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

Optimal. Leaf size=193 \[ -\frac{2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}+\frac{2 \left (A d^3-B c d^2+c^2 C d+c^3 (-D)\right )}{d^3 \sqrt{c+d x} (b c-a d)}+\frac{2 \sqrt{c+d x} (-a d D-b c D+b C d)}{b^2 d^3}+\frac{2 D (c+d x)^{3/2}}{3 b d^3}-\frac{2 c D \sqrt{c+d x}}{b d^3} \]

[Out]

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

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Rubi [A]  time = 0.240137, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1619, 43, 63, 208} \[ -\frac{2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}+\frac{2 \left (A d^3-B c d^2+c^2 C d+c^3 (-D)\right )}{d^3 \sqrt{c+d x} (b c-a d)}+\frac{2 \sqrt{c+d x} (-a d D-b c D+b C d)}{b^2 d^3}+\frac{2 D (c+d x)^{3/2}}{3 b d^3}-\frac{2 c D \sqrt{c+d x}}{b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A]  time = 0.535761, size = 174, normalized size = 0.9 \[ 2 \left (-\frac{\left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}+\frac{A d^3-B c d^2+c^2 C d+c^3 (-D)}{d^3 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{c+d x} (-a d D-2 b c D+b C d)}{b^2 d^3}+\frac{D (c+d x)^{3/2}}{3 b d^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 366, normalized size = 1.9 \begin{align*}{\frac{2\,D}{3\,b{d}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{C\sqrt{dx+c}}{b{d}^{2}}}-2\,{\frac{aD\sqrt{dx+c}}{{b}^{2}{d}^{2}}}-4\,{\frac{cD\sqrt{dx+c}}{b{d}^{3}}}-2\,{\frac{A}{ \left ( ad-bc \right ) \sqrt{dx+c}}}+2\,{\frac{Bc}{ \left ( ad-bc \right ) d\sqrt{dx+c}}}-2\,{\frac{{c}^{2}C}{{d}^{2} \left ( ad-bc \right ) \sqrt{dx+c}}}+2\,{\frac{D{c}^{3}}{{d}^{3} \left ( ad-bc \right ) \sqrt{dx+c}}}-2\,{\frac{Ab}{ \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+2\,{\frac{Ba}{ \left ( ad-bc \right ) \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}}{ \left ( ad-bc \right ) b\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}}{ \left ( ad-bc \right ){b}^{2}\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)^(3/2),x)

[Out]

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

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [A]  time = 50.8723, size = 172, normalized size = 0.89 \begin{align*} \frac{2 D \left (c + d x\right )^{\frac{3}{2}}}{3 b d^{3}} + \frac{2 \left (- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}\right )}{d^{3} \sqrt{c + d x} \left (a d - b c\right )} + \frac{\sqrt{c + d x} \left (2 C b d - 2 D a d - 4 D b c\right )}{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{\sqrt{c + d x}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{b^{3} \sqrt{\frac{a d - b c}{b}} \left (a d - b c\right )} \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)**(3/2),x)

[Out]

2*D*(c + d*x)**(3/2)/(3*b*d**3) + 2*(-A*d**3 + B*c*d**2 - C*c**2*d + D*c**3)/(d**3*sqrt(c + d*x)*(a*d - b*c))
+ sqrt(c + d*x)*(2*C*b*d - 2*D*a*d - 4*D*b*c)/(b**2*d**3) + 2*(-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*atan(sq
rt(c + d*x)/sqrt((a*d - b*c)/b))/(b**3*sqrt((a*d - b*c)/b)*(a*d - b*c))

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Giac [A]  time = 2.41799, size = 270, normalized size = 1.4 \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 )}{{\left (b^{3} c - a b^{2} d\right )} \sqrt{-b^{2} c + a b d}} - \frac{2 \,{\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b c d^{3} - a d^{4}\right )} \sqrt{d x + c}} + \frac{2 \,{\left ({\left (d x + c\right )}^{\frac{3}{2}} D b^{2} d^{6} - 6 \, \sqrt{d x + c} D b^{2} c d^{6} - 3 \, \sqrt{d x + c} D a b d^{7} + 3 \, \sqrt{d x + c} C b^{2} d^{7}\right )}}{3 \, b^{3} d^{9}} \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)^(3/2),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 - a*b^2*d)*sqrt(-b
^2*c + a*b*d)) - 2*(D*c^3 - C*c^2*d + B*c*d^2 - A*d^3)/((b*c*d^3 - a*d^4)*sqrt(d*x + c)) + 2/3*((d*x + c)^(3/2
)*D*b^2*d^6 - 6*sqrt(d*x + c)*D*b^2*c*d^6 - 3*sqrt(d*x + c)*D*a*b*d^7 + 3*sqrt(d*x + c)*C*b^2*d^7)/(b^3*d^9)