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

Optimal. Leaf size=253 \[ \frac{-a^2 b C d^3+a^3 d^3 D+a b^2 B d^3+b^3 \left (-\left (3 A d^3-2 B c d^2+2 c^2 C d-2 c^3 D\right )\right )}{b^3 d^2 \sqrt{c+d x} (b c-a d)^2}-\frac{A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt{c+d x} (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (a^2 b (6 c D+C d)-3 a^3 d D-a b^2 (4 c C-B d)+b^3 (2 B c-3 A d)\right )}{b^{5/2} (b c-a d)^{5/2}}+\frac{2 D \sqrt{c+d x}}{b^2 d^2} \]

[Out]

(a*b^2*B*d^3 - a^2*b*C*d^3 + a^3*d^3*D - b^3*(2*c^2*C*d - 2*B*c*d^2 + 3*A*d^3 - 2*c^3*D))/(b^3*d^2*(b*c - a*d)
^2*Sqrt[c + d*x]) - (A - (a*(b^2*B - a*b*C + a^2*D))/b^3)/((b*c - a*d)*(a + b*x)*Sqrt[c + d*x]) + (2*D*Sqrt[c
+ d*x])/(b^2*d^2) - ((b^3*(2*B*c - 3*A*d) - a*b^2*(4*c*C - B*d) - 3*a^3*d*D + a^2*b*(C*d + 6*c*D))*ArcTanh[(Sq
rt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(5/2)*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.601382, antiderivative size = 253, 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, 1261, 208} \[ \frac{-a^2 b C d^3+a^3 d^3 D+a b^2 B d^3+b^3 \left (-\left (3 A d^3-2 B c d^2+2 c^2 C d-2 c^3 D\right )\right )}{b^3 d^2 \sqrt{c+d x} (b c-a d)^2}-\frac{A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt{c+d x} (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (a^2 b (6 c D+C d)-3 a^3 d D-a b^2 (4 c C-B d)+b^3 (2 B c-3 A d)\right )}{b^{5/2} (b c-a d)^{5/2}}+\frac{2 D \sqrt{c+d x}}{b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2*B*d^3 - a^2*b*C*d^3 + a^3*d^3*D - b^3*(2*c^2*C*d - 2*B*c*d^2 + 3*A*d^3 - 2*c^3*D))/(b^3*d^2*(b*c - a*d)
^2*Sqrt[c + d*x]) - (A - (a*(b^2*B - a*b*C + a^2*D))/b^3)/((b*c - a*d)*(a + b*x)*Sqrt[c + d*x]) + (2*D*Sqrt[c
+ d*x])/(b^2*d^2) - ((b^3*(2*B*c - 3*A*d) - a*b^2*(4*c*C - B*d) - 3*a^3*d*D + a^2*b*(C*d + 6*c*D))*ArcTanh[(Sq
rt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(5/2)*(b*c - a*d)^(5/2))

