### 3.1257 $$\int \frac{1}{(b d+2 c d x)^2 (a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=122 $\frac{256 c^2 \sqrt{a+b x+c x^2}}{3 d^2 \left (b^2-4 a c\right )^3 (b+2 c x)}+\frac{32 c}{3 d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}-\frac{2}{3 d^2 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}$

[Out]

-2/(3*(b^2 - 4*a*c)*d^2*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2)) + (32*c)/(3*(b^2 - 4*a*c)^2*d^2*(b + 2*c*x)*Sqrt[
a + b*x + c*x^2]) + (256*c^2*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*c)^3*d^2*(b + 2*c*x))

________________________________________________________________________________________

Rubi [A]  time = 0.0557404, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {687, 682} $\frac{256 c^2 \sqrt{a+b x+c x^2}}{3 d^2 \left (b^2-4 a c\right )^3 (b+2 c x)}+\frac{32 c}{3 d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}-\frac{2}{3 d^2 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(5/2)),x]

[Out]

-2/(3*(b^2 - 4*a*c)*d^2*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2)) + (32*c)/(3*(b^2 - 4*a*c)^2*d^2*(b + 2*c*x)*Sqrt[
a + b*x + c*x^2]) + (256*c^2*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*c)^3*d^2*(b + 2*c*x))

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 682

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] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2}{3 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}-\frac{(16 c) \int \frac{1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}+\frac{32 c}{3 \left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt{a+b x+c x^2}}+\frac{\left (128 c^2\right ) \int \frac{1}{(b d+2 c d x)^2 \sqrt{a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}+\frac{32 c}{3 \left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt{a+b x+c x^2}}+\frac{256 c^2 \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right )^3 d^2 (b+2 c x)}\\ \end{align*}

Mathematica [A]  time = 0.0516286, size = 108, normalized size = 0.89 $\frac{32 c^2 \left (3 a^2+12 a c x^2+8 c^2 x^4\right )+48 b^2 c \left (a+6 c x^2\right )+128 b c^2 x \left (3 a+4 c x^2\right )+32 b^3 c x-2 b^4}{3 d^2 \left (b^2-4 a c\right )^3 (b+2 c x) (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(-2*b^4 + 32*b^3*c*x + 128*b*c^2*x*(3*a + 4*c*x^2) + 48*b^2*c*(a + 6*c*x^2) + 32*c^2*(3*a^2 + 12*a*c*x^2 + 8*c
^2*x^4))/(3*(b^2 - 4*a*c)^3*d^2*(b + 2*c*x)*(a + x*(b + c*x))^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.048, size = 133, normalized size = 1.1 \begin{align*} -{\frac{256\,{c}^{4}{x}^{4}+512\,b{c}^{3}{x}^{3}+384\,a{c}^{3}{x}^{2}+288\,{b}^{2}{c}^{2}{x}^{2}+384\,ab{c}^{2}x+32\,{b}^{3}cx+96\,{a}^{2}{c}^{2}+48\,ac{b}^{2}-2\,{b}^{4}}{ \left ( 6\,cx+3\,b \right ){d}^{2} \left ( 64\,{a}^{3}{c}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6} \right ) } \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/3*(128*c^4*x^4+256*b*c^3*x^3+192*a*c^3*x^2+144*b^2*c^2*x^2+192*a*b*c^2*x+16*b^3*c*x+48*a^2*c^2+24*a*b^2*c-b
^4)/(2*c*x+b)/d^2/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(c*x^2+b*x+a)^(3/2)

________________________________________________________________________________________

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)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 22.3653, size = 794, normalized size = 6.51 \begin{align*} \frac{2 \,{\left (128 \, c^{4} x^{4} + 256 \, b c^{3} x^{3} - b^{4} + 24 \, a b^{2} c + 48 \, a^{2} c^{2} + 48 \,{\left (3 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} + 16 \,{\left (b^{3} c + 12 \, a b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3 \,{\left (2 \,{\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} d^{2} x^{5} + 5 \,{\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} d^{2} x^{4} + 4 \,{\left (b^{8} c - 11 \, a b^{6} c^{2} + 36 \, a^{2} b^{4} c^{3} - 16 \, a^{3} b^{2} c^{4} - 64 \, a^{4} c^{5}\right )} d^{2} x^{3} +{\left (b^{9} - 6 \, a b^{7} c - 24 \, a^{2} b^{5} c^{2} + 224 \, a^{3} b^{3} c^{3} - 384 \, a^{4} b c^{4}\right )} d^{2} x^{2} + 2 \,{\left (a b^{8} - 11 \, a^{2} b^{6} c + 36 \, a^{3} b^{4} c^{2} - 16 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4}\right )} d^{2} x +{\left (a^{2} b^{7} - 12 \, a^{3} b^{5} c + 48 \, a^{4} b^{3} c^{2} - 64 \, a^{5} b c^{3}\right )} d^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/3*(128*c^4*x^4 + 256*b*c^3*x^3 - b^4 + 24*a*b^2*c + 48*a^2*c^2 + 48*(3*b^2*c^2 + 4*a*c^3)*x^2 + 16*(b^3*c +
12*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a)/(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^
7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*
a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x
^2 + 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c +
48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{a^{2} b^{2} \sqrt{a + b x + c x^{2}} + 4 a^{2} b c x \sqrt{a + b x + c x^{2}} + 4 a^{2} c^{2} x^{2} \sqrt{a + b x + c x^{2}} + 2 a b^{3} x \sqrt{a + b x + c x^{2}} + 10 a b^{2} c x^{2} \sqrt{a + b x + c x^{2}} + 16 a b c^{2} x^{3} \sqrt{a + b x + c x^{2}} + 8 a c^{3} x^{4} \sqrt{a + b x + c x^{2}} + b^{4} x^{2} \sqrt{a + b x + c x^{2}} + 6 b^{3} c x^{3} \sqrt{a + b x + c x^{2}} + 13 b^{2} c^{2} x^{4} \sqrt{a + b x + c x^{2}} + 12 b c^{3} x^{5} \sqrt{a + b x + c x^{2}} + 4 c^{4} x^{6} \sqrt{a + b x + c x^{2}}}\, dx}{d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**2/(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral(1/(a**2*b**2*sqrt(a + b*x + c*x**2) + 4*a**2*b*c*x*sqrt(a + b*x + c*x**2) + 4*a**2*c**2*x**2*sqrt(a +
b*x + c*x**2) + 2*a*b**3*x*sqrt(a + b*x + c*x**2) + 10*a*b**2*c*x**2*sqrt(a + b*x + c*x**2) + 16*a*b*c**2*x**
3*sqrt(a + b*x + c*x**2) + 8*a*c**3*x**4*sqrt(a + b*x + c*x**2) + b**4*x**2*sqrt(a + b*x + c*x**2) + 6*b**3*c*
x**3*sqrt(a + b*x + c*x**2) + 13*b**2*c**2*x**4*sqrt(a + b*x + c*x**2) + 12*b*c**3*x**5*sqrt(a + b*x + c*x**2)
+ 4*c**4*x**6*sqrt(a + b*x + c*x**2)), x)/d**2

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (2 \, c d x + b d\right )}^{2}{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((2*c*d*x + b*d)^2*(c*x^2 + b*x + a)^(5/2)), x)