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

[Out]

2/3*D*(d*x+c)^(3/2)/b^2/d^2-(b^3*(-A*d+2*B*c)-a*b^2*(B*d+4*C*c)-5*a^3*d*D+3*a^2*b*(C*d+2*D*c))*arctanh(b^(1/2)
*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(3/2)+2*(C*b*d-2*D*a*d-D*b*c)*(d*x+c)^(1/2)/b^3/d^2-(A-a*(
B*b^2-C*a*b+D*a^2)/b^3)*(d*x+c)^(1/2)/(-a*d+b*c)/(b*x+a)

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Rubi [A]  time = 0.52, antiderivative size = 201, 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 = {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 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]

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 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]

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*}

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Mathematica [A]  time = 0.44, 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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fricas [B]  time = 0.82, size = 1004, normalized size = 5.00 \[ \left [-\frac {3 \, {\left ({\left (5 \, D a^{4} - 3 \, C a^{3} b + B a^{2} b^{2} + A a b^{3}\right )} d^{3} - 2 \, {\left (3 \, D a^{3} b c - {\left (2 \, C a^{2} b^{2} - B a b^{3}\right )} c\right )} d^{2} + {\left ({\left (5 \, D a^{3} b - 3 \, C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} d^{3} - 2 \, {\left (3 \, D a^{2} b^{2} c - {\left (2 \, C a b^{3} - B b^{4}\right )} c\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^{3} + 3 \, {\left (5 \, D a^{4} b - 3 \, C a^{3} b^{2} + B a^{2} b^{3} - A a b^{4}\right )} d^{3} - {\left (23 \, D a^{3} b^{2} c - 3 \, {\left (5 \, C a^{2} b^{3} - B a b^{4} + A b^{5}\right )} c\right )} d^{2} - 2 \, {\left (D b^{5} c^{2} d - 2 \, D a b^{4} c d^{2} + D a^{2} b^{3} d^{3}\right )} x^{2} + 2 \, {\left (2 \, D a^{2} b^{3} c^{2} - 3 \, C a b^{4} c^{2}\right )} d + 2 \, {\left (2 \, D b^{5} c^{3} + {\left (5 \, D a^{3} b^{2} - 3 \, C a^{2} b^{3}\right )} d^{3} - 2 \, {\left (4 \, D a^{2} b^{3} c - 3 \, C a b^{4} c\right )} d^{2} + {\left (D a b^{4} c^{2} - 3 \, C b^{5} c^{2}\right )} d\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (a b^{6} c^{2} d^{2} - 2 \, a^{2} b^{5} c d^{3} + a^{3} b^{4} d^{4} + {\left (b^{7} c^{2} d^{2} - 2 \, a b^{6} c d^{3} + a^{2} b^{5} d^{4}\right )} x\right )}}, -\frac {3 \, {\left ({\left (5 \, D a^{4} - 3 \, C a^{3} b + B a^{2} b^{2} + A a b^{3}\right )} d^{3} - 2 \, {\left (3 \, D a^{3} b c - {\left (2 \, C a^{2} b^{2} - B a b^{3}\right )} c\right )} d^{2} + {\left ({\left (5 \, D a^{3} b - 3 \, C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} d^{3} - 2 \, {\left (3 \, D a^{2} b^{2} c - {\left (2 \, C a b^{3} - B b^{4}\right )} c\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^{3} + 3 \, {\left (5 \, D a^{4} b - 3 \, C a^{3} b^{2} + B a^{2} b^{3} - A a b^{4}\right )} d^{3} - {\left (23 \, D a^{3} b^{2} c - 3 \, {\left (5 \, C a^{2} b^{3} - B a b^{4} + A b^{5}\right )} c\right )} d^{2} - 2 \, {\left (D b^{5} c^{2} d - 2 \, D a b^{4} c d^{2} + D a^{2} b^{3} d^{3}\right )} x^{2} + 2 \, {\left (2 \, D a^{2} b^{3} c^{2} - 3 \, C a b^{4} c^{2}\right )} d + 2 \, {\left (2 \, D b^{5} c^{3} + {\left (5 \, D a^{3} b^{2} - 3 \, C a^{2} b^{3}\right )} d^{3} - 2 \, {\left (4 \, D a^{2} b^{3} c - 3 \, C a b^{4} c\right )} d^{2} + {\left (D a b^{4} c^{2} - 3 \, C b^{5} c^{2}\right )} d\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (a b^{6} c^{2} d^{2} - 2 \, a^{2} b^{5} c d^{3} + a^{3} b^{4} d^{4} + {\left (b^{7} c^{2} d^{2} - 2 \, a b^{6} c d^{3} + a^{2} b^{5} d^{4}\right )} x\right )}}\right ] \]

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]

