3.6 \(\int \frac{A+B x+C x^2+D x^3}{(a+b x)^2 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=201 \[ -\frac{\sqrt{c+d x} \left (A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (3 a^2 b (2 c D+C d)-5 a^3 d D-a b^2 (B d+4 c C)+b^3 (2 B c-A d)\right )}{b^{7/2} (b c-a d)^{3/2}}+\frac{2 \sqrt{c+d x} (-2 a d D-b c D+b C d)}{b^3 d^2}+\frac{2 D (c+d x)^{3/2}}{3 b^2 d^2} \]

[Out]

(2*(b*C*d - b*c*D - 2*a*d*D)*Sqrt[c + d*x])/(b^3*d^2) - ((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*Sqrt[c + d*x])/
((b*c - a*d)*(a + b*x)) + (2*D*(c + d*x)^(3/2))/(3*b^2*d^2) - ((b^3*(2*B*c - A*d) - a*b^2*(4*c*C + B*d) - 5*a^
3*d*D + 3*a^2*b*(C*d + 2*c*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(7/2)*(b*c - a*d)^(3/2))

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Rubi [A]  time = 0.518193, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1621, 897, 1153, 208} \[ -\frac{\sqrt{c+d x} \left (A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (3 a^2 b (2 c D+C d)-5 a^3 d D-a b^2 (B d+4 c C)+b^3 (2 B c-A d)\right )}{b^{7/2} (b c-a d)^{3/2}}+\frac{2 \sqrt{c+d x} (-2 a d D-b c D+b C d)}{b^3 d^2}+\frac{2 D (c+d x)^{3/2}}{3 b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*C*d - b*c*D - 2*a*d*D)*Sqrt[c + d*x])/(b^3*d^2) - ((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*Sqrt[c + d*x])/
((b*c - a*d)*(a + b*x)) + (2*D*(c + d*x)^(3/2))/(3*b^2*d^2) - ((b^3*(2*B*c - A*d) - a*b^2*(4*c*C + B*d) - 5*a^
3*d*D + 3*a^2*b*(C*d + 2*c*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(7/2)*(b*c - a*d)^(3/2))

Rule 1621

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]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], 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] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2+D x^3}{(a+b x)^2 \sqrt{c+d x}} \, dx &=-\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) \sqrt{c+d x}}{(b c-a d) (a+b x)}+\frac{\int \frac{-\frac{b^3 (2 B c-A d)-a b^2 (2 c C+B d)-a^3 d D+a^2 b (C d+2 c D)}{2 b^3}-\frac{(b c-a d) (b C-a D) x}{b^2}-\left (c-\frac{a d}{b}\right ) D x^2}{(a+b x) \sqrt{c+d 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 ) \sqrt{c+d x}}{(b c-a d) (a+b x)}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{-c^2 \left (c-\frac{a d}{b}\right ) D+\frac{c d (b c-a d) (b C-a D)}{b^2}-\frac{d^2 \left (b^3 (2 B c-A d)-a b^2 (2 c C+B d)-a^3 d D+a^2 b (C d+2 c D)\right )}{2 b^3}}{d^2}-\frac{\left (-2 c \left (c-\frac{a d}{b}\right ) D+\frac{d (b c-a d) (b C-a D)}{b^2}\right ) x^2}{d^2}-\frac{\left (c-\frac{a d}{b}\right ) D x^4}{d^2}}{\frac{-b c+a d}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d (b c-a d)}\\ &=-\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) \sqrt{c+d x}}{(b c-a d) (a+b x)}-\frac{2 \operatorname{Subst}\left (\int \left (-\frac{(b c-a d) (b C d-b c D-2 a d D)}{b^3 d}-\frac{(b c-a d) D x^2}{b^2 d}+\frac{-2 b^3 B c+4 a b^2 c C+A b^3 d+a b^2 B d-3 a^2 b C d-6 a^2 b c D+5 a^3 d D}{2 b^3 \left (a-\frac{b c}{d}+\frac{b x^2}{d}\right )}\right ) \, dx,x,\sqrt{c+d x}\right )}{d (b c-a d)}\\ &=\frac{2 (b C d-b c D-2 a d D) \sqrt{c+d x}}{b^3 d^2}-\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) \sqrt{c+d x}}{(b c-a d) (a+b x)}+\frac{2 D (c+d x)^{3/2}}{3 b^2 d^2}+\frac{\left (b^3 (2 B c-A d)-a b^2 (4 c C+B d)-5 a^3 d D+3 a^2 b (C d+2 c D)\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{b^3 d (b c-a d)}\\ &=\frac{2 (b C d-b c D-2 a d D) \sqrt{c+d x}}{b^3 d^2}-\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) \sqrt{c+d x}}{(b c-a d) (a+b x)}+\frac{2 D (c+d x)^{3/2}}{3 b^2 d^2}-\frac{\left (b^3 (2 B c-A d)-a b^2 (4 c C+B d)-5 a^3 d D+3 a^2 b (C d+2 c D)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.476918, size = 244, normalized size = 1.21 \[ \frac{\sqrt{c+d x} \left (a \left (a^2 D-a b C+b^2 B\right )-A b^3\right )}{b^3 (a+b x) (b c-a d)}+\frac{d \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2} (b c-a d)^{3/2}}-\frac{2 \left (3 a^2 D-2 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2} \sqrt{b c-a d}}+\frac{2 \sqrt{c+d x} (-2 a d D-b c D+b C d)}{b^3 d^2}+\frac{2 D (c+d x)^{3/2}}{3 b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*C*d - b*c*D - 2*a*d*D)*Sqrt[c + d*x])/(b^3*d^2) + ((-(A*b^3) + a*(b^2*B - a*b*C + a^2*D))*Sqrt[c + d*x])
/(b^3*(b*c - a*d)*(a + b*x)) + (2*D*(c + d*x)^(3/2))/(3*b^2*d^2) - (2*(b^2*B - 2*a*b*C + 3*a^2*D)*ArcTanh[(Sqr
t[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(7/2)*Sqrt[b*c - a*d]) + (d*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*ArcTa
nh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(7/2)*(b*c - a*d)^(3/2))

