3.1.15 \(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx\) [15]

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

[Out]

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

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Rubi [A]
time = 0.40, antiderivative size = 253, normalized size of antiderivative = 1.00, 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 = {1635, 911, 1275, 214} \begin {gather*} -\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 {a^3 d^3 D-a^2 b C d^3+a b^2 B d^3-\left (b^3 \left (3 A d^3-2 B c d^2-2 c^3 D+2 c^2 C d\right )\right )}{b^3 d^2 \sqrt {c+d x} (b c-a d)^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-3 a^3 d D+a^2 b (6 c D+C 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} \end {gather*}

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 214

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

Rule 911

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 1275

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 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 {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 \text {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 \text {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 ) \text {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*}

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Mathematica [A]
time = 0.81, size = 259, normalized size = 1.02 \begin {gather*} \frac {3 a^3 d^2 D (c+d x)+a^2 b d (c+d x) (-C d-4 c D+2 d D x)+b^3 \left (-A d^2 (c+3 d x)+2 c x \left (-c C d+B d^2+2 c^2 D+c d D x\right )\right )+a b^2 \left (4 c^3 D+d^3 (-2 A+B x)-2 c^2 d (C+D x)+c d^2 \left (3 B-4 D x^2\right )\right )}{b^2 d^2 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\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 ) \tan ^{-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 {gather*}

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]

(3*a^3*d^2*D*(c + d*x) + a^2*b*d*(c + d*x)*(-(C*d) - 4*c*D + 2*d*D*x) + b^3*(-(A*d^2*(c + 3*d*x)) + 2*c*x*(-(c
*C*d) + B*d^2 + 2*c^2*D + c*d*D*x)) + a*b^2*(4*c^3*D + d^3*(-2*A + B*x) - 2*c^2*d*(C + D*x) + c*d^2*(3*B - 4*D
*x^2)))/(b^2*d^2*(b*c - a*d)^2*(a + b*x)*Sqrt[c + d*x]) + ((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))*ArcTan[(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 [A]
time = 0.10, size = 228, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {\frac {2 D \sqrt {d x +c}}{b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {d x +c}}-\frac {2 d^{2} \left (\frac {\left (\frac {1}{2} A \,b^{3} d -\frac {1}{2} B a \,b^{2} d +\frac {1}{2} C \,a^{2} b d -\frac {1}{2} a^{3} d D\right ) \sqrt {d x +c}}{b \left (d x +c \right )+a d -b c}+\frac {\left (3 A \,b^{3} d -B a \,b^{2} d -2 B \,b^{3} c -C \,a^{2} b d +4 C a \,b^{2} c +3 a^{3} d D-6 D a^{2} b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} b^{2}}}{d^{2}}\)	\(228\)
default	\(\frac {\frac {2 D \sqrt {d x +c}}{b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {d x +c}}-\frac {2 d^{2} \left (\frac {\left (\frac {1}{2} A \,b^{3} d -\frac {1}{2} B a \,b^{2} d +\frac {1}{2} C \,a^{2} b d -\frac {1}{2} a^{3} d D\right ) \sqrt {d x +c}}{b \left (d x +c \right )+a d -b c}+\frac {\left (3 A \,b^{3} d -B a \,b^{2} d -2 B \,b^{3} c -C \,a^{2} b d +4 C a \,b^{2} c +3 a^{3} d D-6 D a^{2} b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} b^{2}}}{d^{2}}\)	\(228\)

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,method=_RETURNVERBOSE)

[Out]

