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

Optimal. Leaf size=438 \[ -\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 (c+d x)^{3/2} (b c-a d)}+\frac {a^3 \left (-d^2\right ) D+a^2 b C d^2+a b^2 \left (-3 B d^2-6 c^2 D+4 c C d\right )+b^3 \left (7 A d^2-4 B c d+2 c^2 C\right )}{b^2 \sqrt {c+d x} (b c-a d)^4}-\frac {3 a^3 d^3 D-3 a^2 b C d^3+3 a b^2 B d^3-\left (b^3 \left (7 A d^3-4 B c d^2-4 c^3 D+4 c^2 C d\right )\right )}{6 b^3 d (c+d x)^{3/2} (b c-a d)^3}-\frac {\sqrt {c+d x} \left (-5 a^3 d D+a^2 b (12 c D+C d)-a b^2 (8 c C-3 B d)+b^3 (4 B c-7 A d)\right )}{4 b (a+b x) (b c-a d)^4}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (a^3 d^2 D+3 a^2 b d (C d-4 c D)+3 a b^2 \left (-5 B d^2-8 c^2 D+8 c C d\right )+b^3 \left (35 A d^2-20 B c d+8 c^2 C\right )\right )}{4 b^{3/2} (b c-a d)^{9/2}} \]

[Out]

1/6*(-3*a*b^2*B*d^3+3*a^2*b*C*d^3-3*a^3*d^3*D+b^3*(7*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)^(3/2)+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)^(3/2)-1/4*(b^3*(35*A*d^2-20*B*
c*d+8*C*c^2)+a^3*d^2*D+3*a^2*b*d*(C*d-4*D*c)+3*a*b^2*(-5*B*d^2+8*C*c*d-8*D*c^2))*arctanh(b^(1/2)*(d*x+c)^(1/2)
/(-a*d+b*c)^(1/2))/b^(3/2)/(-a*d+b*c)^(9/2)+(a^2*b*C*d^2+b^3*(7*A*d^2-4*B*c*d+2*C*c^2)-a^3*d^2*D+a*b^2*(-3*B*d
^2+4*C*c*d-6*D*c^2))/b^2/(-a*d+b*c)^4/(d*x+c)^(1/2)-1/4*(b^3*(-7*A*d+4*B*c)-a*b^2*(-3*B*d+8*C*c)-5*a^3*d*D+a^2
*b*(C*d+12*D*c))*(d*x+c)^(1/2)/b/(-a*d+b*c)^4/(b*x+a)

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Rubi [A]  time = 1.21, antiderivative size = 438, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1621, 897, 1259, 1261, 208} \[ \frac {a^2 b C d^2+a^3 \left (-d^2\right ) D+a b^2 \left (-3 B d^2-6 c^2 D+4 c C d\right )+b^3 \left (7 A d^2-4 B c d+2 c^2 C\right )}{b^2 \sqrt {c+d x} (b c-a d)^4}-\frac {-3 a^2 b C d^3+3 a^3 d^3 D+3 a b^2 B d^3+b^3 \left (-\left (7 A d^3-4 B c d^2+4 c^2 C d-4 c^3 D\right )\right )}{6 b^3 d (c+d x)^{3/2} (b c-a d)^3}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (3 a^2 b d (C d-4 c D)+a^3 d^2 D+3 a b^2 \left (-5 B d^2-8 c^2 D+8 c C d\right )+b^3 \left (35 A d^2-20 B c d+8 c^2 C\right )\right )}{4 b^{3/2} (b c-a d)^{9/2}}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 (c+d x)^{3/2} (b c-a d)}-\frac {\sqrt {c+d x} \left (a^2 b (12 c D+C d)-5 a^3 d D-a b^2 (8 c C-3 B d)+b^3 (4 B c-7 A d)\right )}{4 b (a+b x) (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^3*(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*(4*c^2*C*d - 4*B*c*d^2 + 7*A*d^3 - 4*c^3*D))/(6*b^3*d*(b*c
- a*d)^3*(c + d*x)^(3/2)) - (A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(2*b^3*(b*c - a*d)*(a + b*x)^2*(c + d*x)^(3/2
)) + (a^2*b*C*d^2 + b^3*(2*c^2*C - 4*B*c*d + 7*A*d^2) - a^3*d^2*D + a*b^2*(4*c*C*d - 3*B*d^2 - 6*c^2*D))/(b^2*
(b*c - a*d)^4*Sqrt[c + d*x]) - ((b^3*(4*B*c - 7*A*d) - a*b^2*(8*c*C - 3*B*d) - 5*a^3*d*D + a^2*b*(C*d + 12*c*D
))*Sqrt[c + d*x])/(4*b*(b*c - a*d)^4*(a + b*x)) - ((b^3*(8*c^2*C - 20*B*c*d + 35*A*d^2) + a^3*d^2*D + 3*a^2*b*
d*(C*d - 4*c*D) + 3*a*b^2*(8*c*C*d - 5*B*d^2 - 8*c^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*
b^(3/2)*(b*c - a*d)^(9/2))

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 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 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)^3 (c+d x)^{5/2}} \, 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 (c+d x)^{3/2}}-\frac {\int \frac {-\frac {b^3 (4 B c-7 A d)-a b^2 (4 c C-3 B d)+3 a^3 d D-a^2 b (3 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)^{5/2}} \, 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 (c+d x)^{3/2}}-\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-7 A d)-a b^2 (4 c C-3 B d)+3 a^3 d D-a^2 b (3 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^4 \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 (c+d x)^{3/2}}-\frac {\left (b^3 (4 B c-7 A d)-a b^2 (8 c C-3 B d)-5 a^3 d D+a^2 b (C d+12 c D)\right ) \sqrt {c+d x}}{4 b (b c-a d)^4 (a+b x)}+\frac {d^4 \operatorname {Subst}\left (\int \frac {-\frac {(b c-a d)^2 \left (3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+7 A d^3-4 c^3 D\right )\right )}{b d^6}-\frac {(b c-a d) \left (a^2 b C d^3-a^3 d^3 D-a b^2 d \left (8 c C d-3 B d^2-12 c^2 D\right )+b^3 \left (4 B c d^2-7 A d^3-4 c^3 D\right )\right ) x^2}{d^6}-\frac {b \left (b^3 (4 B c-7 A d)-a b^2 (8 c C-3 B d)-5 a^3 d D+a^2 b (C d+12 c D)\right ) x^4}{2 d^4}}{x^4 \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)^4}\\ &=-\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 (c+d x)^{3/2}}-\frac {\left (b^3 (4 B c-7 A d)-a b^2 (8 c C-3 B d)-5 a^3 d D+a^2 b (C d+12 c D)\right ) \sqrt {c+d x}}{4 b (b c-a d)^4 (a+b x)}+\frac {d^4 \operatorname {Subst}\left (\int \left (\frac {(b c-a d) \left (3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+7 A d^3-4 c^3 D\right )\right )}{b d^5 x^4}+\frac {2 \left (-a^2 b C d^2-b^3 \left (2 c^2 C-4 B c d+7 A d^2\right )+a^3 d^2 D-a b^2 \left (4 c C d-3 B d^2-6 c^2 D\right )\right )}{d^4 x^2}+\frac {b \left (-b^3 \left (8 c^2 C-20 B c d+35 A d^2\right )-a^3 d^2 D-3 a^2 b d (C d-4 c D)-3 a b^2 \left (8 c C d-5 B d^2-8 c^2 D\right )\right )}{2 d^4 \left (b c-a d-b x^2\right )}\right ) \, dx,x,\sqrt {c+d x}\right )}{2 b^2 (b c-a d)^4}\\ &=-\frac {3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+7 A d^3-4 c^3 D\right )}{6 b^3 d (b c-a d)^3 (c+d x)^{3/2}}-\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 (c+d x)^{3/2}}+\frac {a^2 b C d^2+b^3 \left (2 c^2 C-4 B c d+7 A d^2\right )-a^3 d^2 D+a b^2 \left (4 c C d-3 B d^2-6 c^2 D\right )}{b^2 (b c-a d)^4 \sqrt {c+d x}}-\frac {\left (b^3 (4 B c-7 A d)-a b^2 (8 c C-3 B d)-5 a^3 d D+a^2 b (C d+12 c D)\right ) \sqrt {c+d x}}{4 b (b c-a d)^4 (a+b x)}-\frac {\left (b^3 \left (8 c^2 C-20 B c d+35 A d^2\right )+a^3 d^2 D+3 a^2 b d (C d-4 c D)+3 a b^2 \left (8 c C d-5 B d^2-8 c^2 D\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b c-a d-b x^2} \, dx,x,\sqrt {c+d x}\right )}{4 b (b c-a d)^4}\\ &=-\frac {3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+7 A d^3-4 c^3 D\right )}{6 b^3 d (b c-a d)^3 (c+d x)^{3/2}}-\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 (c+d x)^{3/2}}+\frac {a^2 b C d^2+b^3 \left (2 c^2 C-4 B c d+7 A d^2\right )-a^3 d^2 D+a b^2 \left (4 c C d-3 B d^2-6 c^2 D\right )}{b^2 (b c-a d)^4 \sqrt {c+d x}}-\frac {\left (b^3 (4 B c-7 A d)-a b^2 (8 c C-3 B d)-5 a^3 d D+a^2 b (C d+12 c D)\right ) \sqrt {c+d x}}{4 b (b c-a d)^4 (a+b x)}-\frac {\left (b^3 \left (8 c^2 C-20 B c d+35 A d^2\right )+a^3 d^2 D+3 a^2 b d (C d-4 c D)+3 a b^2 \left (8 c C d-5 B d^2-8 c^2 D\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} (b c-a d)^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 2.10, size = 536, normalized size = 1.22 \[ \frac {\sqrt {c+d x} \left (a \left (a^2 D-a b C+b^2 B\right )-A b^3\right )}{2 b (a+b x)^2 (b c-a d)^3}-\frac {3 d \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \left (d (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )-\sqrt {b} \sqrt {c+d x} \sqrt {b c-a d}\right )}{4 b^{3/2} (a+b x) (b c-a d)^{9/2}}-\frac {\sqrt {c+d x} \left (a^3 (-d) D+3 a^2 b c D+a b^2 (B d-2 c C)+b^3 (B c-2 A d)\right )}{b (a+b x) (b c-a d)^4}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (a^3 (-d) D+3 a^2 b c D+a b^2 (B d-2 c C)+b^3 (B c-2 A d)\right )}{b^{3/2} (b c-a d)^{9/2}}+\frac {2 \left (b \left (3 A d^2-2 B c d+c^2 C\right )-a \left (B d^2+3 c^2 D-2 c C d\right )\right )}{\sqrt {c+d x} (b c-a d)^4}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (b \left (3 A d^2-2 B c d+c^2 C\right )-a \left (B d^2+3 c^2 D-2 c C d\right )\right )}{(b c-a d)^{9/2}}+\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{3 d (c+d x)^{3/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(3*d*(b*c - a*d)^3*(c + d*x)^(3/2)) + (2*(b*(c^2*C - 2*B*c*d + 3*A*d^2
) - a*(-2*c*C*d + B*d^2 + 3*c^2*D)))/((b*c - a*d)^4*Sqrt[c + d*x]) + ((-(A*b^3) + a*(b^2*B - a*b*C + a^2*D))*S
qrt[c + d*x])/(2*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)*Sqrt[c + d*x])/(b*(b*c - a*d)^4*(a + b*x)) + (d*(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)^(9/2)) - (2*Sqrt[b]*(b*(c^2*
C - 2*B*c*d + 3*A*d^2) - a*(-2*c*C*d + B*d^2 + 3*c^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*
c - a*d)^(9/2) - (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^(3/2)*(b*c - a*d)^(9/2)*(a + b*x))

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fricas [B]  time = 1.26, size = 3889, normalized size = 8.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(3*(((D*a^3*b^2 + 3*C*a^2*b^3 - 15*B*a*b^4 + 35*A*b^5)*d^5 - 4*(3*D*a^2*b^3*c - (6*C*a*b^4 - 5*B*b^5)*c)
*d^4 - 8*(3*D*a*b^4*c^2 - C*b^5*c^2)*d^3)*x^4 + (D*a^5*c^2 + (3*C*a^4*b - 15*B*a^3*b^2 + 35*A*a^2*b^3)*c^2)*d^
3 + 2*((D*a^4*b + 3*C*a^3*b^2 - 15*B*a^2*b^3 + 35*A*a*b^4)*d^5 - (11*D*a^3*b^2*c - (27*C*a^2*b^3 - 35*B*a*b^4
+ 35*A*b^5)*c)*d^4 - 4*(9*D*a^2*b^3*c^2 - (8*C*a*b^4 - 5*B*b^5)*c^2)*d^3 - 8*(3*D*a*b^4*c^3 - C*b^5*c^3)*d^2)*
x^3 - 4*(3*D*a^4*b*c^3 - (6*C*a^3*b^2 - 5*B*a^2*b^3)*c^3)*d^2 + ((D*a^5 + 3*C*a^4*b - 15*B*a^3*b^2 + 35*A*a^2*
b^3)*d^5 - 4*(2*D*a^4*b*c - (9*C*a^3*b^2 - 20*B*a^2*b^3 + 35*A*a*b^4)*c)*d^4 - (71*D*a^3*b^2*c^2 - (107*C*a^2*
b^3 - 95*B*a*b^4 + 35*A*b^5)*c^2)*d^3 - 4*(27*D*a^2*b^3*c^3 - (14*C*a*b^4 - 5*B*b^5)*c^3)*d^2 - 8*(3*D*a*b^4*c
^4 - C*b^5*c^4)*d)*x^2 - 8*(3*D*a^3*b^2*c^4 - C*a^2*b^3*c^4)*d + 2*((D*a^5*c + (3*C*a^4*b - 15*B*a^3*b^2 + 35*
A*a^2*b^3)*c)*d^4 - (11*D*a^4*b*c^2 - (27*C*a^3*b^2 - 35*B*a^2*b^3 + 35*A*a*b^4)*c^2)*d^3 - 4*(9*D*a^3*b^2*c^3
- (8*C*a^2*b^3 - 5*B*a*b^4)*c^3)*d^2 - 8*(3*D*a^2*b^3*c^4 - C*a*b^4*c^4)*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^5 - 8*A*a^4*b^2*d^5 - 8*(2
*B*a^4*b^2 - 11*A*a^3*b^3)*c*d^4 - (3*D*a^5*b*c^2 - (55*C*a^4*b^2 - 67*B*a^3*b^3 - 41*A*a^2*b^4)*c^2)*d^3 + 3*
((D*a^4*b^2 + 3*C*a^3*b^3 - 15*B*a^2*b^4 + 35*A*a*b^5)*d^5 - (13*D*a^3*b^3*c - (21*C*a^2*b^4 - 5*B*a*b^5 - 35*
A*b^6)*c)*d^4 - 4*(3*D*a^2*b^4*c^2 + (4*C*a*b^5 - 5*B*b^6)*c^2)*d^3 + 8*(3*D*a*b^5*c^3 - C*b^6*c^3)*d^2)*x^3 -
(91*D*a^4*b^2*c^3 + (5*C*a^3*b^3 - 77*B*a^2*b^4 + 45*A*a*b^5)*c^3)*d^2 + (8*D*b^6*c^5 - (3*D*a^5*b - 15*C*a^4
*b^2 + 75*B*a^3*b^3 - 175*A*a^2*b^4)*d^5 - (21*D*a^4*b^2*c - (117*C*a^3*b^3 - 85*B*a^2*b^4 - 35*A*a*b^5)*c)*d^
4 - 4*(48*D*a^3*b^3*c^2 - (C*a^2*b^4 + 20*B*a*b^5 - 35*A*b^6)*c^2)*d^3 + 8*(19*D*a^2*b^4*c^3 - (13*C*a*b^5 - 1
0*B*b^6)*c^3)*d^2 + 8*(7*D*a*b^5*c^4 - 4*C*b^6*c^4)*d)*x^2 + 2*(43*D*a^3*b^3*c^4 - (25*C*a^2*b^4 - 3*B*a*b^5 -
3*A*b^6)*c^4)*d + (16*D*a*b^5*c^5 - 8*(3*B*a^4*b^2 - 7*A*a^3*b^3)*d^5 - 2*(3*D*a^5*b*c - (39*C*a^4*b^2 - 55*B
*a^3*b^3 + 91*A*a^2*b^4)*c)*d^4 - (123*D*a^4*b^2*c^2 - (71*C*a^3*b^3 - 11*B*a^2*b^4 - 217*A*a*b^5)*c^2)*d^3 -
(35*D*a^3*b^3*c^3 + (61*C*a^2*b^4 - 133*B*a*b^5 + 21*A*b^6)*c^3)*d^2 + 4*(37*D*a^2*b^4*c^4 - (22*C*a*b^5 - 3*B
*b^6)*c^4)*d)*x)*sqrt(d*x + c))/(a^2*b^7*c^7*d - 5*a^3*b^6*c^6*d^2 + 10*a^4*b^5*c^5*d^3 - 10*a^5*b^4*c^4*d^4 +
5*a^6*b^3*c^3*d^5 - a^7*b^2*c^2*d^6 + (b^9*c^5*d^3 - 5*a*b^8*c^4*d^4 + 10*a^2*b^7*c^3*d^5 - 10*a^3*b^6*c^2*d^
6 + 5*a^4*b^5*c*d^7 - a^5*b^4*d^8)*x^4 + 2*(b^9*c^6*d^2 - 4*a*b^8*c^5*d^3 + 5*a^2*b^7*c^4*d^4 - 5*a^4*b^5*c^2*
d^6 + 4*a^5*b^4*c*d^7 - a^6*b^3*d^8)*x^3 + (b^9*c^7*d - a*b^8*c^6*d^2 - 9*a^2*b^7*c^5*d^3 + 25*a^3*b^6*c^4*d^4
- 25*a^4*b^5*c^3*d^5 + 9*a^5*b^4*c^2*d^6 + a^6*b^3*c*d^7 - a^7*b^2*d^8)*x^2 + 2*(a*b^8*c^7*d - 4*a^2*b^7*c^6*
d^2 + 5*a^3*b^6*c^5*d^3 - 5*a^5*b^4*c^3*d^5 + 4*a^6*b^3*c^2*d^6 - a^7*b^2*c*d^7)*x), 1/12*(3*(((D*a^3*b^2 + 3*
C*a^2*b^3 - 15*B*a*b^4 + 35*A*b^5)*d^5 - 4*(3*D*a^2*b^3*c - (6*C*a*b^4 - 5*B*b^5)*c)*d^4 - 8*(3*D*a*b^4*c^2 -
C*b^5*c^2)*d^3)*x^4 + (D*a^5*c^2 + (3*C*a^4*b - 15*B*a^3*b^2 + 35*A*a^2*b^3)*c^2)*d^3 + 2*((D*a^4*b + 3*C*a^3*
b^2 - 15*B*a^2*b^3 + 35*A*a*b^4)*d^5 - (11*D*a^3*b^2*c - (27*C*a^2*b^3 - 35*B*a*b^4 + 35*A*b^5)*c)*d^4 - 4*(9*
D*a^2*b^3*c^2 - (8*C*a*b^4 - 5*B*b^5)*c^2)*d^3 - 8*(3*D*a*b^4*c^3 - C*b^5*c^3)*d^2)*x^3 - 4*(3*D*a^4*b*c^3 - (
6*C*a^3*b^2 - 5*B*a^2*b^3)*c^3)*d^2 + ((D*a^5 + 3*C*a^4*b - 15*B*a^3*b^2 + 35*A*a^2*b^3)*d^5 - 4*(2*D*a^4*b*c
- (9*C*a^3*b^2 - 20*B*a^2*b^3 + 35*A*a*b^4)*c)*d^4 - (71*D*a^3*b^2*c^2 - (107*C*a^2*b^3 - 95*B*a*b^4 + 35*A*b^
5)*c^2)*d^3 - 4*(27*D*a^2*b^3*c^3 - (14*C*a*b^4 - 5*B*b^5)*c^3)*d^2 - 8*(3*D*a*b^4*c^4 - C*b^5*c^4)*d)*x^2 - 8
*(3*D*a^3*b^2*c^4 - C*a^2*b^3*c^4)*d + 2*((D*a^5*c + (3*C*a^4*b - 15*B*a^3*b^2 + 35*A*a^2*b^3)*c)*d^4 - (11*D*
