\(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx\) [16]

Optimal result

Integrand size = 32, antiderivative size = 350 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+5 A d^3-4 c^3 D\right )}{2 b^3 d (b c-a d)^3 \sqrt {c+d x}}-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt {c+d x}}-\frac {\left (b^3 (4 B c-5 A d)-a b^2 (8 c C-B d)-7 a^3 d D+3 a^2 b (C d+4 c D)\right ) \sqrt {c+d x}}{4 b^2 (b c-a d)^3 (a+b x)}-\frac {\left (b^3 \left (8 c^2 C-12 B c d+15 A d^2\right )-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (8 c C d-3 B d^2-24 c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{5/2} (b c-a d)^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1635, 911, 1273, 464, 214} \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (-3 B d^2-24 c^2 D+8 c C d\right )+b^3 \left (15 A d^2-12 B c d+8 c^2 C\right )\right )}{4 b^{5/2} (b c-a d)^{7/2}}-\frac {a^3 d^3 D-a^2 b C d^3+a b^2 B d^3-\left (b^3 \left (5 A d^3-4 B c d^2-4 c^3 D+4 c^2 C d\right )\right )}{2 b^3 d \sqrt {c+d x} (b c-a d)^3}-\frac {\sqrt {c+d x} \left (-7 a^3 d D+3 a^2 b (4 c D+C d)-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{4 b^2 (a+b x) (b c-a d)^3} \]

[In]

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

[Out]

-1/2*(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))/(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 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 464

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 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 1273

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/(d + e*x^2))*(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))], 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 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*} \text {integral}& = -\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 {\text {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 \text {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 ) \text {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] (verified)

Time = 1.38 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {\frac {\sqrt {b} \left (-3 a^4 d^2 D (c+d x)-a^3 b d (c+d x) (C d+5 D (-2 c+d x))+4 b^4 c x (2 c (-C d+c D) x+B d (c+3 d x))+a b^3 \left (-8 c x \left (3 c C d-2 c^2 D+C d^2 x\right )+B d \left (2 c^2+21 c d x+3 d^2 x^2\right )\right )-A b^2 d \left (8 a^2 d^2+a b d (9 c+25 d x)+b^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )\right )+a^2 b^2 \left (8 c^3 D+d^3 x (5 B+C x)-2 c^2 d (7 C-6 D x)+c d^2 \left (13 B-5 C x+12 D x^2\right )\right )\right )}{d (-b c+a d)^3 (a+b x)^2 \sqrt {c+d x}}-\frac {\left (b^3 \left (8 c^2 C-12 B c d+15 A d^2\right )-3 a^3 d^2 D+a^2 b d (-C d+12 c D)+a b^2 \left (8 c C d-3 B d^2-24 c^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}}}{4 b^{5/2}} \]

[In]

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

[Out]

((Sqrt[b]*(-3*a^4*d^2*D*(c + d*x) - a^3*b*d*(c + d*x)*(C*d + 5*D*(-2*c + d*x)) + 4*b^4*c*x*(2*c*(-(C*d) + c*D)
*x + B*d*(c + 3*d*x)) + a*b^3*(-8*c*x*(3*c*C*d - 2*c^2*D + C*d^2*x) + B*d*(2*c^2 + 21*c*d*x + 3*d^2*x^2)) - A*
b^2*d*(8*a^2*d^2 + a*b*d*(9*c + 25*d*x) + b^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2)) + a^2*b^2*(8*c^3*D + d^3*x*(5*B
+ C*x) - 2*c^2*d*(7*C - 6*D*x) + c*d^2*(13*B - 5*C*x + 12*D*x^2))))/(d*(-(b*c) + a*d)^3*(a + b*x)^2*Sqrt[c +
d*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))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2))/(4*b^(5/2))

Maple [A] (verified)