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 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 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 (c+d x)^{3/2}} \, dx &=-\frac{A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) \sqrt{c+d x}}+\frac{\int \frac{-\frac{b^3 (2 B c-3 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) (c+d x)^{3/2}} \, dx}{-b c+a d}\\ &=-\frac{A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) \sqrt{c+d 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-3 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}}{x^2 \left (\frac{-b c+a d}{d}+\frac{b x^2}{d}\right )} \, dx,x,\sqrt{c+d x}\right )}{d (b c-a d)}\\ &=-\frac{A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) \sqrt{c+d x}}-\frac{2 \operatorname{Subst}\left (\int \left (-\frac{(b c-a d) D}{b^2 d}+\frac{a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+3 A d^3-2 c^3 D\right )}{2 b^3 d (b c-a d) x^2}+\frac{d \left (b^3 (2 B c-3 A d)-a b^2 (4 c C-B d)-3 a^3 d D+a^2 b (C d+6 c D)\right )}{2 b^2 (b c-a d) \left (b c-a d-b x^2\right )}\right ) \, dx,x,\sqrt{c+d x}\right )}{d (b c-a d)}\\ &=\frac{a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+3 A d^3-2 c^3 D\right )}{b^3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) \sqrt{c+d x}}+\frac{2 D \sqrt{c+d x}}{b^2 d^2}-\frac{\left (b^3 (2 B c-3 A d)-a b^2 (4 c C-B d)-3 a^3 d D+a^2 b (C d+6 c D)\right ) \operatorname{Subst}\left (\int \frac{1}{b c-a d-b x^2} \, dx,x,\sqrt{c+d x}\right )}{b^2 (b c-a d)^2}\\ &=\frac{a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+3 A d^3-2 c^3 D\right )}{b^3 d^2 (b c-a d)^2 \sqrt{c+d x}}-\frac{A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) \sqrt{c+d x}}+\frac{2 D \sqrt{c+d x}}{b^2 d^2}-\frac{\left (b^3 (2 B c-3 A d)-a b^2 (4 c C-B d)-3 a^3 d D+a^2 b (C d+6 c D)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.752097, size = 283, normalized size = 1.12 \[ \frac{\sqrt{c+d x} \left (a \left (a^2 D-a b C+b^2 B\right )-A b^3\right )}{b^2 (a+b x) (b c-a d)^2}+\frac{d \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2} (b c-a d)^{5/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (a^2 b (3 c D+C d)-2 a^3 d D-2 a b^2 c C+b^3 (B c-A d)\right )}{b^{5/2} (b c-a d)^{5/2}}+\frac{2 \left (-A d^3+B c d^2-c^2 C d+c^3 D\right )}{d^2 \sqrt{c+d x} (b c-a d)^2}+\frac{2 D \sqrt{c+d x}}{b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 604, normalized size = 2.4 \begin{align*} 2\,{\frac{D\sqrt{dx+c}}{{b}^{2}{d}^{2}}}-2\,{\frac{Ad}{ \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}+2\,{\frac{Bc}{ \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}-2\,{\frac{{c}^{2}C}{d \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}+2\,{\frac{D{c}^{3}}{{d}^{2} \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}-{\frac{bdA}{ \left ( ad-bc \right ) ^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{Bda}{ \left ( ad-bc \right ) ^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}-{\frac{Cd{a}^{2}}{ \left ( ad-bc \right ) ^{2}b \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{{a}^{3}dD}{ \left ( ad-bc \right ) ^{2}{b}^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}-3\,{\frac{bdA}{ \left ( ad-bc \right ) ^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+{\frac{Bda}{ \left ( ad-bc \right ) ^{2}}\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{bBc}{ \left ( ad-bc \right ) ^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+{\frac{Cd{a}^{2}}{ \left ( ad-bc \right ) ^{2}b}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}-4\,{\frac{Cac}{ \left ( ad-bc \right ) ^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-3\,{\frac{{a}^{3}dD}{ \left ( ad-bc \right ) ^{2}{b}^{2}\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}{ \left ( ad-bc \right ) ^{2}b\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)^(3/2),x)

[Out]

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

[Out]

Timed out

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Giac [A]  time = 2.25407, size = 524, normalized size = 2.07 \begin{align*} \frac{{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - 3 \, D a^{3} d + C a^{2} b d + B a b^{2} d - 3 \, 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^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \,{\left (d x + c\right )} D b^{3} c^{3} - 2 \, D b^{3} c^{4} - 2 \,{\left (d x + c\right )} C b^{3} c^{2} d + 2 \, D a b^{2} c^{3} d + 2 \, C b^{3} c^{3} d + 2 \,{\left (d x + c\right )} B b^{3} c d^{2} - 2 \, C a b^{2} c^{2} d^{2} - 2 \, B b^{3} c^{2} d^{2} +{\left (d x + c\right )} D a^{3} d^{3} -{\left (d x + c\right )} C a^{2} b d^{3} +{\left (d x + c\right )} B a b^{2} d^{3} - 3 \,{\left (d x + c\right )} A b^{3} d^{3} + 2 \, B a b^{2} c d^{3} + 2 \, A b^{3} c d^{3} - 2 \, A a b^{2} d^{4}}{{\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )}{\left ({\left (d x + c\right )}^{\frac{3}{2}} b - \sqrt{d x + c} b c + \sqrt{d x + c} a d\right )}} + \frac{2 \, \sqrt{d x + c} D}{b^{2} d^{2}} \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)^(3/2),x, algorithm="giac")

[Out]

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