a^4*b*c^2 - (27*C*a^3*b^2 - 35*B*a^2*b^3 + 35*A*a*b^4)*c^2)*d^3 - 4*(9*D*a^3*b^2*c^3 - (8*C*a^2*b^3 - 5*B*a*b^
4)*c^3)*d^2 - 8*(3*D*a^2*b^3*c^4 - C*a*b^4*c^4)*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^5 - 8*A*a^4*b^2*d^5 - 8*(2*B*a^4*b^2 - 11*A*a^3*b^3)*c*d^4 - (3*D*a^5*b
*c^2 - (55*C*a^4*b^2 - 67*B*a^3*b^3 - 41*A*a^2*b^4)*c^2)*d^3 + 3*((D*a^4*b^2 + 3*C*a^3*b^3 - 15*B*a^2*b^4 + 35
*A*a*b^5)*d^5 - (13*D*a^3*b^3*c - (21*C*a^2*b^4 - 5*B*a*b^5 - 35*A*b^6)*c)*d^4 - 4*(3*D*a^2*b^4*c^2 + (4*C*a*b
^5 - 5*B*b^6)*c^2)*d^3 + 8*(3*D*a*b^5*c^3 - C*b^6*c^3)*d^2)*x^3 - (91*D*a^4*b^2*c^3 + (5*C*a^3*b^3 - 77*B*a^2*
b^4 + 45*A*a*b^5)*c^3)*d^2 + (8*D*b^6*c^5 - (3*D*a^5*b - 15*C*a^4*b^2 + 75*B*a^3*b^3 - 175*A*a^2*b^4)*d^5 - (2
1*D*a^4*b^2*c - (117*C*a^3*b^3 - 85*B*a^2*b^4 - 35*A*a*b^5)*c)*d^4 - 4*(48*D*a^3*b^3*c^2 - (C*a^2*b^4 + 20*B*a
*b^5 - 35*A*b^6)*c^2)*d^3 + 8*(19*D*a^2*b^4*c^3 - (13*C*a*b^5 - 10*B*b^6)*c^3)*d^2 + 8*(7*D*a*b^5*c^4 - 4*C*b^
6*c^4)*d)*x^2 + 2*(43*D*a^3*b^3*c^4 - (25*C*a^2*b^4 - 3*B*a*b^5 - 3*A*b^6)*c^4)*d + (16*D*a*b^5*c^5 - 8*(3*B*a
^4*b^2 - 7*A*a^3*b^3)*d^5 - 2*(3*D*a^5*b*c - (39*C*a^4*b^2 - 55*B*a^3*b^3 + 91*A*a^2*b^4)*c)*d^4 - (123*D*a^4*
b^2*c^2 - (71*C*a^3*b^3 - 11*B*a^2*b^4 - 217*A*a*b^5)*c^2)*d^3 - (35*D*a^3*b^3*c^3 + (61*C*a^2*b^4 - 133*B*a*b
^5 + 21*A*b^6)*c^3)*d^2 + 4*(37*D*a^2*b^4*c^4 - (22*C*a*b^5 - 3*B*b^6)*c^4)*d)*x)*sqrt(d*x + c))/(a^2*b^7*c^7*
d - 5*a^3*b^6*c^6*d^2 + 10*a^4*b^5*c^5*d^3 - 10*a^5*b^4*c^4*d^4 + 5*a^6*b^3*c^3*d^5 - a^7*b^2*c^2*d^6 + (b^9*c
^5*d^3 - 5*a*b^8*c^4*d^4 + 10*a^2*b^7*c^3*d^5 - 10*a^3*b^6*c^2*d^6 + 5*a^4*b^5*c*d^7 - a^5*b^4*d^8)*x^4 + 2*(b
^9*c^6*d^2 - 4*a*b^8*c^5*d^3 + 5*a^2*b^7*c^4*d^4 - 5*a^4*b^5*c^2*d^6 + 4*a^5*b^4*c*d^7 - a^6*b^3*d^8)*x^3 + (b
^9*c^7*d - a*b^8*c^6*d^2 - 9*a^2*b^7*c^5*d^3 + 25*a^3*b^6*c^4*d^4 - 25*a^4*b^5*c^3*d^5 + 9*a^5*b^4*c^2*d^6 + a
^6*b^3*c*d^7 - a^7*b^2*d^8)*x^2 + 2*(a*b^8*c^7*d - 4*a^2*b^7*c^6*d^2 + 5*a^3*b^6*c^5*d^3 - 5*a^5*b^4*c^3*d^5 +
4*a^6*b^3*c^2*d^6 - a^7*b^2*c*d^7)*x)]

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giac [A]  time = 1.45, size = 767, normalized size = 1.75 \[ -\frac {{\left (24 \, D a b^{2} c^{2} - 8 \, C b^{3} c^{2} + 12 \, D a^{2} b c d - 24 \, C a b^{2} c d + 20 \, B b^{3} c d - D a^{3} d^{2} - 3 \, C a^{2} b d^{2} + 15 \, B a b^{2} d^{2} - 35 \, 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^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (D b c^{4} + 9 \, {\left (d x + c\right )} D a c^{2} d - 3 \, {\left (d x + c\right )} C b c^{2} d - D a c^{3} d - C b c^{3} d - 6 \, {\left (d x + c\right )} C a c d^{2} + 6 \, {\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} - 9 \, {\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^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} - \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 - {\left (d x + c\right )}^{\frac {3}{2}} D a^{3} b d^{2} - 3 \, {\left (d x + c\right )}^{\frac {3}{2}} C a^{2} b^{2} d^{2} + 7 \, {\left (d x + c\right )}^{\frac {3}{2}} B a b^{3} d^{2} - 11 \, {\left (d x + c\right )}^{\frac {3}{2}} A b^{4} d^{2} + 11 \, \sqrt {d x + c} D a^{3} b c d^{2} - 3 \, \sqrt {d x + c} C a^{2} b^{2} c d^{2} - 5 \, \sqrt {d x + c} B a b^{3} c d^{2} + 13 \, \sqrt {d x + c} A b^{4} c d^{2} + \sqrt {d x + c} D a^{4} d^{3} - 5 \, \sqrt {d x + c} C a^{3} b d^{3} + 9 \, \sqrt {d x + c} B a^{2} b^{2} d^{3} - 13 \, \sqrt {d x + c} A a b^{3} d^{3}}{4 \, {\left (b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{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)^3/(d*x+c)^(5/2),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 - 24*C*a*b^2*c*d + 20*B*b^3*c*d - D*a^3*d^2 - 3*C*a^2*b*d^
2 + 15*B*a*b^2*d^2 - 35*A*b^3*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^4 - 4*a*b^4*c^3*d + 6*
a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)*sqrt(-b^2*c + a*b*d)) - 2/3*(D*b*c^4 + 9*(d*x + c)*D*a*c^2*d -
3*(d*x + c)*C*b*c^2*d - D*a*c^3*d - C*b*c^3*d - 6*(d*x + c)*C*a*c*d^2 + 6*(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 - 9*(d*x + c)*A*b*d^3 - B*a*c*d^3 - A*b*c*d^3 + A*a*d^4)/((b^4*c^4*d - 4*a*b
^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)*(d*x + c)^(3/2)) - 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 - (d*x + c)^(3/2)*D*a^3*b*d^2 - 3*(d*x + c)^(3/2)*C*a^2*
b^2*d^2 + 7*(d*x + c)^(3/2)*B*a*b^3*d^2 - 11*(d*x + c)^(3/2)*A*b^4*d^2 + 11*sqrt(d*x + c)*D*a^3*b*c*d^2 - 3*sq
rt(d*x + c)*C*a^2*b^2*c*d^2 - 5*sqrt(d*x + c)*B*a*b^3*c*d^2 + 13*sqrt(d*x + c)*A*b^4*c*d^2 + sqrt(d*x + c)*D*a
^4*d^3 - 5*sqrt(d*x + c)*C*a^3*b*d^3 + 9*sqrt(d*x + c)*B*a^2*b^2*d^3 - 13*sqrt(d*x + c)*A*a*b^3*d^3)/((b^5*c^4
- 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)*((d*x + c)*b - b*c + a*d)^2)

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maple [B]  time = 0.04, size = 1376, normalized size = 3.14 \[ \text {result too large to display} \]

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)^(5/2),x)

[Out]