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

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

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (236) = 472\).
time = 0.94, size = 1583, normalized size = 6.26 \begin {gather*} \left [\frac {{\left ({\left (3 \, D a^{4} c - {\left (C a^{3} b + B a^{2} b^{2} - 3 \, A a b^{3}\right )} c\right )} d^{3} - 2 \, {\left (3 \, D a^{3} b c^{2} - {\left (2 \, C a^{2} b^{2} - B a b^{3}\right )} c^{2}\right )} d^{2} + {\left ({\left (3 \, D a^{3} b - C a^{2} b^{2} - B a b^{3} + 3 \, A b^{4}\right )} d^{4} - 2 \, {\left (3 \, D a^{2} b^{2} c - {\left (2 \, C a b^{3} - B b^{4}\right )} c\right )} d^{3}\right )} x^{2} + {\left ({\left (3 \, D a^{4} - C a^{3} b - B a^{2} b^{2} + 3 \, A a b^{3}\right )} d^{4} - 3 \, {\left (D a^{3} b c - {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} c\right )} d^{3} - 2 \, {\left (3 \, D a^{2} b^{2} c^{2} - {\left (2 \, C a b^{3} - B b^{4}\right )} c^{2}\right )} d^{2}\right )} x\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (4 \, D a b^{4} c^{4} + 2 \, A a^{2} b^{3} d^{4} - {\left (3 \, D a^{4} b c - {\left (C a^{3} b^{2} - 3 \, B a^{2} b^{3} - A a b^{4}\right )} c\right )} d^{3} + {\left (7 \, D a^{3} b^{2} c^{2} + {\left (C a^{2} b^{3} + 3 \, B a b^{4} - A b^{5}\right )} c^{2}\right )} d^{2} + 2 \, {\left (D b^{5} c^{3} d - 3 \, D a b^{4} c^{2} d^{2} + 3 \, D a^{2} b^{3} c d^{3} - D a^{3} b^{2} d^{4}\right )} x^{2} - 2 \, {\left (4 \, D a^{2} b^{3} c^{3} + C a b^{4} c^{3}\right )} d + {\left (4 \, D b^{5} c^{4} + 2 \, {\left (C a b^{4} + B b^{5}\right )} c^{2} d^{2} - {\left (3 \, D a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - 3 \, A a b^{4}\right )} d^{4} + {\left (5 \, D a^{3} b^{2} c - {\left (C a^{2} b^{3} + B a b^{4} + 3 \, A b^{5}\right )} c\right )} d^{3} - 2 \, {\left (3 \, D a b^{4} c^{3} + C b^{5} c^{3}\right )} d\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a b^{6} c^{4} d^{2} - 3 \, a^{2} b^{5} c^{3} d^{3} + 3 \, a^{3} b^{4} c^{2} d^{4} - a^{4} b^{3} c d^{5} + {\left (b^{7} c^{3} d^{3} - 3 \, a b^{6} c^{2} d^{4} + 3 \, a^{2} b^{5} c d^{5} - a^{3} b^{4} d^{6}\right )} x^{2} + {\left (b^{7} c^{4} d^{2} - 2 \, a b^{6} c^{3} d^{3} + 2 \, a^{3} b^{4} c d^{5} - a^{4} b^{3} d^{6}\right )} x\right )}}, -\frac {{\left ({\left (3 \, D a^{4} c - {\left (C a^{3} b + B a^{2} b^{2} - 3 \, A a b^{3}\right )} c\right )} d^{3} - 2 \, {\left (3 \, D a^{3} b c^{2} - {\left (2 \, C a^{2} b^{2} - B a b^{3}\right )} c^{2}\right )} d^{2} + {\left ({\left (3 \, D a^{3} b - C a^{2} b^{2} - B a b^{3} + 3 \, A b^{4}\right )} d^{4} - 2 \, {\left (3 \, D a^{2} b^{2} c - {\left (2 \, C a b^{3} - B b^{4}\right )} c\right )} d^{3}\right )} x^{2} + {\left ({\left (3 \, D a^{4} - C a^{3} b - B a^{2} b^{2} + 3 \, A a b^{3}\right )} d^{4} - 3 \, {\left (D a^{3} b c - {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} c\right )} d^{3} - 2 \, {\left (3 \, D a^{2} b^{2} c^{2} - {\left (2 \, C a b^{3} - B b^{4}\right )} c^{2}\right )} d^{2}\right )} x\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (4 \, D a b^{4} c^{4} + 2 \, A a^{2} b^{3} d^{4} - {\left (3 \, D a^{4} b c - {\left (C a^{3} b^{2} - 3 \, B a^{2} b^{3} - A a b^{4}\right )} c\right )} d^{3} + {\left (7 \, D a^{3} b^{2} c^{2} + {\left (C a^{2} b^{3} + 3 \, B a b^{4} - A b^{5}\right )} c^{2}\right )} d^{2} + 2 \, {\left (D b^{5} c^{3} d - 3 \, D a b^{4} c^{2} d^{2} + 3 \, D a^{2} b^{3} c d^{3} - D a^{3} b^{2} d^{4}\right )} x^{2} - 2 \, {\left (4 \, D a^{2} b^{3} c^{3} + C a b^{4} c^{3}\right )} d + {\left (4 \, D b^{5} c^{4} + 2 \, {\left (C a b^{4} + B b^{5}\right )} c^{2} d^{2} - {\left (3 \, D a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - 3 \, A a b^{4}\right )} d^{4} + {\left (5 \, D a^{3} b^{2} c - {\left (C a^{2} b^{3} + B a b^{4} + 3 \, A b^{5}\right )} c\right )} d^{3} - 2 \, {\left (3 \, D a b^{4} c^{3} + C b^{5} c^{3}\right )} d\right )} x\right )} \sqrt {d x + c}}{a b^{6} c^{4} d^{2} - 3 \, a^{2} b^{5} c^{3} d^{3} + 3 \, a^{3} b^{4} c^{2} d^{4} - a^{4} b^{3} c d^{5} + {\left (b^{7} c^{3} d^{3} - 3 \, a b^{6} c^{2} d^{4} + 3 \, a^{2} b^{5} c d^{5} - a^{3} b^{4} d^{6}\right )} x^{2} + {\left (b^{7} c^{4} d^{2} - 2 \, a b^{6} c^{3} d^{3} + 2 \, a^{3} b^{4} c d^{5} - a^{4} b^{3} d^{6}\right )} x}\right ] \end {gather*}

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]

