### 3.1241 $$\int \frac{1}{(b d+2 c d x)^4 \sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=79 $\frac{4 \sqrt{a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac{2 \sqrt{a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3}$

[Out]

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

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Rubi [A]  time = 0.0319232, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {693, 682} $\frac{4 \sqrt{a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac{2 \sqrt{a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 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)^4 \sqrt{a+b x+c x^2}} \, dx &=\frac{2 \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right ) d^4 (b+2 c x)^3}+\frac{2 \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 ) d^2}\\ &=\frac{2 \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right ) d^4 (b+2 c x)^3}+\frac{4 \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right )^2 d^4 (b+2 c x)}\\ \end{align*}

Mathematica [A]  time = 0.0275424, size = 60, normalized size = 0.76 $\frac{2 \sqrt{a+x (b+c x)} \left (-4 c \left (a-2 c x^2\right )+3 b^2+8 b c x\right )}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[a + x*(b + c*x)]*(3*b^2 + 8*b*c*x - 4*c*(a - 2*c*x^2)))/(3*(b^2 - 4*a*c)^2*d^4*(b + 2*c*x)^3)

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Maple [A]  time = 0.044, size = 70, normalized size = 0.9 \begin{align*} -{\frac{-16\,{c}^{2}{x}^{2}-16\,bcx+8\,ac-6\,{b}^{2}}{3\, \left ( 2\,cx+b \right ) ^{3}{d}^{4} \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) }\sqrt{c{x}^{2}+bx+a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x)

[Out]

-2/3*(-8*c^2*x^2-8*b*c*x+4*a*c-3*b^2)*(c*x^2+b*x+a)^(1/2)/(2*c*x+b)^3/d^4/(16*a^2*c^2-8*a*b^2*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(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.41221, size = 346, normalized size = 4.38 \begin{align*} \frac{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}}{3 \,{\left (8 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{4} x^{3} + 12 \,{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{4} x^{2} + 6 \,{\left (b^{6} c - 8 \, a b^{4} c^{2} + 16 \, a^{2} b^{2} c^{3}\right )} d^{4} x +{\left (b^{7} - 8 \, a b^{5} c + 16 \, a^{2} b^{3} c^{2}\right )} d^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2/3*(8*c^2*x^2 + 8*b*c*x + 3*b^2 - 4*a*c)*sqrt(c*x^2 + b*x + a)/(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^
3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7
- 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{b^{4} \sqrt{a + b x + c x^{2}} + 8 b^{3} c x \sqrt{a + b x + c x^{2}} + 24 b^{2} c^{2} x^{2} \sqrt{a + b x + c x^{2}} + 32 b c^{3} x^{3} \sqrt{a + b x + c x^{2}} + 16 c^{4} x^{4} \sqrt{a + b x + c x^{2}}}\, dx}{d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**4/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral(1/(b**4*sqrt(a + b*x + c*x**2) + 8*b**3*c*x*sqrt(a + b*x + c*x**2) + 24*b**2*c**2*x**2*sqrt(a + b*x +
c*x**2) + 32*b*c**3*x**3*sqrt(a + b*x + c*x**2) + 16*c**4*x**4*sqrt(a + b*x + c*x**2)), x)/d**4

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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(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError