3.1.30 \(\int \frac {(c+d x)^n (A+B x+C x^2+D x^3)}{(a+b x)^2} \, dx\) [30]

Optimal. Leaf size=220 \[ \frac {(b C d-b c D-2 a d D) (c+d x)^{1+n}}{b^3 d^2 (1+n)}-\frac {\left (A-\frac {a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) (c+d x)^{1+n}}{(b c-a d) (a+b x)}+\frac {D (c+d x)^{2+n}}{b^2 d^2 (2+n)}+\frac {\left (a^3 d D (3+n)-b^3 (B c+A d n)+a b^2 (2 c C+B d (1+n))-a^2 b (3 c D+C d (2+n))\right ) (c+d x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {b (c+d x)}{b c-a d}\right )}{b^3 (b c-a d)^2 (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1635, 965, 81, 70} \begin {gather*} -\frac {(c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}+\frac {(c+d x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b (c+d x)}{b c-a d}\right ) \left (a^3 d D (n+3)-a^2 b (3 c D+C d (n+2))+a b^2 (B d (n+1)+2 c C)-b^3 (A d n+B c)\right )}{b^3 (n+1) (b c-a d)^2}+\frac {(c+d x)^{n+1} (-2 a d D-b c D+b C d)}{b^3 d^2 (n+1)}+\frac {D (c+d x)^{n+2}}{b^2 d^2 (n+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^2,x]

[Out]

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 81

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

Rule 965

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 1635

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(
b*c - a*d))), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rubi steps

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

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Mathematica [F]
time = 0.78, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^2, x)

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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((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**n*(D*x**3+C*x**2+B*x+A)/(b*x+a)**2,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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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((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^n\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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