\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^2} \, dx\) [1208]

Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2} \]

[Out]

-1/2*(c*x^2+b*x+a)^(3/2)/c/d^2/(2*c*x+b)-3/32*(-4*a*c+b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/
c^(5/2)/d^2+3/16*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2/d^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {698, 626, 635, 212} \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=-\frac {3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2}+\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)} \]

[In]

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

[Out]

(3*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(16*c^2*d^2) - (a + b*x + c*x^2)^(3/2)/(2*c*d^2*(b + 2*c*x)) - (3*(b^2 -
4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(32*c^(5/2)*d^2)

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 626

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

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 698

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}+\frac {3 \int \sqrt {a+b x+c x^2} \, dx}{4 c d^2} \\ & = \frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{32 c^2 d^2} \\ & = \frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {\left (3 \left (b^2-4 a c\right )\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 )}{16 c^2 d^2} \\ & = \frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(244\) vs. \(2(113)=226\).

Time = 1.82 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.16 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (3 b^2+4 b c x+4 c \left (-2 a+c x^2\right )\right )}{b+2 c x}+\frac {2 \left (b^2-4 a c\right )^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {b^2-4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b}-3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )-\frac {2 \left (-b^2+4 a c\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b}}{16 c^{5/2} d^2} \]

[In]

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

[Out]

((Sqrt[c]*Sqrt[a + x*(b + c*x)]*(3*b^2 + 4*b*c*x + 4*c*(-2*a + c*x^2)))/(b + 2*c*x) + (2*(b^2 - 4*a*c)^(3/2)*A
rcTan[(Sqrt[c]*Sqrt[b^2 - 4*a*c]*x)/(Sqrt[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c*x)])])/b - 3*(b^2 - 4*a*c)*ArcT
anh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])] - (2*(-b^2 + 4*a*c)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[-b^2 + 4*a
*c]*x)/(Sqrt[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c*x)])])/b)/(16*c^(5/2)*d^2)

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.58

method	result	size
risch	\(\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{16 c^{2} d^{2}}+\frac {-\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+12 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )-\frac {\left (32 a^{2} c^{2}-16 a \,b^{2} c +2 b^{4}\right ) \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{c \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}}{32 c^{2} d^{2}}\)	\(179\)
default	\(\frac {-\frac {4 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}+\frac {16 c^{2} \left (\frac {\left (x +\frac {b}{2 c}\right ) \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{4}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (x +\frac {b}{2 c}\right ) \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{2}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\sqrt {c}\, \left (x +\frac {b}{2 c}\right )+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{4 a c -b^{2}}}{4 d^{2} c^{2}}\)	\(238\)

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^2,x,method=_RETURNVERBOSE)

[Out]

1/16*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2/d^2+1/32/c^2*(-3*b^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2
)+12*a*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-(32*a^2*c^2-16*a*b^2*c+2*b^4)/c/(4*a*c-b^2)/(x+1/2/
c*b)*((x+1/2/c*b)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))/d^2

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.49 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\left [-\frac {3 \, {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (4 \, c^{3} x^{2} + 4 \, b c^{2} x + 3 \, b^{2} c - 8 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{64 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}}, \frac {3 \, {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (4 \, c^{3} x^{2} + 4 \, b c^{2} x + 3 \, b^{2} c - 8 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{32 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}}\right ] \]

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^2,x, algorithm="fricas")

[Out]

[-1/64*(3*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*sqrt(c)*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*(4*c^3*x^2 + 4*b*c^2*x + 3*b^2*c - 8*a*c^2)*sqrt(c*x^2 + b*x + a))/(2*c^
4*d^2*x + b*c^3*d^2), 1/32*(3*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a
)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(4*c^3*x^2 + 4*b*c^2*x + 3*b^2*c - 8*a*c^2)*sqrt(c*x^2 + b
*x + a))/(2*c^4*d^2*x + b*c^3*d^2)]

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx}{d^{2}} \]

[In]

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**2,x)

[Out]

(Integral(a*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + Integral(b*x*sqrt(a + b*x + c*x**2)/(b
**2 + 4*b*c*x + 4*c**2*x**2), x) + Integral(c*x**2*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x))/
d**2

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (95) = 190\).

Time = 0.39 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.89 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {1}{32} \, d^{2} {\left (\frac {3 \, {\left (b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )\right )} \arctan \left (\frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2} d^{4} {\left | c \right |}} + \frac {2 \, {\left (\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )\right )}}{c^{3} d^{4} {\left | c \right |}} + \frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{{\left (\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}}\right )} c^{2} d^{4} {\left | c \right |}}\right )} {\left | c \right |} \]

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^2,x, algorithm="giac")

[Out]

1/32*d^2*(3*(b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))*arctan(sqr
t(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)/sqrt(-c))/(sqrt(-c)*c^2*d^4*abs(c)) + 2*(s
qrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)
- 4*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*
sgn(d))/(c^3*d^4*abs(c)) + (sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*b^2*sgn(1/(
2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a*c*s
gn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))/((b^2*c*d^2/(2*c*d*x + b*d)^2 - 4*a*c^2*d^2/(2*c*d*x + b*d)^2)*c^2*d^4*ab
s(c)))*abs(c)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^2} \,d x \]

[In]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^2,x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^2, x)