3.1.5 \(\int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx\) [5]

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

[Out]

2/3*(C*b*d-D*a*d-2*D*b*c)*(d*x+c)^(3/2)/b^2/d^3+2/5*D*(d*x+c)^(5/2)/b/d^3-2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*arct
anh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(1/2)+2*(a^2*d^2*D-a*b*d*(C*d-D*c)-b^2*(-B*d^2+
C*c*d-D*c^2))*(d*x+c)^(1/2)/b^3/d^3

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Rubi [A]
time = 0.14, antiderivative size = 188, 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 = {1634, 65, 214} \begin {gather*} -\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)-\left (b^2 \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} \end {gather*}

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 1634

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.31, size = 161, normalized size = 0.86 \begin {gather*} \frac {2 \sqrt {c+d x} \left (15 a^2 d^2 D-5 a b d (3 C d-2 c D+d D x)+b^2 \left (8 c^2 D-2 c d (5 C+2 D x)+d^2 \left (15 B+5 C x+3 D x^2\right )\right )\right )}{15 b^3 d^3}+\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{7/2} \sqrt {-b c+a d}} \end {gather*}

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*Sqrt[c + d*x]*(15*a^2*d^2*D - 5*a*b*d*(3*C*d - 2*c*D + d*D*x) + b^2*(8*c^2*D - 2*c*d*(5*C + 2*D*x) + d^2*(1
5*B + 5*C*x + 3*D*x^2))))/(15*b^3*d^3) + (2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])
/Sqrt[-(b*c) + a*d]])/(b^(7/2)*Sqrt[-(b*c) + a*d])

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Maple [A]
time = 0.10, size = 220, normalized size = 1.17

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\frac {D \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {C \,b^{2} d \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {D a b d \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 D b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+B \,b^{2} d^{2} \sqrt {d x +c}-C a b \,d^{2} \sqrt {d x +c}-C \,b^{2} c d \sqrt {d x +c}+D a^{2} d^{2} \sqrt {d x +c}+D a b c d \sqrt {d x +c}+D b^{2} c^{2} \sqrt {d x +c}\right )}{b^{3}}+\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\)	\(220\)
default	\(\frac {\frac {2 \left (\frac {D \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {C \,b^{2} d \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {D a b d \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 D b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+B \,b^{2} d^{2} \sqrt {d x +c}-C a b \,d^{2} \sqrt {d x +c}-C \,b^{2} c d \sqrt {d x +c}+D a^{2} d^{2} \sqrt {d x +c}+D a b c d \sqrt {d x +c}+D b^{2} c^{2} \sqrt {d x +c}\right )}{b^{3}}+\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\)	\(220\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> 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 detail

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Fricas [A]
time = 1.70, size = 565, normalized size = 3.01 \begin {gather*} \left [\frac {15 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \sqrt {b^{2} c - a b d} d^{3} \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^{3} - 15 \, {\left (D a^{3} b - C a^{2} b^{2} + B a b^{3}\right )} d^{3} + 5 \, {\left (D a^{2} b^{2} c - {\left (C a b^{3} - 3 \, B b^{4}\right )} c\right )} d^{2} + 3 \, {\left (D b^{4} c d^{2} - D a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (D a b^{3} c^{2} - 5 \, C b^{4} c^{2}\right )} d - {\left (4 \, D b^{4} c^{2} d - 5 \, {\left (D a^{2} b^{2} - C a b^{3}\right )} d^{3} + {\left (D a b^{3} c - 5 \, C b^{4} c\right )} d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )}}, -\frac {2 \, {\left (15 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \sqrt {-b^{2} c + a b d} d^{3} \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^{3} - 15 \, {\left (D a^{3} b - C a^{2} b^{2} + B a b^{3}\right )} d^{3} + 5 \, {\left (D a^{2} b^{2} c - {\left (C a b^{3} - 3 \, B b^{4}\right )} c\right )} d^{2} + 3 \, {\left (D b^{4} c d^{2} - D a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (D a b^{3} c^{2} - 5 \, C b^{4} c^{2}\right )} d - {\left (4 \, D b^{4} c^{2} d - 5 \, {\left (D a^{2} b^{2} - C a b^{3}\right )} d^{3} + {\left (D a b^{3} c - 5 \, C b^{4} c\right )} d^{2}\right )} x\right )} \sqrt {d x + c}\right )}}{15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )}}\right ] \end {gather*}

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]

[1/15*(15*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*sqrt(b^2*c - a*b*d)*d^3*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^3 - 15*(D*a^3*b - C*a^2*b^2 + B*a*b^3)*d^3 + 5*(D*a^2*b^2*c
- (C*a*b^3 - 3*B*b^4)*c)*d^2 + 3*(D*b^4*c*d^2 - D*a*b^3*d^3)*x^2 + 2*(D*a*b^3*c^2 - 5*C*b^4*c^2)*d - (4*D*b^4*
c^2*d - 5*(D*a^2*b^2 - C*a*b^3)*d^3 + (D*a*b^3*c - 5*C*b^4*c)*d^2)*x)*sqrt(d*x + c))/(b^5*c*d^3 - a*b^4*d^4),
-2/15*(15*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*sqrt(-b^2*c + a*b*d)*d^3*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x +
c)/(b*d*x + b*c)) - (8*D*b^4*c^3 - 15*(D*a^3*b - C*a^2*b^2 + B*a*b^3)*d^3 + 5*(D*a^2*b^2*c - (C*a*b^3 - 3*B*b^
4)*c)*d^2 + 3*(D*b^4*c*d^2 - D*a*b^3*d^3)*x^2 + 2*(D*a*b^3*c^2 - 5*C*b^4*c^2)*d - (4*D*b^4*c^2*d - 5*(D*a^2*b^
2 - C*a*b^3)*d^3 + (D*a*b^3*c - 5*C*b^4*c)*d^2)*x)*sqrt(d*x + c))/(b^5*c*d^3 - a*b^4*d^4)]

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Sympy [A]
time = 26.93, size = 192, normalized size = 1.02 \begin {gather*} \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 {gather*}

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
)

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Giac [A]
time = 0.55, size = 248, normalized size = 1.32 \begin {gather*} -\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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x+C\,x^2+x^3\,D}{\left (a+b\,x\right )\,\sqrt {c+d\,x}} \,d x \end {gather*}

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)^(1/2)),x)

[Out]

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

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