3.23 \(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=336 \[ -\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}-\frac {-a^3 d^3 D+a^2 b C d^3+a b^2 d \left (-3 B d^2-6 c^2 D+4 c C d\right )-\left (b^3 \left (-5 A d^3+2 B c d^2-2 c^3 D\right )\right )}{b^2 d^2 \sqrt {c+d x} (b c-a d)^3}+\frac {3 a^3 d^3 D-3 a^2 b C d^3+3 a b^2 B d^3-\left (b^3 \left (5 A d^3-2 B c d^2-2 c^3 D+2 c^2 C d\right )\right )}{3 b^3 d^2 (c+d x)^{3/2} (b c-a d)^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (a^3 (-d) D-a^2 b (C d-6 c D)-a b^2 (4 c C-3 B d)+b^3 (2 B c-5 A d)\right )}{b^{3/2} (b c-a d)^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.79, antiderivative size = 336, 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, 1261, 208} \[ -\frac {a^2 b C d^3-a^3 d^3 D+a b^2 d \left (-3 B d^2-6 c^2 D+4 c C d\right )+b^3 \left (-\left (-5 A d^3+2 B c d^2-2 c^3 D\right )\right )}{b^2 d^2 \sqrt {c+d x} (b c-a d)^3}+\frac {-3 a^2 b C d^3+3 a^3 d^3 D+3 a b^2 B d^3+b^3 \left (-\left (5 A d^3-2 B c d^2+2 c^2 C d-2 c^3 D\right )\right )}{3 b^3 d^2 (c+d x)^{3/2} (b c-a d)^2}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-a^2 b (C d-6 c D)+a^3 (-d) D-a b^2 (4 c C-3 B d)+b^3 (2 B c-5 A d)\right )}{b^{3/2} (b c-a d)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 (c+d x)^{5/2}} \, 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) (c+d x)^{3/2}}+\frac {\int \frac {-\frac {b^3 (2 B c-5 A d)-a b^2 (2 c C-3 B d)+3 a^3 d D-a^2 b (3 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)^{5/2}} \, 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) (c+d x)^{3/2}}-\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-5 A d)-a b^2 (2 c C-3 B d)+3 a^3 d D-a^2 b (3 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^4 \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) (c+d x)^{3/2}}-\frac {2 \operatorname {Subst}\left (\int \left (\frac {3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+5 A d^3-2 c^3 D\right )}{2 b^3 d (b c-a d) x^4}+\frac {-a^2 b C d^3+a^3 d^3 D-a b^2 d \left (4 c C d-3 B d^2-6 c^2 D\right )+b^3 \left (2 B c d^2-5 A d^3-2 c^3 D\right )}{2 b^2 d (b c-a d)^2 x^2}+\frac {d \left (b^3 (2 B c-5 A d)-a b^2 (4 c C-3 B d)-a^3 d D-a^2 b (C d-6 c D)\right )}{2 b (b c-a d)^2 \left (b c-a d-b x^2\right )}\right ) \, dx,x,\sqrt {c+d x}\right )}{d (b c-a d)}\\ &=\frac {3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+5 A d^3-2 c^3 D\right )}{3 b^3 d^2 (b c-a d)^2 (c+d x)^{3/2}}-\frac {A-\frac {a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) (c+d x)^{3/2}}-\frac {a^2 b C d^3-a^3 d^3 D+a b^2 d \left (4 c C d-3 B d^2-6 c^2 D\right )-b^3 \left (2 B c d^2-5 A d^3-2 c^3 D\right )}{b^2 d^2 (b c-a d)^3 \sqrt {c+d x}}-\frac {\left (b^3 (2 B c-5 A d)-a b^2 (4 c C-3 B d)-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 (b c-a d)^3}\\ &=\frac {3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+5 A d^3-2 c^3 D\right )}{3 b^3 d^2 (b c-a d)^2 (c+d x)^{3/2}}-\frac {A-\frac {a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) (c+d x)^{3/2}}-\frac {a^2 b C d^3-a^3 d^3 D+a b^2 d \left (4 c C d-3 B d^2-6 c^2 D\right )-b^3 \left (2 B c d^2-5 A d^3-2 c^3 D\right )}{b^2 d^2 (b c-a d)^3 \sqrt {c+d x}}-\frac {\left (b^3 (2 B c-5 A d)-a b^2 (4 c C-3 B d)-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^{3/2} (b c-a d)^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.05, size = 334, normalized size = 0.99 \[ \frac {\sqrt {c+d x} \left (a \left (a^2 D-a b C+b^2 B\right )-A b^3\right )}{b (a+b x) (b c-a d)^3}+\frac {d \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{7/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (a^3 (-d) D+3 a^2 b c D+a b^2 (B d-2 c C)+b^3 (B c-2 A d)\right )}{b^{3/2} (b c-a d)^{7/2}}+\frac {2 \left (-A d^3+B c d^2+c^3 D-c^2 C d\right )}{3 d^2 (c+d x)^{3/2} (b c-a d)^2}+\frac {b \left (4 A d^3-2 B c d^2+2 c^3 D\right )-2 a d \left (B d^2+3 c^2 D-2 c C d\right )}{d^2 \sqrt {c+d x} (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.06, size = 2444, normalized size = 7.27 \[ \text {result too large to display} \]

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)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*((D*a^4*c^2 + (C*a^3*b - 3*B*a^2*b^2 + 5*A*a*b^3)*c^2)*d^3 + ((D*a^3*b + C*a^2*b^2 - 3*B*a*b^3 + 5*A*
b^4)*d^5 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*d^4)*x^3 - 2*(3*D*a^3*b*c^3 - (2*C*a^2*b^2 - B*a*b^3)*c^3
)*d^2 + ((D*a^4 + C*a^3*b - 3*B*a^2*b^2 + 5*A*a*b^3)*d^5 - 2*(2*D*a^3*b*c - (3*C*a^2*b^2 - 4*B*a*b^3 + 5*A*b^4
)*c)*d^4 - 4*(3*D*a^2*b^2*c^2 - (2*C*a*b^3 - B*b^4)*c^2)*d^3)*x^2 + (2*(D*a^4*c + (C*a^3*b - 3*B*a^2*b^2 + 5*A
*a*b^3)*c)*d^4 - (11*D*a^3*b*c^2 - (9*C*a^2*b^2 - 7*B*a*b^3 + 5*A*b^4)*c^2)*d^3 - 2*(3*D*a^2*b^2*c^3 - (2*C*a*
b^3 - B*b^4)*c^3)*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^5 + 2*A*a^3*b^2*d^5 + 4*(B*a^3*b^2 - 4*A*a^2*b^3)*c*d^4 + (3*D*a^4*b*c^2 - (13*C*a
^3*b^2 - 7*B*a^2*b^3 - 11*A*a*b^4)*c^2)*d^3 + (13*D*a^3*b^2*c^3 + (11*C*a^2*b^3 - 11*B*a*b^4 + 3*A*b^5)*c^3)*d
^2 + 3*(2*D*b^5*c^4*d - 8*D*a*b^4*c^3*d^2 + (D*a^4*b - C*a^3*b^2 + 3*B*a^2*b^3 - 5*A*a*b^4)*d^5 - (D*a^3*b^2*c
+ (3*C*a^2*b^3 + B*a*b^4 - 5*A*b^5)*c)*d^4 + 2*(3*D*a^2*b^3*c^2 + (2*C*a*b^4 - B*b^5)*c^2)*d^3)*x^2 - 2*(10*D
*a^2*b^3*c^4 - C*a*b^4*c^4)*d + 2*(2*D*b^5*c^5 + (3*B*a^3*b^2 - 5*A*a^2*b^3)*d^5 + (3*D*a^4*b*c - (9*C*a^3*b^2
- 5*B*a^2*b^3 + 5*A*a*b^4)*c)*d^4 + 2*(3*D*a^3*b^2*c^2 + (2*C*a^2*b^3 - 2*B*a*b^4 + 5*A*b^5)*c^2)*d^3 - 4*(D*
a^2*b^3*c^3 - (C*a*b^4 - B*b^5)*c^3)*d^2 - (7*D*a*b^4*c^4 - C*b^5*c^4)*d)*x)*sqrt(d*x + c))/(a*b^6*c^6*d^2 - 4
*a^2*b^5*c^5*d^3 + 6*a^3*b^4*c^4*d^4 - 4*a^4*b^3*c^3*d^5 + a^5*b^2*c^2*d^6 + (b^7*c^4*d^4 - 4*a*b^6*c^3*d^5 +
6*a^2*b^5*c^2*d^6 - 4*a^3*b^4*c*d^7 + a^4*b^3*d^8)*x^3 + (2*b^7*c^5*d^3 - 7*a*b^6*c^4*d^4 + 8*a^2*b^5*c^3*d^5
- 2*a^3*b^4*c^2*d^6 - 2*a^4*b^3*c*d^7 + a^5*b^2*d^8)*x^2 + (b^7*c^6*d^2 - 2*a*b^6*c^5*d^3 - 2*a^2*b^5*c^4*d^4
+ 8*a^3*b^4*c^3*d^5 - 7*a^4*b^3*c^2*d^6 + 2*a^5*b^2*c*d^7)*x), -1/3*(3*((D*a^4*c^2 + (C*a^3*b - 3*B*a^2*b^2 +
5*A*a*b^3)*c^2)*d^3 + ((D*a^3*b + C*a^2*b^2 - 3*B*a*b^3 + 5*A*b^4)*d^5 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4
)*c)*d^4)*x^3 - 2*(3*D*a^3*b*c^3 - (2*C*a^2*b^2 - B*a*b^3)*c^3)*d^2 + ((D*a^4 + C*a^3*b - 3*B*a^2*b^2 + 5*A*a*
b^3)*d^5 - 2*(2*D*a^3*b*c - (3*C*a^2*b^2 - 4*B*a*b^3 + 5*A*b^4)*c)*d^4 - 4*(3*D*a^2*b^2*c^2 - (2*C*a*b^3 - B*b
^4)*c^2)*d^3)*x^2 + (2*(D*a^4*c + (C*a^3*b - 3*B*a^2*b^2 + 5*A*a*b^3)*c)*d^4 - (11*D*a^3*b*c^2 - (9*C*a^2*b^2
- 7*B*a*b^3 + 5*A*b^4)*c^2)*d^3 - 2*(3*D*a^2*b^2*c^3 - (2*C*a*b^3 - B*b^4)*c^3)*d^2)*x)*sqrt(-b^2*c + a*b*d)*a
rctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (4*D*a*b^4*c^5 + 2*A*a^3*b^2*d^5 + 4*(B*a^3*b^2 - 4*
A*a^2*b^3)*c*d^4 + (3*D*a^4*b*c^2 - (13*C*a^3*b^2 - 7*B*a^2*b^3 - 11*A*a*b^4)*c^2)*d^3 + (13*D*a^3*b^2*c^3 + (
11*C*a^2*b^3 - 11*B*a*b^4 + 3*A*b^5)*c^3)*d^2 + 3*(2*D*b^5*c^4*d - 8*D*a*b^4*c^3*d^2 + (D*a^4*b - C*a^3*b^2 +
3*B*a^2*b^3 - 5*A*a*b^4)*d^5 - (D*a^3*b^2*c + (3*C*a^2*b^3 + B*a*b^4 - 5*A*b^5)*c)*d^4 + 2*(3*D*a^2*b^3*c^2 +
(2*C*a*b^4 - B*b^5)*c^2)*d^3)*x^2 - 2*(10*D*a^2*b^3*c^4 - C*a*b^4*c^4)*d + 2*(2*D*b^5*c^5 + (3*B*a^3*b^2 - 5*A
*a^2*b^3)*d^5 + (3*D*a^4*b*c - (9*C*a^3*b^2 - 5*B*a^2*b^3 + 5*A*a*b^4)*c)*d^4 + 2*(3*D*a^3*b^2*c^2 + (2*C*a^2*
b^3 - 2*B*a*b^4 + 5*A*b^5)*c^2)*d^3 - 4*(D*a^2*b^3*c^3 - (C*a*b^4 - B*b^5)*c^3)*d^2 - (7*D*a*b^4*c^4 - C*b^5*c
^4)*d)*x)*sqrt(d*x + c))/(a*b^6*c^6*d^2 - 4*a^2*b^5*c^5*d^3 + 6*a^3*b^4*c^4*d^4 - 4*a^4*b^3*c^3*d^5 + a^5*b^2*
c^2*d^6 + (b^7*c^4*d^4 - 4*a*b^6*c^3*d^5 + 6*a^2*b^5*c^2*d^6 - 4*a^3*b^4*c*d^7 + a^4*b^3*d^8)*x^3 + (2*b^7*c^5
*d^3 - 7*a*b^6*c^4*d^4 + 8*a^2*b^5*c^3*d^5 - 2*a^3*b^4*c^2*d^6 - 2*a^4*b^3*c*d^7 + a^5*b^2*d^8)*x^2 + (b^7*c^6
*d^2 - 2*a*b^6*c^5*d^3 - 2*a^2*b^5*c^4*d^4 + 8*a^3*b^4*c^3*d^5 - 7*a^4*b^3*c^2*d^6 + 2*a^5*b^2*c*d^7)*x)]

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giac [A]  time = 1.42, size = 439, normalized size = 1.31 \[ \frac {{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - D a^{3} d - C a^{2} b d + 3 \, B a b^{2} d - 5 \, A b^{3} d\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {\sqrt {d x + c} D a^{3} d - \sqrt {d x + c} C a^{2} b d + \sqrt {d x + c} B a b^{2} d - \sqrt {d x + c} A b^{3} d}{{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} - \frac {2 \, {\left (3 \, {\left (d x + c\right )} D b c^{3} - D b c^{4} - 9 \, {\left (d x + c\right )} D a c^{2} d + D a c^{3} d + C b c^{3} d + 6 \, {\left (d x + c\right )} C a c d^{2} - 3 \, {\left (d x + c\right )} B b c d^{2} - C a c^{2} d^{2} - B b c^{2} d^{2} - 3 \, {\left (d x + c\right )} B a d^{3} + 6 \, {\left (d x + c\right )} A b d^{3} + B a c d^{3} + A b c d^{3} - A a d^{4}\right )}}{3 \, {\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} \]

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)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.03, size = 730, normalized size = 2.17 \[ \frac {5 A \,b^{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 )^{3} \sqrt {\left (a d -b c \right ) b}}-\frac {3 B a b d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3} \sqrt {\left (a d -b c \right ) b}}-\frac {2 B \,b^{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 )^{3} \sqrt {\left (a d -b c \right ) b}}+\frac {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 )^{3} \sqrt {\left (a d -b c \right ) b}}+\frac {4 C a 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 )^{3} \sqrt {\left (a d -b c \right ) b}}+\frac {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 )^{3} \sqrt {\left (a d -b c \right ) b}\, b}-\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 )^{3} \sqrt {\left (a d -b c \right ) b}}+\frac {\sqrt {d x +c}\, A \,b^{2} d}{\left (a d -b c \right )^{3} \left (b d x +a d \right )}-\frac {\sqrt {d x +c}\, B a b d}{\left (a d -b c \right )^{3} \left (b d x +a d \right )}+\frac {\sqrt {d x +c}\, C \,a^{2} d}{\left (a d -b c \right )^{3} \left (b d x +a d \right )}-\frac {\sqrt {d x +c}\, D a^{3} d}{\left (a d -b c \right )^{3} \left (b d x +a d \right ) b}+\frac {4 A b d}{\left (a d -b c \right )^{3} \sqrt {d x +c}}-\frac {2 B a d}{\left (a d -b c \right )^{3} \sqrt {d x +c}}-\frac {2 B b c}{\left (a d -b c \right )^{3} \sqrt {d x +c}}+\frac {4 C a c}{\left (a d -b c \right )^{3} \sqrt {d x +c}}-\frac {6 D a \,c^{2}}{\left (a d -b c \right )^{3} \sqrt {d x +c}\, d}+\frac {2 D b \,c^{3}}{\left (a d -b c \right )^{3} \sqrt {d x +c}\, d^{2}}-\frac {2 A d}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B c}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 C \,c^{2}}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}} d}+\frac {2 D c^{3}}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{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)^(5/2),x)

[Out]

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

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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)^(5/2),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 \[ \int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{5/2}} \,d x \]

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)^(5/2)),x)

[Out]

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^2*(c + d*x)^(5/2)), 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)**(5/2),x)

[Out]

Timed out

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