(c + d*x)^3)/x^2, x)

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