-2/3*d^2/(a*d-b*c)^3/(d*x+c)^(3/2)*A-2/3/(a*d-b*c)^3/(d*x+c)^(3/2)*C*c^2+2/3/d/(a*d-b*c)^3/(d*x+c)^(3/2)*D*c^3
+35/4*d^2/(a*d-b*c)^4*b^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*A+11/4*d^2/(a*d-b*c)
^4/(b*d*x+a*d)^2*(d*x+c)^(3/2)*A*b^3+1/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)^2*(d*x+c)^(3/2)*a^3*D+5/4*d^3/(a*d-b*c)^4
/(b*d*x+a*d)^2*(d*x+c)^(1/2)*C*a^3+2/(a*d-b*c)^4*b^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1
/2)*b)*C*c^2-4*d/(a*d-b*c)^4/(d*x+c)^(1/2)*B*b*c+4*d/(a*d-b*c)^4/(d*x+c)^(1/2)*C*a*c+3/4*d^2/(a*d-b*c)^4/((a*d
-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a^2*C-2*d^2/(a*d-b*c)^4/(d*x+c)^(1/2)*B*a+2/3*d/(a*
d-b*c)^3/(d*x+c)^(3/2)*B*c+6*d^2/(a*d-b*c)^4/(d*x+c)^(1/2)*A*b+2/(a*d-b*c)^4/(d*x+c)^(1/2)*c^2*b*C-6/(a*d-b*c)
^4/(d*x+c)^(1/2)*a*c^2*D+2*d/(a*d-b*c)^4/(b*d*x+a*d)^2*(d*x+c)^(3/2)*C*a*b^2*c-3*d/(a*d-b*c)^4/(b*d*x+a*d)^2*(
d*x+c)^(3/2)*D*a^2*b*c+3*d/(a*d-b*c)^4/(b*d*x+a*d)^2*b*(d*x+c)^(1/2)*a^2*c^2*D+5/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)
^2*b^2*(d*x+c)^(1/2)*B*a*c+3/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)^2*b*(d*x+c)^(1/2)*C*a^2*c-2*d/(a*d-b*c)^4/(b*d*x+a*
d)^2*b^2*(d*x+c)^(1/2)*a*c^2*C+6*d/(a*d-b*c)^4*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+13/4*d^3/(a*d-b*c)^4/(b*d*x+a*d)^2*b^2*(d*x+c)^(1/2)*A*a-13/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)^2*b^3*(d*x+
c)^(1/2)*A*c-9/4*d^3/(a*d-b*c)^4/(b*d*x+a*d)^2*b*(d*x+c)^(1/2)*B*a^2-6/(a*d-b*c)^4*b/((a*d-b*c)*b)^(1/2)*arcta
n((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*D*a*c^2+d/(a*d-b*c)^4/(b*d*x+a*d)^2*b^3*(d*x+c)^(1/2)*c^2*B-1/4*d^3/(a*
d-b*c)^4/(b*d*x+a*d)^2/b*(d*x+c)^(1/2)*D*a^4-11/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)^2*(d*x+c)^(1/2)*D*a^3*c+1/4*d^2/
(a*d-b*c)^4/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-3*d/(a*d-b*c)^4/((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-5*d/(a*d-b*c)^4*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-15/4*d^2/(a*d-b*c)^4*b/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-
b*c)*b)^(1/2)*b)*B*a-7/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)^2*(d*x+c)^(3/2)*B*a*b^2-d/(a*d-b*c)^4/(b*d*x+a*d)^2*(d*x+
c)^(3/2)*B*b^3*c+3/4*d^2/(a*d-b*c)^4/(b*d*x+a*d)^2*(d*x+c)^(3/2)*a^2*b*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)^3/(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 \[ \int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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