3.1207 \(\int \frac{(a+b x+c x^2)^{3/2}}{b d+2 c d x} \, dx\)

Optimal. Leaf size=115 \[ -\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{8 c^2 d}+\frac{\left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{16 c^{5/2} d}+\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d} \]

[Out]

-((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])/(8*c^2*d) + (a + b*x + c*x^2)^(3/2)/(6*c*d) + ((b^2 - 4*a*c)^(3/2)*ArcT
an[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(16*c^(5/2)*d)

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Rubi [A]  time = 0.0827866, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {685, 688, 205} \[ -\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{8 c^2 d}+\frac{\left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{16 c^{5/2} d}+\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])/(8*c^2*d) + (a + b*x + c*x^2)^(3/2)/(6*c*d) + ((b^2 - 4*a*c)^(3/2)*ArcT
an[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(16*c^(5/2)*d)

Rule 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 688

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{3/2}}{b d+2 c d x} \, dx &=\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d}-\frac{\left (b^2-4 a c\right ) \int \frac{\sqrt{a+b x+c x^2}}{b d+2 c d x} \, dx}{4 c}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{8 c^2 d}+\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d}+\frac{\left (b^2-4 a c\right )^2 \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{16 c^2}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{8 c^2 d}+\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d}+\frac{\left (b^2-4 a c\right )^2 \operatorname{Subst}\left (\int \frac{1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt{a+b x+c x^2}\right )}{4 c}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{8 c^2 d}+\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d}+\frac{\left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{16 c^{5/2} d}\\ \end{align*}

Mathematica [A]  time = 0.105074, size = 103, normalized size = 0.9 \[ \frac{2 \sqrt{c} \sqrt{a+x (b+c x)} \left (4 c \left (4 a+c x^2\right )-3 b^2+4 b c x\right )+3 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+x (b+c x)}}{\sqrt{b^2-4 a c}}\right )}{48 c^{5/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-3*b^2 + 4*b*c*x + 4*c*(4*a + c*x^2)) + 3*(b^2 - 4*a*c)^(3/2)*ArcTan[(2*Sqrt
[c]*Sqrt[a + x*(b + c*x)])/Sqrt[b^2 - 4*a*c]])/(48*c^(5/2)*d)

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Maple [B]  time = 0.19, size = 430, normalized size = 3.7 \begin{align*}{\frac{1}{6\,cd} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}}}+{\frac{a}{4\,cd}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{{b}^{2}}{16\,{c}^{2}d}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{{a}^{2}}{cd}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}}+{\frac{{b}^{2}a}{2\,{c}^{2}d}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}}-{\frac{{b}^{4}}{16\,d{c}^{3}}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/c*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+1/4/d/c*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a-1/16/d/c^2
*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^2-1/d/c/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2
)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*a^2+1/2/d/c^2/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(
4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*a*b^2-1/16/d/c^3/
((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))
/(x+1/2*b/c))*b^4

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.69747, size = 570, normalized size = 4.96 \begin{align*} \left [-\frac{3 \,{\left (b^{2} - 4 \, a c\right )} \sqrt{-\frac{b^{2} - 4 \, a c}{c}} \log \left (-\frac{4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt{c x^{2} + b x + a} c \sqrt{-\frac{b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) - 4 \,{\left (4 \, c^{2} x^{2} + 4 \, b c x - 3 \, b^{2} + 16 \, a c\right )} \sqrt{c x^{2} + b x + a}}{96 \, c^{2} d}, -\frac{3 \,{\left (b^{2} - 4 \, a c\right )} \sqrt{\frac{b^{2} - 4 \, a c}{c}} \arctan \left (\frac{\sqrt{\frac{b^{2} - 4 \, a c}{c}}}{2 \, \sqrt{c x^{2} + b x + a}}\right ) - 2 \,{\left (4 \, c^{2} x^{2} + 4 \, b c x - 3 \, b^{2} + 16 \, a c\right )} \sqrt{c x^{2} + b x + a}}{48 \, c^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(b^2 - 4*a*c)*sqrt(-(b^2 - 4*a*c)/c)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x +
a)*c*sqrt(-(b^2 - 4*a*c)/c))/(4*c^2*x^2 + 4*b*c*x + b^2)) - 4*(4*c^2*x^2 + 4*b*c*x - 3*b^2 + 16*a*c)*sqrt(c*x^
2 + b*x + a))/(c^2*d), -1/48*(3*(b^2 - 4*a*c)*sqrt((b^2 - 4*a*c)/c)*arctan(1/2*sqrt((b^2 - 4*a*c)/c)/sqrt(c*x^
2 + b*x + a)) - 2*(4*c^2*x^2 + 4*b*c*x - 3*b^2 + 16*a*c)*sqrt(c*x^2 + b*x + a))/(c^2*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a \sqrt{a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac{b x \sqrt{a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac{c x^{2} \sqrt{a + b x + c x^{2}}}{b + 2 c x}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15905, size = 201, normalized size = 1.75 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x + a}{\left (4 \, x{\left (\frac{x}{d} + \frac{b}{c d}\right )} - \frac{3 \, b^{2} c^{3} d^{3} - 16 \, a c^{4} d^{3}}{c^{5} d^{4}}\right )} + \frac{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac{2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} c + b \sqrt{c}}{\sqrt{b^{2} c - 4 \, a c^{2}}}\right )}{8 \, \sqrt{b^{2} c - 4 \, a c^{2}} c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(4*x*(x/d + b/(c*d)) - (3*b^2*c^3*d^3 - 16*a*c^4*d^3)/(c^5*d^4)) + 1/8*(b^4 - 8*a*b
^2*c + 16*a^2*c^2)*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^2))/(sqrt(
b^2*c - 4*a*c^2)*c^2*d)