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

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 193, normalized size of antiderivative = 1.00, 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 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]

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)^{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*}

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Mathematica [A]  time = 0.53, size = 174, normalized size = 0.90 \[ 2 \left (-\frac {\left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}+\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{d^3 \sqrt {c+d x} (b c-a d)}+\frac {\sqrt {c+d x} (-a d D-2 b c D+b C d)}{b^2 d^3}+\frac {D (c+d x)^{3/2}}{3 b d^3}\right ) \]

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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fricas [B]  time = 0.97, size = 866, normalized size = 4.49 \[ \left [-\frac {3 \, {\left ({\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} d^{4} x + {\left (D a^{3} c - {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} c\right )} d^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (8 \, D b^{4} c^{4} + 3 \, A a b^{3} d^{4} + 3 \, {\left (D a^{3} b c - {\left (C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} c\right )} d^{3} - {\left (D a^{2} b^{2} c^{2} - 3 \, {\left (3 \, C a b^{3} + B b^{4}\right )} c^{2}\right )} d^{2} - {\left (D b^{4} c^{2} d^{2} - 2 \, D a b^{3} c d^{3} + D a^{2} b^{2} d^{4}\right )} x^{2} - 2 \, {\left (5 \, D a b^{3} c^{3} + 3 \, C b^{4} c^{3}\right )} d + {\left (4 \, D b^{4} c^{3} d + 3 \, {\left (D a^{3} b - C a^{2} b^{2}\right )} d^{4} - 2 \, {\left (D a^{2} b^{2} c - 3 \, C a b^{3} c\right )} d^{3} - {\left (5 \, D a b^{3} c^{2} + 3 \, C b^{4} c^{2}\right )} d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (b^{5} c^{3} d^{3} - 2 \, a b^{4} c^{2} d^{4} + a^{2} b^{3} c d^{5} + {\left (b^{5} c^{2} d^{4} - 2 \, a b^{4} c d^{5} + a^{2} b^{3} d^{6}\right )} x\right )}}, -\frac {2 \, {\left (3 \, {\left ({\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} d^{4} x + {\left (D a^{3} c - {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} c\right )} d^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) + {\left (8 \, D b^{4} c^{4} + 3 \, A a b^{3} d^{4} + 3 \, {\left (D a^{3} b c - {\left (C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} c\right )} d^{3} - {\left (D a^{2} b^{2} c^{2} - 3 \, {\left (3 \, C a b^{3} + B b^{4}\right )} c^{2}\right )} d^{2} - {\left (D b^{4} c^{2} d^{2} - 2 \, D a b^{3} c d^{3} + D a^{2} b^{2} d^{4}\right )} x^{2} - 2 \, {\left (5 \, D a b^{3} c^{3} + 3 \, C b^{4} c^{3}\right )} d + {\left (4 \, D b^{4} c^{3} d + 3 \, {\left (D a^{3} b - C a^{2} b^{2}\right )} d^{4} - 2 \, {\left (D a^{2} b^{2} c - 3 \, C a b^{3} c\right )} d^{3} - {\left (5 \, D a b^{3} c^{2} + 3 \, C b^{4} c^{2}\right )} d^{2}\right )} x\right )} \sqrt {d x + c}\right )}}{3 \, {\left (b^{5} c^{3} d^{3} - 2 \, a b^{4} c^{2} d^{4} + a^{2} b^{3} c d^{5} + {\left (b^{5} c^{2} d^{4} - 2 \, a b^{4} c d^{5} + a^{2} b^{3} d^{6}\right )} x\right )}}\right ] \]

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]

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

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giac [A]  time = 1.33, size = 200, normalized size = 1.04 \[ -\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c - a b^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b c d^{3} - a d^{4}\right )} \sqrt {d x + c}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} D b^{2} d^{6} - 6 \, \sqrt {d x + c} D b^{2} c d^{6} - 3 \, \sqrt {d x + c} D a b d^{7} + 3 \, \sqrt {d x + c} C b^{2} d^{7}\right )}}{3 \, b^{3} d^{9}} \]

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)

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maple [B]  time = 0.02, size = 366, normalized size = 1.90 \[ -\frac {2 A b \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}+\frac {2 B a \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}-\frac {2 C \,a^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b}+\frac {2 D a^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b^{2}}-\frac {2 A}{\left (a d -b c \right ) \sqrt {d x +c}}+\frac {2 B c}{\left (a d -b c \right ) \sqrt {d x +c}\, d}-\frac {2 C \,c^{2}}{\left (a d -b c \right ) \sqrt {d x +c}\, d^{2}}+\frac {2 D c^{3}}{\left (a d -b c \right ) \sqrt {d x +c}\, d^{3}}+\frac {2 \sqrt {d x +c}\, C}{b \,d^{2}}-\frac {2 \sqrt {d x +c}\, D a}{b^{2} d^{2}}-\frac {4 \sqrt {d x +c}\, D c}{b \,d^{3}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} D}{3 b \,d^{3}} \]

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)^(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)*b/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*A+2/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arct
an((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*B*a-2/(a*d-b*c)/b/((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/(a*d-b*c)/b^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/(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

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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)/(d*x+c)^(3/2),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 = -0.01 \[ \int \frac {A+B\,x+C\,x^2+x^3\,D}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

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)^(3/2)),x)

[Out]

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

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sympy [A]  time = 85.73, size = 172, normalized size = 0.89 \[ \frac {2 D \left (c + d x\right )^{\frac {3}{2}}}{3 b d^{3}} + \frac {2 \left (- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}\right )}{d^{3} \sqrt {c + d x} \left (a d - b c\right )} + \frac {\sqrt {c + d x} \left (2 C b d - 2 D a d - 4 D b c\right )}{b^{2} d^{3}} + \frac {2 \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{3} \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )} \]

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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