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Maple [B]  time = 0.019, size = 566, normalized size = 2.8 \begin{align*}{\frac{2\,D}{3\,{b}^{2}{d}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{C\sqrt{dx+c}}{{b}^{2}d}}-4\,{\frac{Da\sqrt{dx+c}}{{b}^{3}d}}-2\,{\frac{cD\sqrt{dx+c}}{{b}^{2}{d}^{2}}}+{\frac{Ad}{ \left ( ad-bc \right ) \left ( bdx+ad \right ) }\sqrt{dx+c}}-{\frac{Bda}{ \left ( ad-bc \right ) b \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{Cd{a}^{2}}{{b}^{2} \left ( ad-bc \right ) \left ( bdx+ad \right ) }\sqrt{dx+c}}-{\frac{{a}^{3}dD}{{b}^{3} \left ( ad-bc \right ) \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{Ad}{ad-bc}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}+{\frac{Bda}{ \left ( ad-bc \right ) b}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}-2\,{\frac{Bc}{ \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-3\,{\frac{Cd{a}^{2}}{{b}^{2} \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+4\,{\frac{Cac}{ \left ( ad-bc \right ) b\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+5\,{\frac{{a}^{3}dD}{{b}^{3} \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{D{a}^{2}c}{{b}^{2} \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(1/2),x)

[Out]

2/3*D*(d*x+c)^(3/2)/b^2/d^2+2/d/b^2*C*(d*x+c)^(1/2)-4/d/b^3*D*a*(d*x+c)^(1/2)-2/d^2/b^2*D*c*(d*x+c)^(1/2)+d/(a
*d-b*c)*(d*x+c)^(1/2)/(b*d*x+a*d)*A-d/b/(a*d-b*c)*(d*x+c)^(1/2)/(b*d*x+a*d)*B*a+d/b^2/(a*d-b*c)*(d*x+c)^(1/2)/
(b*d*x+a*d)*C*a^2-d/b^3/(a*d-b*c)*(d*x+c)^(1/2)/(b*d*x+a*d)*a^3*D+d/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*
x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*A+d/b/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)
)*B*a-2/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*B*c-3*d/b^2/(a*d-b*c)/((a*d-
b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*C*a^2+4/b/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*
x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*C*a*c+5*d/b^3/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*
b)^(1/2))*a^3*D-6/b^2/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*D*a^2*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**2/(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.27904, size = 366, normalized size = 1.82 \begin{align*} \frac{{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - 5 \, D a^{3} d + 3 \, C a^{2} b d - B a b^{2} d - A b^{3} d\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{4} c - a b^{3} d\right )} \sqrt{-b^{2} c + a b d}} + \frac{\sqrt{d x + c} D a^{3} d - \sqrt{d x + c} C a^{2} b d + \sqrt{d x + c} B a b^{2} d - \sqrt{d x + c} A b^{3} d}{{\left (b^{4} c - a b^{3} d\right )}{\left ({\left (d x + c\right )} b - b c + a d\right )}} + \frac{2 \,{\left ({\left (d x + c\right )}^{\frac{3}{2}} D b^{4} d^{4} - 3 \, \sqrt{d x + c} D b^{4} c d^{4} - 6 \, \sqrt{d x + c} D a b^{3} d^{5} + 3 \, \sqrt{d x + c} C b^{4} d^{5}\right )}}{3 \, b^{6} d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

(6*D*a^2*b*c - 4*C*a*b^2*c + 2*B*b^3*c - 5*D*a^3*d + 3*C*a^2*b*d - B*a*b^2*d - A*b^3*d)*arctan(sqrt(d*x + c)*b
/sqrt(-b^2*c + a*b*d))/((b^4*c - a*b^3*d)*sqrt(-b^2*c + a*b*d)) + (sqrt(d*x + c)*D*a^3*d - sqrt(d*x + c)*C*a^2
*b*d + sqrt(d*x + c)*B*a*b^2*d - sqrt(d*x + c)*A*b^3*d)/((b^4*c - a*b^3*d)*((d*x + c)*b - b*c + a*d)) + 2/3*((
d*x + c)^(3/2)*D*b^4*d^4 - 3*sqrt(d*x + c)*D*b^4*c*d^4 - 6*sqrt(d*x + c)*D*a*b^3*d^5 + 3*sqrt(d*x + c)*C*b^4*d
^5)/(b^6*d^6)