\(\int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx\) [1234]

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {3}{4} \left (b^2-4 a c\right ) d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}+\frac {3 \left (b^2-4 a c\right )^2 d^4 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}} \]

[Out]

3/8*(-4*a*c+b^2)^2*d^4*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(1/2)+3/4*(-4*a*c+b^2)*d^4*(2*c*x+
b)*(c*x^2+b*x+a)^(1/2)+1/2*d^4*(2*c*x+b)^3*(c*x^2+b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {706, 635, 212} \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {3 d^4 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}+\frac {3}{4} d^4 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2} \]

[In]

Int[(b*d + 2*c*d*x)^4/Sqrt[a + b*x + c*x^2],x]

[Out]

(3*(b^2 - 4*a*c)*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/4 + (d^4*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/2 + (3*(
b^2 - 4*a*c)^2*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}+\frac {1}{4} \left (3 \left (b^2-4 a c\right ) d^2\right ) \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx \\ & = \frac {3}{4} \left (b^2-4 a c\right ) d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}+\frac {1}{8} \left (3 \left (b^2-4 a c\right )^2 d^4\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx \\ & = \frac {3}{4} \left (b^2-4 a c\right ) d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}+\frac {1}{4} \left (3 \left (b^2-4 a c\right )^2 d^4\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right ) \\ & = \frac {3}{4} \left (b^2-4 a c\right ) d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\frac {1}{2} d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}+\frac {3 \left (b^2-4 a c\right )^2 d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{4} d^4 \left ((b+2 c x) \sqrt {a+x (b+c x)} \left (5 b^2+8 b c x+4 c \left (-3 a+2 c x^2\right )\right )+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}\right ) \]

[In]

Integrate[(b*d + 2*c*d*x)^4/Sqrt[a + b*x + c*x^2],x]

[Out]

(d^4*((b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(5*b^2 + 8*b*c*x + 4*c*(-3*a + 2*c*x^2)) + (3*(b^2 - 4*a*c)^2*ArcTanh[
(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/Sqrt[c]))/4

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {\left (-16 c^{3} x^{3}-24 b \,c^{2} x^{2}+24 a \,c^{2} x -18 b^{2} c x +12 a b c -5 b^{3}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{4}}{4}+\frac {\left (\frac {3}{8} b^{4}+6 a^{2} c^{2}-3 a \,b^{2} c \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{4}}{\sqrt {c}}\)	\(114\)
default	\(d^{4} \left (\frac {b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+16 c^{4} \left (\frac {x^{3} \sqrt {c \,x^{2}+b x +a}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )}{8 c}-\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+8 b^{3} c \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+32 c^{3} b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+24 b^{2} c^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\right )\)	\(722\)

[In]

int((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(-16*c^3*x^3-24*b*c^2*x^2+24*a*c^2*x-18*b^2*c*x+12*a*b*c-5*b^3)*(c*x^2+b*x+a)^(1/2)*d^4+(3/8*b^4+6*a^2*c^
2-3*a*b^2*c)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)*d^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.68 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} d^{4} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} d^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x + {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{16 \, c}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} d^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} d^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x + {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{8 \, c}\right ] \]

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*d^4*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(
2*c*x + b)*sqrt(c) - 4*a*c) + 4*(16*c^4*d^4*x^3 + 24*b*c^3*d^4*x^2 + 6*(3*b^2*c^2 - 4*a*c^3)*d^4*x + (5*b^3*c
- 12*a*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a))/c, -1/8*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*d^4*arctan(1/2*sq
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(16*c^4*d^4*x^3 + 24*b*c^3*d^4*x^2 + 6*(
3*b^2*c^2 - 4*a*c^3)*d^4*x + (5*b^3*c - 12*a*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a))/c]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 549 vs. \(2 (110) = 220\).

Time = 0.75 (sec) , antiderivative size = 549, normalized size of antiderivative = 4.69 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \cdot \left (6 b c^{2} d^{4} x^{2} + 4 c^{3} d^{4} x^{3} + \frac {x \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{2 c} + \frac {- 12 a b c^{2} d^{4} + 8 b^{3} c d^{4} - \frac {3 b \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{4 c}}{c}\right ) + \left (- \frac {a \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{2 c} + b^{4} d^{4} - \frac {b \left (- 12 a b c^{2} d^{4} + 8 b^{3} c d^{4} - \frac {3 b \left (- 12 a c^{3} d^{4} + 9 b^{2} c^{2} d^{4}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \cdot \left (\frac {16 c^{4} d^{4} \left (a + b x\right )^{\frac {9}{2}}}{9 b^{4}} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (- 64 a c^{4} d^{4} + 32 b^{2} c^{3} d^{4}\right )}{7 b^{4}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \cdot \left (96 a^{2} c^{4} d^{4} - 96 a b^{2} c^{3} d^{4} + 24 b^{4} c^{2} d^{4}\right )}{5 b^{4}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 64 a^{3} c^{4} d^{4} + 96 a^{2} b^{2} c^{3} d^{4} - 48 a b^{4} c^{2} d^{4} + 8 b^{6} c d^{4}\right )}{3 b^{4}} + \frac {\sqrt {a + b x} \left (16 a^{4} c^{4} d^{4} - 32 a^{3} b^{2} c^{3} d^{4} + 24 a^{2} b^{4} c^{2} d^{4} - 8 a b^{6} c d^{4} + b^{8} d^{4}\right )}{b^{4}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {16 c^{4} d^{4} x^{5}}{5 \sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

integrate((2*c*d*x+b*d)**4/(c*x**2+b*x+a)**(1/2),x)

[Out]

Piecewise((sqrt(a + b*x + c*x**2)*(6*b*c**2*d**4*x**2 + 4*c**3*d**4*x**3 + x*(-12*a*c**3*d**4 + 9*b**2*c**2*d*
*4)/(2*c) + (-12*a*b*c**2*d**4 + 8*b**3*c*d**4 - 3*b*(-12*a*c**3*d**4 + 9*b**2*c**2*d**4)/(4*c))/c) + (-a*(-12
*a*c**3*d**4 + 9*b**2*c**2*d**4)/(2*c) + b**4*d**4 - b*(-12*a*b*c**2*d**4 + 8*b**3*c*d**4 - 3*b*(-12*a*c**3*d*
*4 + 9*b**2*c**2*d**4)/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(
a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)), Ne(c, 0)), (2*(16*c**4*
d**4*(a + b*x)**(9/2)/(9*b**4) + (a + b*x)**(7/2)*(-64*a*c**4*d**4 + 32*b**2*c**3*d**4)/(7*b**4) + (a + b*x)**
(5/2)*(96*a**2*c**4*d**4 - 96*a*b**2*c**3*d**4 + 24*b**4*c**2*d**4)/(5*b**4) + (a + b*x)**(3/2)*(-64*a**3*c**4
*d**4 + 96*a**2*b**2*c**3*d**4 - 48*a*b**4*c**2*d**4 + 8*b**6*c*d**4)/(3*b**4) + sqrt(a + b*x)*(16*a**4*c**4*d
**4 - 32*a**3*b**2*c**3*d**4 + 24*a**2*b**4*c**2*d**4 - 8*a*b**6*c*d**4 + b**8*d**4)/b**4)/b, Ne(b, 0)), (16*c
**4*d**4*x**5/(5*sqrt(a)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(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(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34 \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c^{3} d^{4} x + 3 \, b c^{2} d^{4}\right )} x + \frac {3 \, {\left (3 \, b^{2} c^{4} d^{4} - 4 \, a c^{5} d^{4}\right )}}{c^{3}}\right )} x + \frac {5 \, b^{3} c^{3} d^{4} - 12 \, a b c^{4} d^{4}}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, \sqrt {c}} \]

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c^3*d^4*x + 3*b*c^2*d^4)*x + 3*(3*b^2*c^4*d^4 - 4*a*c^5*d^4)/c^3)*x + (5*b^
3*c^3*d^4 - 12*a*b*c^4*d^4)/c^3) - 3/8*(b^4*d^4 - 8*a*b^2*c*d^4 + 16*a^2*c^2*d^4)*log(abs(2*(sqrt(c)*x - sqrt(
c*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^4}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

[In]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(1/2),x)

[Out]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(1/2), x)