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

Optimal. Leaf size=210 \[ -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{5/2}}+\frac {2 \left (a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (-A d^3-2 c^3 D+c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{3 d^3 (c+d x)^{3/2} (b c-a d)}+\frac {2 D \sqrt {c+d x}}{b d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1619, 63, 208} \[ -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{5/2}}+\frac {2 \left (a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (-A d^3+c^2 C d-2 c^3 D\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \left (A d^3-B c d^2+c^2 C d+c^3 (-D)\right )}{3 d^3 (c+d x)^{3/2} (b c-a d)}+\frac {2 D \sqrt {c+d x}}{b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 1619

Int[((Px_)*((c_.) + (d_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[c + d*x],
(Px*(c + d*x)^(n + 1/2))/(a + b*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[n + 1/2, 0] &
& GtQ[Expon[Px, x], 2]

Rubi steps

\begin {align*} \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx &=\int \left (\frac {c^2 C d-B c d^2+A d^3-c^3 D}{d^2 (-b c+a d) (c+d x)^{5/2}}+\frac {-a d \left (2 c C d-B d^2-3 c^2 D\right )+b \left (c^2 C d-A d^3-2 c^3 D\right )}{d^2 (b c-a d)^2 (c+d x)^{3/2}}+\frac {D}{b d^2 \sqrt {c+d x}}+\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{b (b c-a d)^2 (a+b x) \sqrt {c+d x}}\right ) \, dx\\ &=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^3 (b c-a d) (c+d x)^{3/2}}+\frac {2 \left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (c^2 C d-A d^3-2 c^3 D\right )\right )}{d^3 (b c-a d)^2 \sqrt {c+d x}}+\frac {2 D \sqrt {c+d x}}{b d^3}+\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{b (b c-a d)^2}\\ &=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^3 (b c-a d) (c+d x)^{3/2}}+\frac {2 \left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (c^2 C d-A d^3-2 c^3 D\right )\right )}{d^3 (b c-a d)^2 \sqrt {c+d x}}+\frac {2 D \sqrt {c+d x}}{b d^3}+\frac {\left (2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right )\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 d (b c-a d)^2}\\ &=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^3 (b c-a d) (c+d x)^{3/2}}+\frac {2 \left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (c^2 C d-A d^3-2 c^3 D\right )\right )}{d^3 (b c-a d)^2 \sqrt {c+d x}}+\frac {2 D \sqrt {c+d x}}{b d^3}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\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)^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.66, size = 210, normalized size = 1.00 \[ 2 \left (-\frac {\left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{5/2}}+\frac {a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (-A d^3-2 c^3 D+c^2 C d\right )}{d^3 \sqrt {c+d x} (b c-a d)^2}+\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{3 d^3 (c+d x)^{3/2} (b c-a d)}+\frac {D \sqrt {c+d x}}{b d^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.77, size = 1287, normalized size = 6.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*((D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*d^5*x^2 + 2*(D*a^3*c - (C*a^2*b - B*a*b^2 + A*b^3)*c)*d^4*x + (D*
a^3*c^2 - (C*a^2*b - B*a*b^2 + A*b^3)*c^2)*d^3)*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*b^4*c^5 + A*a^2*b^2*d^5 + (2*B*a^2*b^2 - 5*A*a*b^3)*c*d^4 - (3*D*a^3
*b*c^2 + (5*C*a^2*b^2 + B*a*b^3 - 4*A*b^4)*c^2)*d^3 + (17*D*a^2*b^2*c^3 + (7*C*a*b^3 - B*b^4)*c^3)*d^2 + 3*(D*
b^4*c^3*d^2 - 3*D*a*b^3*c^2*d^3 + 3*D*a^2*b^2*c*d^4 - D*a^3*b*d^5)*x^2 - 2*(11*D*a*b^3*c^4 + C*b^4*c^4)*d + 3*
(4*D*b^4*c^4*d + (B*a^2*b^2 - A*a*b^3)*d^5 - (2*D*a^3*b*c + (2*C*a^2*b^2 + B*a*b^3 - A*b^4)*c)*d^4 + 3*(3*D*a^
2*b^2*c^2 + C*a*b^3*c^2)*d^3 - (11*D*a*b^3*c^3 + C*b^4*c^3)*d^2)*x)*sqrt(d*x + c))/(b^5*c^5*d^3 - 3*a*b^4*c^4*
d^4 + 3*a^2*b^3*c^3*d^5 - a^3*b^2*c^2*d^6 + (b^5*c^3*d^5 - 3*a*b^4*c^2*d^6 + 3*a^2*b^3*c*d^7 - a^3*b^2*d^8)*x^
2 + 2*(b^5*c^4*d^4 - 3*a*b^4*c^3*d^5 + 3*a^2*b^3*c^2*d^6 - a^3*b^2*c*d^7)*x), -2/3*(3*((D*a^3 - C*a^2*b + B*a*
b^2 - A*b^3)*d^5*x^2 + 2*(D*a^3*c - (C*a^2*b - B*a*b^2 + A*b^3)*c)*d^4*x + (D*a^3*c^2 - (C*a^2*b - B*a*b^2 + A
*b^3)*c^2)*d^3)*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*b^4*c^5 +
A*a^2*b^2*d^5 + (2*B*a^2*b^2 - 5*A*a*b^3)*c*d^4 - (3*D*a^3*b*c^2 + (5*C*a^2*b^2 + B*a*b^3 - 4*A*b^4)*c^2)*d^3
+ (17*D*a^2*b^2*c^3 + (7*C*a*b^3 - B*b^4)*c^3)*d^2 + 3*(D*b^4*c^3*d^2 - 3*D*a*b^3*c^2*d^3 + 3*D*a^2*b^2*c*d^4
- D*a^3*b*d^5)*x^2 - 2*(11*D*a*b^3*c^4 + C*b^4*c^4)*d + 3*(4*D*b^4*c^4*d + (B*a^2*b^2 - A*a*b^3)*d^5 - (2*D*a
^3*b*c + (2*C*a^2*b^2 + B*a*b^3 - A*b^4)*c)*d^4 + 3*(3*D*a^2*b^2*c^2 + C*a*b^3*c^2)*d^3 - (11*D*a*b^3*c^3 + C*
b^4*c^3)*d^2)*x)*sqrt(d*x + c))/(b^5*c^5*d^3 - 3*a*b^4*c^4*d^4 + 3*a^2*b^3*c^3*d^5 - a^3*b^2*c^2*d^6 + (b^5*c^
3*d^5 - 3*a*b^4*c^2*d^6 + 3*a^2*b^3*c*d^7 - a^3*b^2*d^8)*x^2 + 2*(b^5*c^4*d^4 - 3*a*b^4*c^3*d^5 + 3*a^2*b^3*c^
2*d^6 - a^3*b^2*c*d^7)*x)]

________________________________________________________________________________________

giac [A]  time = 1.31, size = 281, normalized size = 1.34 \[ -\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (6 \, {\left (d x + c\right )} D b c^{3} - D b c^{4} - 9 \, {\left (d x + c\right )} D a c^{2} d - 3 \, {\left (d x + c\right )} C b c^{2} d + D a c^{3} d + C b c^{3} d + 6 \, {\left (d x + c\right )} C a c d^{2} - C a c^{2} d^{2} - B b c^{2} d^{2} - 3 \, {\left (d x + c\right )} B a d^{3} + 3 \, {\left (d x + c\right )} A b d^{3} + B a c d^{3} + A b c d^{3} - A a d^{4}\right )}}{3 \, {\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} + \frac {2 \, \sqrt {d x + c} D}{b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 464, normalized size = 2.21 \[ \frac {2 A \,b^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {2 B a b \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}+\frac {2 C \,a^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {2 D a^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}\, b}+\frac {2 A b}{\left (a d -b c \right )^{2} \sqrt {d x +c}}-\frac {2 B a}{\left (a d -b c \right )^{2} \sqrt {d x +c}}+\frac {4 C a c}{\left (a d -b c \right )^{2} \sqrt {d x +c}\, d}-\frac {2 C b \,c^{2}}{\left (a d -b c \right )^{2} \sqrt {d x +c}\, d^{2}}-\frac {6 D a \,c^{2}}{\left (a d -b c \right )^{2} \sqrt {d x +c}\, d^{2}}+\frac {4 D b \,c^{3}}{\left (a d -b c \right )^{2} \sqrt {d x +c}\, d^{3}}-\frac {2 A}{3 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B c}{3 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}} d}-\frac {2 C \,c^{2}}{3 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}} d^{2}}+\frac {2 D c^{3}}{3 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}} d^{3}}+\frac {2 \sqrt {d x +c}\, D}{b \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)/(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?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x+C\,x^2+x^3\,D}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 153.50, size = 214, normalized size = 1.02 \[ \frac {2 D \sqrt {c + d x}}{b d^{3}} - \frac {2 \left (- A b d^{3} + B a d^{3} - 2 C a c d^{2} + C b c^{2} d + 3 D a c^{2} d - 2 D b c^{3}\right )}{d^{3} \sqrt {c + d x} \left (a d - b c\right )^{2}} + \frac {2 \left (- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}\right )}{3 d^{3} \left (c + d x\right )^{\frac {3}{2}} \left (a d - b c\right )} - \frac {2 \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{2} \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*D*sqrt(c + d*x)/(b*d**3) - 2*(-A*b*d**3 + B*a*d**3 - 2*C*a*c*d**2 + C*b*c**2*d + 3*D*a*c**2*d - 2*D*b*c**3)/
(d**3*sqrt(c + d*x)*(a*d - b*c)**2) + 2*(-A*d**3 + B*c*d**2 - C*c**2*d + D*c**3)/(3*d**3*(c + d*x)**(3/2)*(a*d
- b*c)) - 2*(-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**2*sqrt((a*d
- b*c)/b)*(a*d - b*c)**2)

________________________________________________________________________________________