[1/2*(((3*D*a^4*c - (C*a^3*b + B*a^2*b^2 - 3*A*a*b^3)*c)*d^3 - 2*(3*D*a^3*b*c^2 - (2*C*a^2*b^2 - B*a*b^3)*c^2)
*d^2 + ((3*D*a^3*b - C*a^2*b^2 - B*a*b^3 + 3*A*b^4)*d^4 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*d^3)*x^2 +
((3*D*a^4 - C*a^3*b - B*a^2*b^2 + 3*A*a*b^3)*d^4 - 3*(D*a^3*b*c - (C*a^2*b^2 - B*a*b^3 + A*b^4)*c)*d^3 - 2*(3
*D*a^2*b^2*c^2 - (2*C*a*b^3 - B*b^4)*c^2)*d^2)*x)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d + 2*sqrt(b^2*c
- a*b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(4*D*a*b^4*c^4 + 2*A*a^2*b^3*d^4 - (3*D*a^4*b*c - (C*a^3*b^2 - 3*B*a^2*
b^3 - A*a*b^4)*c)*d^3 + (7*D*a^3*b^2*c^2 + (C*a^2*b^3 + 3*B*a*b^4 - A*b^5)*c^2)*d^2 + 2*(D*b^5*c^3*d - 3*D*a*b
^4*c^2*d^2 + 3*D*a^2*b^3*c*d^3 - D*a^3*b^2*d^4)*x^2 - 2*(4*D*a^2*b^3*c^3 + C*a*b^4*c^3)*d + (4*D*b^5*c^4 + 2*(
C*a*b^4 + B*b^5)*c^2*d^2 - (3*D*a^4*b - C*a^3*b^2 + B*a^2*b^3 - 3*A*a*b^4)*d^4 + (5*D*a^3*b^2*c - (C*a^2*b^3 +
B*a*b^4 + 3*A*b^5)*c)*d^3 - 2*(3*D*a*b^4*c^3 + C*b^5*c^3)*d)*x)*sqrt(d*x + c))/(a*b^6*c^4*d^2 - 3*a^2*b^5*c^3
*d^3 + 3*a^3*b^4*c^2*d^4 - a^4*b^3*c*d^5 + (b^7*c^3*d^3 - 3*a*b^6*c^2*d^4 + 3*a^2*b^5*c*d^5 - a^3*b^4*d^6)*x^2
+ (b^7*c^4*d^2 - 2*a*b^6*c^3*d^3 + 2*a^3*b^4*c*d^5 - a^4*b^3*d^6)*x), -(((3*D*a^4*c - (C*a^3*b + B*a^2*b^2 -
3*A*a*b^3)*c)*d^3 - 2*(3*D*a^3*b*c^2 - (2*C*a^2*b^2 - B*a*b^3)*c^2)*d^2 + ((3*D*a^3*b - C*a^2*b^2 - B*a*b^3 +
3*A*b^4)*d^4 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*d^3)*x^2 + ((3*D*a^4 - C*a^3*b - B*a^2*b^2 + 3*A*a*b^
3)*d^4 - 3*(D*a^3*b*c - (C*a^2*b^2 - B*a*b^3 + A*b^4)*c)*d^3 - 2*(3*D*a^2*b^2*c^2 - (2*C*a*b^3 - B*b^4)*c^2)*d
^2)*x)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (4*D*a*b^4*c^4 + 2*A*a^
2*b^3*d^4 - (3*D*a^4*b*c - (C*a^3*b^2 - 3*B*a^2*b^3 - A*a*b^4)*c)*d^3 + (7*D*a^3*b^2*c^2 + (C*a^2*b^3 + 3*B*a*
b^4 - A*b^5)*c^2)*d^2 + 2*(D*b^5*c^3*d - 3*D*a*b^4*c^2*d^2 + 3*D*a^2*b^3*c*d^3 - D*a^3*b^2*d^4)*x^2 - 2*(4*D*a
^2*b^3*c^3 + C*a*b^4*c^3)*d + (4*D*b^5*c^4 + 2*(C*a*b^4 + B*b^5)*c^2*d^2 - (3*D*a^4*b - C*a^3*b^2 + B*a^2*b^3
- 3*A*a*b^4)*d^4 + (5*D*a^3*b^2*c - (C*a^2*b^3 + B*a*b^4 + 3*A*b^5)*c)*d^3 - 2*(3*D*a*b^4*c^3 + C*b^5*c^3)*d)*
x)*sqrt(d*x + c))/(a*b^6*c^4*d^2 - 3*a^2*b^5*c^3*d^3 + 3*a^3*b^4*c^2*d^4 - a^4*b^3*c*d^5 + (b^7*c^3*d^3 - 3*a*
b^6*c^2*d^4 + 3*a^2*b^5*c*d^5 - a^3*b^4*d^6)*x^2 + (b^7*c^4*d^2 - 2*a*b^6*c^3*d^3 + 2*a^3*b^4*c*d^5 - a^4*b^3*
d^6)*x)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.61, size = 388, normalized size = 1.53 \begin {gather*} \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 {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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