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

Optimal. Leaf size=188 \[ -\frac{2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2} \sqrt{b c-a d}}+\frac{2 \sqrt{c+d x} \left (a^2 d^2 D-a b d (C d-c D)+b^2 \left (-\left (-B d^2+c^2 (-D)+c C d\right )\right )\right )}{b^3 d^3}+\frac{2 (c+d x)^{3/2} (-a d D-2 b c D+b C d)}{3 b^2 d^3}+\frac{2 D (c+d x)^{5/2}}{5 b d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.205511, antiderivative size = 188, normalized size of antiderivative = 1., 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 = {1620, 63, 208} \[ -\frac{2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2} \sqrt{b c-a d}}+\frac{2 \sqrt{c+d x} \left (a^2 d^2 D-a b d (C d-c D)+b^2 \left (-\left (-B d^2+c^2 (-D)+c C d\right )\right )\right )}{b^3 d^3}+\frac{2 (c+d x)^{3/2} (-a d D-2 b c D+b C d)}{3 b^2 d^3}+\frac{2 D (c+d x)^{5/2}}{5 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 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]

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2+D x^3}{(a+b x) \sqrt{c+d x}} \, dx &=\int \left (\frac{a^2 d^2 D-a b d (C d-c D)-b^2 \left (c C d-B d^2-c^2 D\right )}{b^3 d^2 \sqrt{c+d x}}+\frac{A b^3-a \left (b^2 B-a b C+a^2 D\right )}{b^3 (a+b x) \sqrt{c+d x}}+\frac{(b C d-2 b c D-a d D) \sqrt{c+d x}}{b^2 d^2}+\frac{D (c+d x)^{3/2}}{b d^2}\right ) \, dx\\ &=\frac{2 \left (a^2 d^2 D-a b d (C d-c D)-b^2 \left (c C d-B d^2-c^2 D\right )\right ) \sqrt{c+d x}}{b^3 d^3}+\frac{2 (b C d-2 b c D-a d D) (c+d x)^{3/2}}{3 b^2 d^3}+\frac{2 D (c+d x)^{5/2}}{5 b d^3}+\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx\\ &=\frac{2 \left (a^2 d^2 D-a b d (C d-c D)-b^2 \left (c C d-B d^2-c^2 D\right )\right ) \sqrt{c+d x}}{b^3 d^3}+\frac{2 (b C d-2 b c D-a d D) (c+d x)^{3/2}}{3 b^2 d^3}+\frac{2 D (c+d x)^{5/2}}{5 b d^3}+\frac{\left (2 \left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\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 )}{d}\\ &=\frac{2 \left (a^2 d^2 D-a b d (C d-c D)-b^2 \left (c C d-B d^2-c^2 D\right )\right ) \sqrt{c+d x}}{b^3 d^3}+\frac{2 (b C d-2 b c D-a d D) (c+d x)^{3/2}}{3 b^2 d^3}+\frac{2 D (c+d x)^{5/2}}{5 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^{7/2} \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.25427, size = 185, normalized size = 0.98 \[ -\frac{2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2} \sqrt{b c-a d}}+\frac{2 \sqrt{c+d x} \left (a^2 d^2 D+a b d (c D-C d)+b^2 \left (B d^2+c^2 D-c C d\right )\right )}{b^3 d^3}+\frac{2 (c+d x)^{3/2} (-a d D-2 b c D+b C d)}{3 b^2 d^3}+\frac{2 D (c+d x)^{5/2}}{5 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.011, size = 338, normalized size = 1.8 \begin{align*}{\frac{2\,D}{5\,b{d}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{2\,C}{3\,b{d}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{2\,Da}{3\,{b}^{2}{d}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{4\,cD}{3\,b{d}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{B\sqrt{dx+c}}{bd}}-2\,{\frac{Ca\sqrt{dx+c}}{{b}^{2}d}}-2\,{\frac{cC\sqrt{dx+c}}{b{d}^{2}}}+2\,{\frac{D{a}^{2}\sqrt{dx+c}}{d{b}^{3}}}+2\,{\frac{acD\sqrt{dx+c}}{{b}^{2}{d}^{2}}}+2\,{\frac{{c}^{2}D\sqrt{dx+c}}{b{d}^{3}}}+2\,{\frac{A}{\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-2\,{\frac{Ba}{b\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+2\,{\frac{C{a}^{2}}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-2\,{\frac{D{a}^{3}}{{b}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

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)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [A]  time = 50.2713, size = 192, normalized size = 1.02 \begin{align*} \frac{2 D \left (c + d x\right )^{\frac{5}{2}}}{5 b d^{3}} - \frac{2 \left (c + d x\right )^{\frac{3}{2}} \left (- C b d + D a d + 2 D b c\right )}{3 b^{2} d^{3}} + \frac{2 \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname{atan}{\left (\frac{1}{\sqrt{\frac{b}{a d - b c}} \sqrt{c + d x}} \right )}}{b^{3} \sqrt{\frac{b}{a d - b c}} \left (a d - b c\right )} + \frac{2 \sqrt{c + d x} \left (B b^{2} d^{2} - C a b d^{2} - C b^{2} c d + D a^{2} d^{2} + D a b c d + D b^{2} c^{2}\right )}{b^{3} d^{3}} \end{align*}

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)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 2.52301, size = 335, normalized size = 1.78 \begin{align*} -\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 )}{\sqrt{-b^{2} c + a b d} b^{3}} + \frac{2 \,{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} D b^{4} d^{12} - 10 \,{\left (d x + c\right )}^{\frac{3}{2}} D b^{4} c d^{12} + 15 \, \sqrt{d x + c} D b^{4} c^{2} d^{12} - 5 \,{\left (d x + c\right )}^{\frac{3}{2}} D a b^{3} d^{13} + 5 \,{\left (d x + c\right )}^{\frac{3}{2}} C b^{4} d^{13} + 15 \, \sqrt{d x + c} D a b^{3} c d^{13} - 15 \, \sqrt{d x + c} C b^{4} c d^{13} + 15 \, \sqrt{d x + c} D a^{2} b^{2} d^{14} - 15 \, \sqrt{d x + c} C a b^{3} d^{14} + 15 \, \sqrt{d x + c} B b^{4} d^{14}\right )}}{15 \, b^{5} d^{15}} \end{align*}

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)^(1/2),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))/(sqrt(-b^2*c + a*b*d)*b^3)
+ 2/15*(3*(d*x + c)^(5/2)*D*b^4*d^12 - 10*(d*x + c)^(3/2)*D*b^4*c*d^12 + 15*sqrt(d*x + c)*D*b^4*c^2*d^12 - 5*
(d*x + c)^(3/2)*D*a*b^3*d^13 + 5*(d*x + c)^(3/2)*C*b^4*d^13 + 15*sqrt(d*x + c)*D*a*b^3*c*d^13 - 15*sqrt(d*x +
c)*C*b^4*c*d^13 + 15*sqrt(d*x + c)*D*a^2*b^2*d^14 - 15*sqrt(d*x + c)*C*a*b^3*d^14 + 15*sqrt(d*x + c)*B*b^4*d^1
4)/(b^5*d^15)