### 3.1209 $$\int \frac{(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^3} \, dx$$

Optimal. Leaf size=115 $-\frac{3 \sqrt{b^2-4 a c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{32 c^{5/2} d^3}+\frac{3 \sqrt{a+b x+c x^2}}{16 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}$

[Out]

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

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Rubi [A]  time = 0.0711629, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {684, 685, 688, 205} $-\frac{3 \sqrt{b^2-4 a c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{32 c^{5/2} d^3}+\frac{3 \sqrt{a+b x+c x^2}}{16 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 684

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]

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)^3} \, dx &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}+\frac{3 \int \frac{\sqrt{a+b x+c x^2}}{b d+2 c d x} \, dx}{8 c d^2}\\ &=\frac{3 \sqrt{a+b x+c x^2}}{16 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{32 c^2 d^2}\\ &=\frac{3 \sqrt{a+b x+c x^2}}{16 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \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 )}{8 c d^2}\\ &=\frac{3 \sqrt{a+b x+c x^2}}{16 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}-\frac{3 \sqrt{b^2-4 a c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{32 c^{5/2} d^3}\\ \end{align*}

Mathematica [C]  time = 0.032334, size = 62, normalized size = 0.54 $\frac{2 (a+x (b+c x))^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{5 d^3 \left (b^2-4 a c\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(a + x*(b + c*x))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(5*(b^2 - 4
*a*c)^2*d^3)

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Maple [B]  time = 0.191, size = 562, normalized size = 4.9 \begin{align*} -{\frac{1}{4\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{5}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-2}}+{\frac{1}{4\,c{d}^{3} \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,a}{8\,c{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{3\,{b}^{2}}{32\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{3\,{a}^{2}}{2\,c{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\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{3\,{b}^{2}a}{4\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\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{3\,{b}^{4}}{32\,{c}^{3}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\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)^3,x)

[Out]

-1/4/d^3/c^2/(4*a*c-b^2)/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+1/4/d^3/c/(4*a*c-b^2)*((x+1/2
*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+3/8/d^3/c/(4*a*c-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a-3/32/d^3/c^
2/(4*a*c-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^2-3/2/d^3/c/(4*a*c-b^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^2+3/4/d^3/c^2/
(4*a*c-b^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^2-3/32/d^3/c^3/(4*a*c-b^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))*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.63833, size = 705, normalized size = 6.13 \begin{align*} \left [\frac{3 \,{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\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 (8 \, c^{2} x^{2} + 8 \, b c x + 3 \, b^{2} - 4 \, a c\right )} \sqrt{c x^{2} + b x + a}}{64 \,{\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}}, \frac{3 \,{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\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 (8 \, c^{2} x^{2} + 8 \, b c x + 3 \, b^{2} - 4 \, a c\right )} \sqrt{c x^{2} + b x + a}}{32 \,{\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}}\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)^3,x, algorithm="fricas")

[Out]

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

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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^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac{b x \sqrt{a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac{c x^{2} \sqrt{a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(b*x*sqrt(
a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(c*x**2*sqrt(a + b*x + c*x*
*2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x))/d**3

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError