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

Optimal. Leaf size=350 \[ -\frac{-a^2 b C d^3+a^3 d^3 D+a b^2 B d^3+b^3 \left (-\left (5 A d^3-4 B c d^2+4 c^2 C d-4 c^3 D\right )\right )}{2 b^3 d \sqrt{c+d x} (b c-a d)^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (-a^2 b d (C d-12 c D)-3 a^3 d^2 D+a b^2 \left (-3 B d^2-24 c^2 D+8 c C d\right )+b^3 \left (15 A d^2-12 B c d+8 c^2 C\right )\right )}{4 b^{5/2} (b c-a d)^{7/2}}-\frac{A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt{c+d x} (b c-a d)}-\frac{\sqrt{c+d x} \left (3 a^2 b (4 c D+C d)-7 a^3 d D-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{4 b^2 (a+b x) (b c-a d)^3} \]

[Out]

-(a*b^2*B*d^3 - a^2*b*C*d^3 + a^3*d^3*D - b^3*(4*c^2*C*d - 4*B*c*d^2 + 5*A*d^3 - 4*c^3*D))/(2*b^3*d*(b*c - a*d
)^3*Sqrt[c + d*x]) - (A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(2*b^3*(b*c - a*d)*(a + b*x)^2*Sqrt[c + d*x]) - ((b^3
*(4*B*c - 5*A*d) - a*b^2*(8*c*C - B*d) - 7*a^3*d*D + 3*a^2*b*(C*d + 4*c*D))*Sqrt[c + d*x])/(4*b^2*(b*c - a*d)^
3*(a + b*x)) - ((b^3*(8*c^2*C - 12*B*c*d + 15*A*d^2) - 3*a^3*d^2*D - a^2*b*d*(C*d - 12*c*D) + a*b^2*(8*c*C*d -
3*B*d^2 - 24*c^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*b^(5/2)*(b*c - a*d)^(7/2))

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Rubi [A]  time = 0.832386, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1621, 897, 1259, 453, 208} \[ -\frac{-a^2 b C d^3+a^3 d^3 D+a b^2 B d^3+b^3 \left (-\left (5 A d^3-4 B c d^2+4 c^2 C d-4 c^3 D\right )\right )}{2 b^3 d \sqrt{c+d x} (b c-a d)^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (-a^2 b d (C d-12 c D)-3 a^3 d^2 D+a b^2 \left (-3 B d^2-24 c^2 D+8 c C d\right )+b^3 \left (15 A d^2-12 B c d+8 c^2 C\right )\right )}{4 b^{5/2} (b c-a d)^{7/2}}-\frac{A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt{c+d x} (b c-a d)}-\frac{\sqrt{c+d x} \left (3 a^2 b (4 c D+C d)-7 a^3 d D-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{4 b^2 (a+b x) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^3*(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*(4*c^2*C*d - 4*B*c*d^2 + 5*A*d^3 - 4*c^3*D))/(2*b^3*d*(b*c - a*d
)^3*Sqrt[c + d*x]) - (A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(2*b^3*(b*c - a*d)*(a + b*x)^2*Sqrt[c + d*x]) - ((b^3
*(4*B*c - 5*A*d) - a*b^2*(8*c*C - B*d) - 7*a^3*d*D + 3*a^2*b*(C*d + 4*c*D))*Sqrt[c + d*x])/(4*b^2*(b*c - a*d)^
3*(a + b*x)) - ((b^3*(8*c^2*C - 12*B*c*d + 15*A*d^2) - 3*a^3*d^2*D - a^2*b*d*(C*d - 12*c*D) + a*b^2*(8*c*C*d -
3*B*d^2 - 24*c^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*b^(5/2)*(b*c - a*d)^(7/2))

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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 453