Time = 1.96 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(-\frac {15 \left (\left (\left (A \,d^{2}-\frac {4}{5} B c d +\frac {8}{15} C \,c^{2}\right ) b^{3}-\frac {a \left (B \,d^{2}-\frac {8}{3} C c d +8 D c^{2}\right ) b^{2}}{5}-\frac {a^{2} b d \left (C d -12 D c \right )}{15}-\frac {a^{3} d^{2} D}{5}\right ) \sqrt {d x +c}\, \left (b x +a \right )^{2} d \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\frac {8 \sqrt {\left (a d -b c \right ) b}\, \left (\left (\frac {15 A \,d^{3} x^{2}}{8}+\frac {5 \left (-\frac {12 B x}{5}+A \right ) x c \,d^{2}}{8}-\frac {c^{2} \left (-4 C \,x^{2}+2 B x +A \right ) d}{4}-D c^{3} x^{2}\right ) b^{4}+\frac {9 a \left (\left (-\frac {1}{3} x^{2} B +\frac {25}{9} A x \right ) d^{3}+c \left (\frac {8}{9} C \,x^{2}-\frac {7}{3} B x +A \right ) d^{2}-\frac {2 c^{2} \left (-12 C x +B \right ) d}{9}-\frac {16 D c^{3} x}{9}\right ) b^{3}}{8}+a^{2} \left (\left (-\frac {5}{8} B x +A -\frac {1}{8} C \,x^{2}\right ) d^{3}-\frac {13 c \left (\frac {12}{13} D x^{2}-\frac {5}{13} C x +B \right ) d^{2}}{8}+\frac {7 c^{2} \left (-\frac {6 D x}{7}+C \right ) d}{4}-D c^{3}\right ) b^{2}+\frac {a^{3} \left (d x +c \right ) \left (\left (5 D x +C \right ) d -10 D c \right ) d b}{8}+\frac {3 D a^{4} d^{2} \left (d x +c \right )}{8}\right )}{15}\right )}{4 \sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}\, \left (a d -b c \right )^{3} \left (b x +a \right )^{2} b^{2} d}\)	\(379\)
derivativedivides	\(\frac {-\frac {2 d \left (\frac {\frac {d \left (7 A \,b^{3} d -3 B a \,b^{2} d -4 B \,b^{3} c -C \,a^{2} b d +8 C a \,b^{2} c +5 a^{3} d D-12 D a^{2} b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{8 b}+\frac {d \left (9 A a \,b^{3} d^{2}-9 A \,b^{4} c d -5 B \,a^{2} b^{2} d^{2}+B a \,b^{3} c d +4 B \,b^{4} c^{2}+C \,a^{3} b \,d^{2}+7 C \,a^{2} b^{2} c d -8 C a \,b^{3} c^{2}+3 D a^{4} d^{2}-15 D a^{3} b c d +12 D a^{2} b^{2} c^{2}\right ) \sqrt {d x +c}}{8 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\left (15 A \,b^{3} d^{2}-3 B a \,b^{2} d^{2}-12 B \,b^{3} c d -a^{2} b C \,d^{2}+8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-3 a^{3} d^{2} D+12 D a^{2} b c d -24 D a \,b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{3} \sqrt {d x +c}}}{d}\)	\(395\)
default	\(\frac {-\frac {2 d \left (\frac {\frac {d \left (7 A \,b^{3} d -3 B a \,b^{2} d -4 B \,b^{3} c -C \,a^{2} b d +8 C a \,b^{2} c +5 a^{3} d D-12 D a^{2} b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{8 b}+\frac {d \left (9 A a \,b^{3} d^{2}-9 A \,b^{4} c d -5 B \,a^{2} b^{2} d^{2}+B a \,b^{3} c d +4 B \,b^{4} c^{2}+C \,a^{3} b \,d^{2}+7 C \,a^{2} b^{2} c d -8 C a \,b^{3} c^{2}+3 D a^{4} d^{2}-15 D a^{3} b c d +12 D a^{2} b^{2} c^{2}\right ) \sqrt {d x +c}}{8 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\left (15 A \,b^{3} d^{2}-3 B a \,b^{2} d^{2}-12 B \,b^{3} c d -a^{2} b C \,d^{2}+8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-3 a^{3} d^{2} D+12 D a^{2} b c d -24 D a \,b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{3} \sqrt {d x +c}}}{d}\)	\(395\)

