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

[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))

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Rubi [A]  time = 0.313043, antiderivative size = 210, 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 = {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 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 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)^{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 = 0.653803, size = 210, normalized size = 1. \[ 2 \left (-\frac{\left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{3/2} (b c-a d)^{5/2}}+\frac{a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (-A d^3+c^2 C d-2 c^3 D\right )}{d^3 \sqrt{c+d x} (b c-a d)^2}+\frac{A d^3-B c d^2+c^2 C d+c^3 (-D)}{3 d^3 (c+d x)^{3/2} (b c-a d)}+\frac{D \sqrt{c+d x}}{b d^3}\right ) \]

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)))

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Maple [B]  time = 0.017, size = 464, normalized size = 2.2 \begin{align*} 2\,{\frac{D\sqrt{dx+c}}{b{d}^{3}}}-{\frac{2\,A}{3\,ad-3\,bc} \left ( dx+c \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Bc}{3\, \left ( ad-bc \right ) d} \left ( dx+c \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{c}^{2}C}{3\,{d}^{2} \left ( ad-bc \right ) } \left ( dx+c \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,D{c}^{3}}{3\,{d}^{3} \left ( ad-bc \right ) } \left ( dx+c \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{Ab}{ \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}-2\,{\frac{Ba}{ \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}+4\,{\frac{Cac}{d \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}-2\,{\frac{Cb{c}^{2}}{{d}^{2} \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}-6\,{\frac{aD{c}^{2}}{{d}^{2} \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}+4\,{\frac{Db{c}^{3}}{{d}^{3} \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}+2\,{\frac{{b}^{2}A}{ \left ( ad-bc \right ) ^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-2\,{\frac{bBa}{ \left ( ad-bc \right ) ^{2}\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 ) ^{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 \left ( ad-bc \right ) ^{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)^(5/2),x)

[Out]

2*D*(d*x+c)^(1/2)/b/d^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+2*b^2/(a*d-b*c)^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)^2/((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)^2/((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/b/(a*d-b*c)^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)^(5/2),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)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [A]  time = 87.2755, size = 214, normalized size = 1.02 \begin{align*} \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}} \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)**(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)

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Giac [A]  time = 2.59033, size = 379, normalized size = 1.8 \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^{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}} \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)^(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)