[-1/6*(3*((5*D*a^4 - 3*C*a^3*b + B*a^2*b^2 + A*a*b^3)*d^3 - 2*(3*D*a^3*b*c - (2*C*a^2*b^2 - B*a*b^3)*c)*d^2 +
((5*D*a^3*b - 3*C*a^2*b^2 + B*a*b^3 + A*b^4)*d^3 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*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^3 + 3*(
5*D*a^4*b - 3*C*a^3*b^2 + B*a^2*b^3 - A*a*b^4)*d^3 - (23*D*a^3*b^2*c - 3*(5*C*a^2*b^3 - B*a*b^4 + A*b^5)*c)*d^
2 - 2*(D*b^5*c^2*d - 2*D*a*b^4*c*d^2 + D*a^2*b^3*d^3)*x^2 + 2*(2*D*a^2*b^3*c^2 - 3*C*a*b^4*c^2)*d + 2*(2*D*b^5
*c^3 + (5*D*a^3*b^2 - 3*C*a^2*b^3)*d^3 - 2*(4*D*a^2*b^3*c - 3*C*a*b^4*c)*d^2 + (D*a*b^4*c^2 - 3*C*b^5*c^2)*d)*
x)*sqrt(d*x + c))/(a*b^6*c^2*d^2 - 2*a^2*b^5*c*d^3 + a^3*b^4*d^4 + (b^7*c^2*d^2 - 2*a*b^6*c*d^3 + a^2*b^5*d^4)
*x), -1/3*(3*((5*D*a^4 - 3*C*a^3*b + B*a^2*b^2 + A*a*b^3)*d^3 - 2*(3*D*a^3*b*c - (2*C*a^2*b^2 - B*a*b^3)*c)*d^
2 + ((5*D*a^3*b - 3*C*a^2*b^2 + B*a*b^3 + A*b^4)*d^3 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*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^3 + 3*(5*D*a^4*b - 3*C
*a^3*b^2 + B*a^2*b^3 - A*a*b^4)*d^3 - (23*D*a^3*b^2*c - 3*(5*C*a^2*b^3 - B*a*b^4 + A*b^5)*c)*d^2 - 2*(D*b^5*c^
2*d - 2*D*a*b^4*c*d^2 + D*a^2*b^3*d^3)*x^2 + 2*(2*D*a^2*b^3*c^2 - 3*C*a*b^4*c^2)*d + 2*(2*D*b^5*c^3 + (5*D*a^3
*b^2 - 3*C*a^2*b^3)*d^3 - 2*(4*D*a^2*b^3*c - 3*C*a*b^4*c)*d^2 + (D*a*b^4*c^2 - 3*C*b^5*c^2)*d)*x)*sqrt(d*x + c
))/(a*b^6*c^2*d^2 - 2*a^2*b^5*c*d^3 + a^3*b^4*d^4 + (b^7*c^2*d^2 - 2*a*b^6*c*d^3 + a^2*b^5*d^4)*x)]

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giac [A]  time = 1.32, size = 271, normalized size = 1.35 \[ \frac {{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - 5 \, D a^{3} d + 3 \, C a^{2} b d - B a b^{2} d - A b^{3} d\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{4} c - a b^{3} d\right )} \sqrt {-b^{2} c + a b d}} + \frac {\sqrt {d x + c} D a^{3} d - \sqrt {d x + c} C a^{2} b d + \sqrt {d x + c} B a b^{2} d - \sqrt {d x + c} A b^{3} d}{{\left (b^{4} c - a b^{3} d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} D b^{4} d^{4} - 3 \, \sqrt {d x + c} D b^{4} c d^{4} - 6 \, \sqrt {d x + c} D a b^{3} d^{5} + 3 \, \sqrt {d x + c} C b^{4} d^{5}\right )}}{3 \, b^{6} d^{6}} \]

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)

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maple [B]  time = 0.02, size = 566, normalized size = 2.82 \[ \frac {A d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}+\frac {B a d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b}-\frac {2 B c \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}-\frac {3 C \,a^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b^{2}}+\frac {4 C a c \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b}+\frac {5 D a^{3} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b^{3}}-\frac {6 D a^{2} c \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b^{2}}+\frac {\sqrt {d x +c}\, A d}{\left (a d -b c \right ) \left (b d x +a d \right )}-\frac {\sqrt {d x +c}\, B a d}{\left (a d -b c \right ) \left (b d x +a d \right ) b}+\frac {\sqrt {d x +c}\, C \,a^{2} d}{\left (a d -b c \right ) \left (b d x +a d \right ) b^{2}}-\frac {\sqrt {d x +c}\, D a^{3} d}{\left (a d -b c \right ) \left (b d x +a d \right ) b^{3}}+\frac {2 \sqrt {d x +c}\, C}{b^{2} d}-\frac {4 \sqrt {d x +c}\, D a}{b^{3} d}-\frac {2 \sqrt {d x +c}\, D c}{b^{2} d^{2}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} D}{3 b^{2} d^{2}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 details)Is a*d-b*c positive or negative?

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

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)^(1/2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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