[In]

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

[Out]

-15/4/((a*d-b*c)*b)^(1/2)*(((A*d^2-4/5*B*c*d+8/15*C*c^2)*b^3-1/5*a*(B*d^2-8/3*C*c*d+8*D*c^2)*b^2-1/15*a^2*b*d*
(C*d-12*D*c)-1/5*a^3*d^2*D)*(d*x+c)^(1/2)*(b*x+a)^2*d*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))+8/15*((a*d-b
*c)*b)^(1/2)*((15/8*A*d^3*x^2+5/8*(-12/5*B*x+A)*x*c*d^2-1/4*c^2*(-4*C*x^2+2*B*x+A)*d-D*c^3*x^2)*b^4+9/8*a*((-1
/3*x^2*B+25/9*A*x)*d^3+c*(8/9*C*x^2-7/3*B*x+A)*d^2-2/9*c^2*(-12*C*x+B)*d-16/9*D*c^3*x)*b^3+a^2*((-5/8*B*x+A-1/
8*C*x^2)*d^3-13/8*c*(12/13*D*x^2-5/13*C*x+B)*d^2+7/4*c^2*(-6/7*D*x+C)*d-D*c^3)*b^2+1/8*a^3*(d*x+c)*((5*D*x+C)*
d-10*D*c)*d*b+3/8*D*a^4*d^2*(d*x+c)))/(d*x+c)^(1/2)/(a*d-b*c)^3/(b*x+a)^2/b^2/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1290 vs. \(2 (327) = 654\).