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

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)^3 (c+d x)^{3/2}} \, dx &=-\frac{A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt{c+d x}}-\frac{\int \frac{-\frac{b^3 (4 B c-5 A d)-a b^2 (4 c C-B d)+a^3 d D-a^2 b (C d-4 c D)}{2 b^3}-\frac{2 (b c-a d) (b C-a D) x}{b^2}-2 \left (c-\frac{a d}{b}\right ) D x^2}{(a+b x)^2 (c+d x)^{3/2}} \, dx}{2 (b c-a d)}\\ &=-\frac{A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt{c+d x}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{-2 c^2 \left (c-\frac{a d}{b}\right ) D+\frac{2 c d (b c-a d) (b C-a D)}{b^2}-\frac{d^2 \left (b^3 (4 B c-5 A d)-a b^2 (4 c C-B d)+a^3 d D-a^2 b (C d-4 c D)\right )}{2 b^3}}{d^2}-\frac{\left (-4 c \left (c-\frac{a d}{b}\right ) D+\frac{2 d (b c-a d) (b C-a D)}{b^2}\right ) x^2}{d^2}-\frac{2 \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 )^2} \, dx,x,\sqrt{c+d x}\right )}{d (b c-a d)}\\ &=-\frac{A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt{c+d x}}-\frac{\left (b^3 (4 B c-5 A d)-a b^2 (8 c C-B d)-7 a^3 d D+3 a^2 b (C d+4 c D)\right ) \sqrt{c+d x}}{4 b^2 (b c-a d)^3 (a+b x)}+\frac{d^3 \operatorname{Subst}\left (\int \frac{-\frac{(b c-a d) \left (a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+5 A d^3-4 c^3 D\right )\right )}{b d^5}-\frac{\left (a^3 d^3 D+3 a^2 b d^2 (C d-4 c D)-a b^2 d \left (8 c C d-B d^2-24 c^2 D\right )+b^3 \left (4 B c d^2-5 A d^3-8 c^3 D\right )\right ) x^2}{2 d^5}}{x^2 \left (\frac{-b c+a d}{d}+\frac{b x^2}{d}\right )} \, dx,x,\sqrt{c+d x}\right )}{2 b^2 (b c-a d)^3}\\ &=-\frac{a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+5 A d^3-4 c^3 D\right )}{2 b^3 d (b c-a d)^3 \sqrt{c+d x}}-\frac{A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt{c+d x}}-\frac{\left (b^3 (4 B c-5 A d)-a b^2 (8 c C-B d)-7 a^3 d D+3 a^2 b (C d+4 c D)\right ) \sqrt{c+d x}}{4 b^2 (b c-a d)^3 (a+b x)}+\frac{\left (b^3 \left (8 c^2 C-12 B c d+15 A d^2\right )-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (8 c C d-3 B d^2-24 c^2 D\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{-b c+a d}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{4 b^2 d (b c-a d)^3}\\ &=-\frac{a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+5 A d^3-4 c^3 D\right )}{2 b^3 d (b c-a d)^3 \sqrt{c+d x}}-\frac{A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt{c+d x}}-\frac{\left (b^3 (4 B c-5 A d)-a b^2 (8 c C-B d)-7 a^3 d D+3 a^2 b (C d+4 c D)\right ) \sqrt{c+d x}}{4 b^2 (b c-a d)^3 (a+b x)}-\frac{\left (b^3 \left (8 c^2 C-12 B c d+15 A d^2\right )-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (8 c C d-3 B d^2-24 c^2 D\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.43742, size = 482, normalized size = 1.38 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (3 a^2 b c d D+a^3 \left (-d^2\right ) D-3 a b^2 c^2 D+b^3 \left (A d^2-B c d+c^2 C\right )\right )}{b^{5/2} (b c-a d)^{7/2}}-\frac{\sqrt{c+d x} \left (a^2 b (3 c D+C d)-2 a^3 d D-2 a b^2 c C+b^3 (B c-A d)\right )}{b^2 (a+b x) (b c-a d)^3}+\frac{\sqrt{c+d x} \left (a \left (a^2 D-a b C+b^2 B\right )-A b^3\right )}{2 b^2 (a+b x)^2 (b c-a d)^2}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (a^2 b (3 c D+C d)-2 a^3 d D-2 a b^2 c C+b^3 (B c-A d)\right )}{b^{5/2} (b c-a d)^{7/2}}-\frac{3 d \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \left (d (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )-\sqrt{b} \sqrt{c+d x} \sqrt{b c-a d}\right )}{4 b^{5/2} (a+b x) (b c-a d)^{7/2}}+\frac{2 \left (A d^3-B c d^2+c^2 C d+c^3 (-D)\right )}{d \sqrt{c+d x} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.028, size = 1225, normalized size = 3.5 \begin{align*} \text{result too large to display} \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)^3/(d*x+c)^(3/2),x)

