3.934 \(\int \frac {(a+b x)^n (c+d x)^3}{x^2} \, dx\)
Optimal. Leaf size=143 \[ -\frac {c^2 (a+b x)^{n+1} (3 a d+b c n) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 (n+1)}-\frac {(a+b x)^{n+1} \left (b c^2 (n+1) (a d+b c (n+2))+a d^2 x (a d-b c (n+4))\right )}{a b^2 (n+1) (n+2) x}+\frac {d (c+d x)^2 (a+b x)^{n+1}}{b (n+2) x} \]
[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 = 0.13, 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 =
{100, 146, 65} \[ -\frac {c^2 (a+b x)^{n+1} (3 a d+b c n) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 (n+1)}-\frac {(a+b x)^{n+1} \left (b c^2 (n+1) (a d+b c (n+2))+a d^2 x (a d-b c (n+4))\right )}{a b^2 (n+1) (n+2) x}+\frac {d (c+d x)^2 (a+b x)^{n+1}}{b (n+2) x} \]
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 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 100
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 146
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)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), 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 = 0.08, size = 114, normalized size = 0.80 \[ -\frac {(a+b x)^{n+1} \left (a \left (a^2 d^3 x-a b d^2 x (3 c (n+2)+d (n+1) x)+b^2 c^3 \left (n^2+3 n+2\right )\right )+b^2 c^2 (n+2) x (3 a d+b c n) \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )\right )}{a^2 b^2 (n+1) (n+2) x} \]
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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fricas [F] time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} {\left (b x + a\right )}^{n}}{x^{2}}, x\right ) \]
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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{3} {\left (b x + a\right )}^{n}}{x^{2}}\,{d x} \]
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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maple [F] time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{3} \left (b x +a \right )^{n}}{x^{2}}\, dx \]
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] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3 \, {\left (b x + a\right )}^{n + 1} c d^{2}}{b {\left (n + 1\right )}} + \int \frac {{\left (d^{3} x^{3} + 3 \, c^{2} d x + c^{3}\right )} {\left (b x + a\right )}^{n}}{x^{2}}\,{d x} \]
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]
3*(b*x + a)^(n + 1)*c*d^2/(b*(n + 1)) + integrate((d^3*x^3 + 3*c^2*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 \[ \int \frac {{\left (a+b\,x\right )}^n\,{\left (c+d\,x\right )}^3}{x^2} \,d x \]
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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sympy [A] time = 7.17, size = 770, normalized size = 5.38 \[ \text {result too large to display} \]
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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