Time = 0.39 (sec) , antiderivative size = 2594, normalized size of antiderivative = 7.41 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/8*(((3*D*a^5*c + (C*a^4*b + 3*B*a^3*b^2 - 15*A*a^2*b^3)*c)*d^3 + ((3*D*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 1
5*A*b^5)*d^4 - 4*(3*D*a^2*b^3*c + (2*C*a*b^4 - 3*B*b^5)*c)*d^3 + 8*(3*D*a*b^4*c^2 - C*b^5*c^2)*d^2)*x^3 - 4*(3
*D*a^4*b*c^2 + (2*C*a^3*b^2 - 3*B*a^2*b^3)*c^2)*d^2 + (2*(3*D*a^4*b + C*a^3*b^2 + 3*B*a^2*b^3 - 15*A*a*b^4)*d^
4 - 3*(7*D*a^3*b^2*c + (5*C*a^2*b^3 - 9*B*a*b^4 + 5*A*b^5)*c)*d^3 + 12*(3*D*a^2*b^3*c^2 - (2*C*a*b^4 - B*b^5)*
c^2)*d^2 + 8*(3*D*a*b^4*c^3 - C*b^5*c^3)*d)*x^2 + 8*(3*D*a^3*b^2*c^3 - C*a^2*b^3*c^3)*d - (24*(C*a^2*b^3 - B*a
*b^4)*c^2*d^2 - (3*D*a^5 + C*a^4*b + 3*B*a^3*b^2 - 15*A*a^2*b^3)*d^4 + 6*(D*a^4*b*c + (C*a^3*b^2 - 3*B*a^2*b^3
+ 5*A*a*b^4)*c)*d^3 - 16*(3*D*a^2*b^3*c^3 - C*a*b^4*c^3)*d)*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*(8*D*a^2*b^4*c^4 + 8*A*a^3*b^3*d^4 + (3*D*a^5*b*c + (C*a^
4*b^2 - 13*B*a^3*b^3 + A*a^2*b^4)*c)*d^3 - (13*D*a^4*b^2*c^2 - (13*C*a^3*b^3 + 11*B*a^2*b^4 - 11*A*a*b^5)*c^2)
*d^2 + (8*D*b^6*c^4 + (5*D*a^4*b^2 - C*a^3*b^3 - 3*B*a^2*b^4 + 15*A*a*b^5)*d^4 - (17*D*a^3*b^3*c - 3*(3*C*a^2*
b^4 - 3*B*a*b^5 - 5*A*b^6)*c)*d^3 + 12*(D*a^2*b^4*c^2 + B*b^6*c^2)*d^2 - 8*(D*a*b^5*c^3 + C*b^6*c^3)*d)*x^2 +
2*(D*a^3*b^3*c^3 - (7*C*a^2*b^4 - B*a*b^5 - A*b^6)*c^3)*d + (16*D*a*b^5*c^4 + (3*D*a^5*b + C*a^4*b^2 - 5*B*a^3
*b^3 + 25*A*a^2*b^4)*d^4 - 4*(2*D*a^4*b^2*c - (C*a^3*b^3 - 4*B*a^2*b^4 - 5*A*a*b^5)*c)*d^3 - (7*D*a^3*b^3*c^2
- (19*C*a^2*b^4 + 17*B*a*b^5 - 5*A*b^6)*c^2)*d^2 - 4*(D*a^2*b^4*c^3 + (6*C*a*b^5 - B*b^6)*c^3)*d)*x)*sqrt(d*x
+ c))/(a^2*b^7*c^5*d - 4*a^3*b^6*c^4*d^2 + 6*a^4*b^5*c^3*d^3 - 4*a^5*b^4*c^2*d^4 + a^6*b^3*c*d^5 + (b^9*c^4*d^
2 - 4*a*b^8*c^3*d^3 + 6*a^2*b^7*c^2*d^4 - 4*a^3*b^6*c*d^5 + a^4*b^5*d^6)*x^3 + (b^9*c^5*d - 2*a*b^8*c^4*d^2 -
2*a^2*b^7*c^3*d^3 + 8*a^3*b^6*c^2*d^4 - 7*a^4*b^5*c*d^5 + 2*a^5*b^4*d^6)*x^2 + (2*a*b^8*c^5*d - 7*a^2*b^7*c^4*
d^2 + 8*a^3*b^6*c^3*d^3 - 2*a^4*b^5*c^2*d^4 - 2*a^5*b^4*c*d^5 + a^6*b^3*d^6)*x), -1/4*(((3*D*a^5*c + (C*a^4*b
+ 3*B*a^3*b^2 - 15*A*a^2*b^3)*c)*d^3 + ((3*D*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 15*A*b^5)*d^4 - 4*(3*D*a^2*b^3*
c + (2*C*a*b^4 - 3*B*b^5)*c)*d^3 + 8*(3*D*a*b^4*c^2 - C*b^5*c^2)*d^2)*x^3 - 4*(3*D*a^4*b*c^2 + (2*C*a^3*b^2 -
3*B*a^2*b^3)*c^2)*d^2 + (2*(3*D*a^4*b + C*a^3*b^2 + 3*B*a^2*b^3 - 15*A*a*b^4)*d^4 - 3*(7*D*a^3*b^2*c + (5*C*a^
2*b^3 - 9*B*a*b^4 + 5*A*b^5)*c)*d^3 + 12*(3*D*a^2*b^3*c^2 - (2*C*a*b^4 - B*b^5)*c^2)*d^2 + 8*(3*D*a*b^4*c^3 -
C*b^5*c^3)*d)*x^2 + 8*(3*D*a^3*b^2*c^3 - C*a^2*b^3*c^3)*d - (24*(C*a^2*b^3 - B*a*b^4)*c^2*d^2 - (3*D*a^5 + C*a
^4*b + 3*B*a^3*b^2 - 15*A*a^2*b^3)*d^4 + 6*(D*a^4*b*c + (C*a^3*b^2 - 3*B*a^2*b^3 + 5*A*a*b^4)*c)*d^3 - 16*(3*D
*a^2*b^3*c^3 - C*a*b^4*c^3)*d)*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)
) + (8*D*a^2*b^4*c^4 + 8*A*a^3*b^3*d^4 + (3*D*a^5*b*c + (C*a^4*b^2 - 13*B*a^3*b^3 + A*a^2*b^4)*c)*d^3 - (13*D*
a^4*b^2*c^2 - (13*C*a^3*b^3 + 11*B*a^2*b^4 - 11*A*a*b^5)*c^2)*d^2 + (8*D*b^6*c^4 + (5*D*a^4*b^2 - C*a^3*b^3 -
3*B*a^2*b^4 + 15*A*a*b^5)*d^4 - (17*D*a^3*b^3*c - 3*(3*C*a^2*b^4 - 3*B*a*b^5 - 5*A*b^6)*c)*d^3 + 12*(D*a^2*b^4
*c^2 + B*b^6*c^2)*d^2 - 8*(D*a*b^5*c^3 + C*b^6*c^3)*d)*x^2 + 2*(D*a^3*b^3*c^3 - (7*C*a^2*b^4 - B*a*b^5 - A*b^6
)*c^3)*d + (16*D*a*b^5*c^4 + (3*D*a^5*b + C*a^4*b^2 - 5*B*a^3*b^3 + 25*A*a^2*b^4)*d^4 - 4*(2*D*a^4*b^2*c - (C*
a^3*b^3 - 4*B*a^2*b^4 - 5*A*a*b^5)*c)*d^3 - (7*D*a^3*b^3*c^2 - (19*C*a^2*b^4 + 17*B*a*b^5 - 5*A*b^6)*c^2)*d^2
- 4*(D*a^2*b^4*c^3 + (6*C*a*b^5 - B*b^6)*c^3)*d)*x)*sqrt(d*x + c))/(a^2*b^7*c^5*d - 4*a^3*b^6*c^4*d^2 + 6*a^4*
b^5*c^3*d^3 - 4*a^5*b^4*c^2*d^4 + a^6*b^3*c*d^5 + (b^9*c^4*d^2 - 4*a*b^8*c^3*d^3 + 6*a^2*b^7*c^2*d^4 - 4*a^3*b
^6*c*d^5 + a^4*b^5*d^6)*x^3 + (b^9*c^5*d - 2*a*b^8*c^4*d^2 - 2*a^2*b^7*c^3*d^3 + 8*a^3*b^6*c^2*d^4 - 7*a^4*b^5
*c*d^5 + 2*a^5*b^4*d^6)*x^2 + (2*a*b^8*c^5*d - 7*a^2*b^7*c^4*d^2 + 8*a^3*b^6*c^3*d^3 - 2*a^4*b^5*c^2*d^4 - 2*a
^5*b^4*c*d^5 + a^6*b^3*d^6)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[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 >> 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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 617, normalized size of antiderivative = 1.76 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {{\left (24 \, D a b^{2} c^{2} - 8 \, C b^{3} c^{2} - 12 \, D a^{2} b c d - 8 \, C a b^{2} c d + 12 \, B b^{3} c d + 3 \, D a^{3} d^{2} + C a^{2} b d^{2} + 3 \, B a b^{2} d^{2} - 15 \, A b^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} \sqrt {d x + c}} - \frac {12 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{2} b^{2} c d - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} C a b^{3} c d + 4 \, {\left (d x + c\right )}^{\frac {3}{2}} B b^{4} c d - 12 \, \sqrt {d x + c} D a^{2} b^{2} c^{2} d + 8 \, \sqrt {d x + c} C a b^{3} c^{2} d - 4 \, \sqrt {d x + c} B b^{4} c^{2} d - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{3} b d^{2} + {\left (d x + c\right )}^{\frac {3}{2}} C a^{2} b^{2} d^{2} + 3 \, {\left (d x + c\right )}^{\frac {3}{2}} B a b^{3} d^{2} - 7 \, {\left (d x + c\right )}^{\frac {3}{2}} A b^{4} d^{2} + 15 \, \sqrt {d x + c} D a^{3} b c d^{2} - 7 \, \sqrt {d x + c} C a^{2} b^{2} c d^{2} - \sqrt {d x + c} B a b^{3} c d^{2} + 9 \, \sqrt {d x + c} A b^{4} c d^{2} - 3 \, \sqrt {d x + c} D a^{4} d^{3} - \sqrt {d x + c} C a^{3} b d^{3} + 5 \, \sqrt {d x + c} B a^{2} b^{2} d^{3} - 9 \, \sqrt {d x + c} A a b^{3} d^{3}}{4 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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