[Out]

-3*d/(a*d-b*c)^3/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-1/4*d^2/(a*d-b*c)^3
/(b*d*x+a*d)^2*b*(d*x+c)^(1/2)*B*a*c+2*d/(a*d-b*c)^3/(b*d*x+a*d)^2*b*(d*x+c)^(1/2)*C*a*c^2+3*d/(a*d-b*c)^3/(b*
d*x+a*d)^2*(d*x+c)^(3/2)*D*a^2*c-9/4*d^3/(a*d-b*c)^3/(b*d*x+a*d)^2*b*(d*x+c)^(1/2)*A*a-3/4*d^3/(a*d-b*c)^3/(b*
d*x+a*d)^2/b^2*(d*x+c)^(1/2)*D*a^4-2*d/(a*d-b*c)^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1
/2))*C*a*c+3/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2*b*(d*x+c)^(3/2)*B*a+3*d/(a*d-b*c)^3*b/((a*d-b*c)*b)^(1/2)*arctan(
b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*B*c+1/4*d^2/(a*d-b*c)^3/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*
d-b*c)*b)^(1/2))*a^2*C+3/4*d^2/(a*d-b*c)^3/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-5/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2/b*(d*x+c)^(3/2)*a^3*D+9/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2*b^2*(d*x+c)^(
1/2)*A*c-1/4*d^3/(a*d-b*c)^3/(b*d*x+a*d)^2/b*(d*x+c)^(1/2)*C*a^3-7/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2*(d*x+c)^(1/
2)*C*a^2*c+d/(a*d-b*c)^3/(b*d*x+a*d)^2*b^2*(d*x+c)^(3/2)*B*c+2/d/(a*d-b*c)^3/(d*x+c)^(1/2)*D*c^3-d/(a*d-b*c)^3
/(b*d*x+a*d)^2*b^2*(d*x+c)^(1/2)*B*c^2-3*d/(a*d-b*c)^3/(b*d*x+a*d)^2*(d*x+c)^(1/2)*D*a^2*c^2+2*d/(a*d-b*c)^3/(
d*x+c)^(1/2)*B*c-2*d/(a*d-b*c)^3/(b*d*x+a*d)^2*b*(d*x+c)^(3/2)*C*a*c+15/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2/b*(d*x
+c)^(1/2)*D*a^3*c-2/(a*d-b*c)^3/(d*x+c)^(1/2)*C*c^2-2*d^2/(a*d-b*c)^3/(d*x+c)^(1/2)*A+5/4*d^3/(a*d-b*c)^3/(b*d
*x+a*d)^2*(d*x+c)^(1/2)*B*a^2-15/4*d^2/(a*d-b*c)^3*b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^
(1/2))*A-7/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2*b^2*(d*x+c)^(3/2)*A+1/4*d^2/(a*d-b*c)^3/(b*d*x+a*d)^2*(d*x+c)^(3/2)
*C*a^2-2/(a*d-b*c)^3*b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*C*c^2+6/(a*d-b*c)^3/((a
*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*D*a*c^2+3/4*d^2/(a*d-b*c)^3/((a*d-b*c)*b)^(1/2)*a
rctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*B*a

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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)^3/(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)^3/(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)**3/(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 2.02575, size = 833, normalized size = 2.38 \begin{align*} -\frac{{\left (24 \, D a b^{2} c^{2} - 8 \, C b^{3} c^{2} - 12 \, D a^{2} b c d - 8 \, C a b^{2} c d + 12 \, B b^{3} c d + 3 \, D a^{3} d^{2} + C a^{2} b d^{2} + 3 \, B a b^{2} d^{2} - 15 \, A b^{3} d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{4 \,{\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} \sqrt{-b^{2} c + a b d}} - \frac{2 \,{\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} \sqrt{d x + c}} - \frac{12 \,{\left (d x + c\right )}^{\frac{3}{2}} D a^{2} b^{2} c d - 8 \,{\left (d x + c\right )}^{\frac{3}{2}} C a b^{3} c d + 4 \,{\left (d x + c\right )}^{\frac{3}{2}} B b^{4} c d - 12 \, \sqrt{d x + c} D a^{2} b^{2} c^{2} d + 8 \, \sqrt{d x + c} C a b^{3} c^{2} d - 4 \, \sqrt{d x + c} B b^{4} c^{2} d - 5 \,{\left (d x + c\right )}^{\frac{3}{2}} D a^{3} b d^{2} +{\left (d x + c\right )}^{\frac{3}{2}} C a^{2} b^{2} d^{2} + 3 \,{\left (d x + c\right )}^{\frac{3}{2}} B a b^{3} d^{2} - 7 \,{\left (d x + c\right )}^{\frac{3}{2}} A b^{4} d^{2} + 15 \, \sqrt{d x + c} D a^{3} b c d^{2} - 7 \, \sqrt{d x + c} C a^{2} b^{2} c d^{2} - \sqrt{d x + c} B a b^{3} c d^{2} + 9 \, \sqrt{d x + c} A b^{4} c d^{2} - 3 \, \sqrt{d x + c} D a^{4} d^{3} - \sqrt{d x + c} C a^{3} b d^{3} + 5 \, \sqrt{d x + c} B a^{2} b^{2} d^{3} - 9 \, \sqrt{d x + c} A a b^{3} d^{3}}{4 \,{\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )}{\left ({\left (d x + c\right )} b - b c + a d\right )}^{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)^3/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-1/4*(24*D*a*b^2*c^2 - 8*C*b^3*c^2 - 12*D*a^2*b*c*d - 8*C*a*b^2*c*d + 12*B*b^3*c*d + 3*D*a^3*d^2 + C*a^2*b*d^2
+ 3*B*a*b^2*d^2 - 15*A*b^3*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^
2*b^3*c*d^2 - a^3*b^2*d^3)*sqrt(-b^2*c + a*b*d)) - 2*(D*c^3 - C*c^2*d + B*c*d^2 - A*d^3)/((b^3*c^3*d - 3*a*b^2
*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*sqrt(d*x + c)) - 1/4*(12*(d*x + c)^(3/2)*D*a^2*b^2*c*d - 8*(d*x + c)^(3/2)
*C*a*b^3*c*d + 4*(d*x + c)^(3/2)*B*b^4*c*d - 12*sqrt(d*x + c)*D*a^2*b^2*c^2*d + 8*sqrt(d*x + c)*C*a*b^3*c^2*d
- 4*sqrt(d*x + c)*B*b^4*c^2*d - 5*(d*x + c)^(3/2)*D*a^3*b*d^2 + (d*x + c)^(3/2)*C*a^2*b^2*d^2 + 3*(d*x + c)^(3
/2)*B*a*b^3*d^2 - 7*(d*x + c)^(3/2)*A*b^4*d^2 + 15*sqrt(d*x + c)*D*a^3*b*c*d^2 - 7*sqrt(d*x + c)*C*a^2*b^2*c*d
^2 - sqrt(d*x + c)*B*a*b^3*c*d^2 + 9*sqrt(d*x + c)*A*b^4*c*d^2 - 3*sqrt(d*x + c)*D*a^4*d^3 - sqrt(d*x + c)*C*a
^3*b*d^3 + 5*sqrt(d*x + c)*B*a^2*b^2*d^3 - 9*sqrt(d*x + c)*A*a*b^3*d^3)/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*
c*d^2 - a^3*b^2*d^3)*((d*x + c)*b - b*c + a*d)^2)