### 3.1248 $$\int \frac{1}{(b d+2 c d x)^3 (a+b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=132 $-\frac{12 c \sqrt{a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2}-\frac{2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{6 \sqrt{c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{5/2}}$

[Out]

-2/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*Sqrt[a + b*x + c*x^2]) - (12*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2*d^3
*(b + 2*c*x)^2) - (6*Sqrt[c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)
*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0845561, antiderivative size = 132, 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 = {687, 693, 688, 205} $-\frac{12 c \sqrt{a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2}-\frac{2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{6 \sqrt{c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x]

[Out]

-2/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*Sqrt[a + b*x + c*x^2]) - (12*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2*d^3
*(b + 2*c*x)^2) - (6*Sqrt[c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)
*d^3)

Rule 687

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

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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{1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{(12 c) \int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{12 c \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac{(6 c) \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{12 c \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac{\left (24 c^2\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 )}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{12 c \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac{6 \sqrt{c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d^3}\\ \end{align*}

Mathematica [C]  time = 0.0285623, size = 60, normalized size = 0.45 $-\frac{2 \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{d^3 \left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 2, 1/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/((b^2 - 4*a*c)^2*d^3*Sqrt[a + x*(
b + c*x)])

________________________________________________________________________________________

Maple [A]  time = 0.196, size = 218, normalized size = 1.7 \begin{align*} -{\frac{1}{4\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) } \left ( x+{\frac{b}{2\,c}} \right ) ^{-2}{\frac{1}{\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}}}-3\,{\frac{1}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{ \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+1/4\,{\frac{4\,ac-{b}^{2}}{c}}}}}}+6\,{\frac{1}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ({ \left ( 1/2\,{\frac{4\,ac-{b}^{2}}{c}}+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+1/2\,{\frac{b}{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(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),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)^(1/2)-3/d^3/(4*a*c-b^2)^2/((x+1/2*b
/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)+6/d^3/(4*a*c-b^2)^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))

________________________________________________________________________________________

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(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 13.6228, size = 1496, normalized size = 11.33 \begin{align*} \left [\frac{3 \,{\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} +{\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} +{\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt{-\frac{c}{b^{2} - 4 \, a 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}{\left (b^{2} - 4 \, a c\right )} \sqrt{-\frac{c}{b^{2} - 4 \, a c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) - 2 \,{\left (6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c\right )} \sqrt{c x^{2} + b x + a}}{4 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \,{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} +{\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} +{\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x +{\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}}, -\frac{2 \,{\left (3 \,{\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} +{\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} +{\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt{\frac{c}{b^{2} - 4 \, a c}} \arctan \left (-\frac{\sqrt{c x^{2} + b x + a}{\left (b^{2} - 4 \, a c\right )} \sqrt{\frac{c}{b^{2} - 4 \, a c}}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) +{\left (6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c\right )} \sqrt{c x^{2} + b x + a}\right )}}{4 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \,{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} +{\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} +{\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x +{\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[(3*(4*c^3*x^4 + 8*b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4*a*b*c)*x)*sqrt(-c/(b^2 - 4*a*c))*log
(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^
2 + 4*b*c*x + b^2)) - 2*(6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)*sqrt(c*x^2 + b*x + a))/(4*(b^4*c^3 - 8*a*b^2*c^4 +
16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2
*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c +
16*a^3*b^2*c^2)*d^3), -2*(3*(4*c^3*x^4 + 8*b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4*a*b*c)*x)*sq
rt(c/(b^2 - 4*a*c))*arctan(-1/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(c/(b^2 - 4*a*c))/(c^2*x^2 + b*c*x + a
*c)) + (6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)*sqrt(c*x^2 + b*x + a))/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*
x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4
)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d
^3)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{a b^{3} \sqrt{a + b x + c x^{2}} + 6 a b^{2} c x \sqrt{a + b x + c x^{2}} + 12 a b c^{2} x^{2} \sqrt{a + b x + c x^{2}} + 8 a c^{3} x^{3} \sqrt{a + b x + c x^{2}} + b^{4} x \sqrt{a + b x + c x^{2}} + 7 b^{3} c x^{2} \sqrt{a + b x + c x^{2}} + 18 b^{2} c^{2} x^{3} \sqrt{a + b x + c x^{2}} + 20 b c^{3} x^{4} \sqrt{a + b x + c x^{2}} + 8 c^{4} x^{5} \sqrt{a + b x + c x^{2}}}\, dx}{d^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/(a*b**3*sqrt(a + b*x + c*x**2) + 6*a*b**2*c*x*sqrt(a + b*x + c*x**2) + 12*a*b*c**2*x**2*sqrt(a + b*
x + c*x**2) + 8*a*c**3*x**3*sqrt(a + b*x + c*x**2) + b**4*x*sqrt(a + b*x + c*x**2) + 7*b**3*c*x**2*sqrt(a + b*
x + c*x**2) + 18*b**2*c**2*x**3*sqrt(a + b*x + c*x**2) + 20*b*c**3*x**4*sqrt(a + b*x + c*x**2) + 8*c**4*x**5*s
qrt(a + b*x + c*x**2)), x)/d**3

________________________________________________________________________________________

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(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError