3.1223 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^2} \, dx\)

Optimal. Leaf size=153 \[ -\frac{15 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{256 c^3 d^2}+\frac{15 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{512 c^{7/2} d^2}+\frac{5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)} \]

[Out]

(-15*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(256*c^3*d^2) + (5*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/
(32*c^2*d^2) - (a + b*x + c*x^2)^(5/2)/(2*c*d^2*(b + 2*c*x)) + (15*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt
[c]*Sqrt[a + b*x + c*x^2])])/(512*c^(7/2)*d^2)

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Rubi [A]  time = 0.064322, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {684, 612, 621, 206} \[ -\frac{15 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{256 c^3 d^2}+\frac{15 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{512 c^{7/2} d^2}+\frac{5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(256*c^3*d^2) + (5*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/
(32*c^2*d^2) - (a + b*x + c*x^2)^(5/2)/(2*c*d^2*(b + 2*c*x)) + (15*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt
[c]*Sqrt[a + b*x + c*x^2])])/(512*c^(7/2)*d^2)

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^2} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}+\frac{5 \int \left (a+b x+c x^2\right )^{3/2} \, dx}{4 c d^2}\\ &=\frac{5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}-\frac{\left (15 \left (b^2-4 a c\right )\right ) \int \sqrt{a+b x+c x^2} \, dx}{64 c^2 d^2}\\ &=-\frac{15 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{256 c^3 d^2}+\frac{5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}+\frac{\left (15 \left (b^2-4 a c\right )^2\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{512 c^3 d^2}\\ &=-\frac{15 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{256 c^3 d^2}+\frac{5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}+\frac{\left (15 \left (b^2-4 a c\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{256 c^3 d^2}\\ &=-\frac{15 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{256 c^3 d^2}+\frac{5 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{32 c^2 d^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 c d^2 (b+2 c x)}+\frac{15 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{512 c^{7/2} d^2}\\ \end{align*}

Mathematica [C]  time = 0.0519634, size = 97, normalized size = 0.63 \[ -\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 d^2 (b+2 c x) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -1/2, 1/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64*c
^3*d^2*(b + 2*c*x)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.192, size = 961, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c/d^2/(4*a*c-b^2)/(x+1/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+1/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+
1/4*(4*a*c-b^2)/c)^(5/2)*x+1/2/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)*b+5/4/d^2/(4*a*c-b^
2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*x*a-5/16/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3
/2)*x*b^2+5/8/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b*a-5/32/c^2/d^2/(4*a*c-b^2)*((x+1/2
*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^3+15/8/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*a^2-15
/16/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*a*b^2+15/128/c^2/d^2/(4*a*c-b^2)*((x+1/2*b/c
)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*b^4+15/16/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b*a^2-1
5/32/c^2/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b^3*a+15/256/c^3/d^2/(4*a*c-b^2)*((x+1/2*b/
c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b^5+15/8/c^(1/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(
4*a*c-b^2)/c)^(1/2))*a^3-45/32/c^(3/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)
/c)^(1/2))*b^2*a^2+45/128/c^(5/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(
1/2))*b^4*a-15/512/c^(7/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*b
^6

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.96526, size = 977, normalized size = 6.39 \begin{align*} \left [\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2} + 2 \,{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (32 \, c^{5} x^{4} + 64 \, b c^{4} x^{3} - 15 \, b^{4} c + 100 \, a b^{2} c^{2} - 128 \, a^{2} c^{3} + 12 \,{\left (b^{2} c^{3} + 12 \, a c^{4}\right )} x^{2} - 4 \,{\left (5 \, b^{3} c^{2} - 36 \, a b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{1024 \,{\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}}, -\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2} + 2 \,{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \,{\left (32 \, c^{5} x^{4} + 64 \, b c^{4} x^{3} - 15 \, b^{4} c + 100 \, a b^{2} c^{2} - 128 \, a^{2} c^{3} + 12 \,{\left (b^{2} c^{3} + 12 \, a c^{4}\right )} x^{2} - 4 \,{\left (5 \, b^{3} c^{2} - 36 \, a b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{512 \,{\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1024*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2 + 2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x)*sqrt(c)*log(-8*c^2*x^2 -
8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(32*c^5*x^4 + 64*b*c^4*x^3 - 15*b^4*
c + 100*a*b^2*c^2 - 128*a^2*c^3 + 12*(b^2*c^3 + 12*a*c^4)*x^2 - 4*(5*b^3*c^2 - 36*a*b*c^3)*x)*sqrt(c*x^2 + b*x
+ a))/(2*c^5*d^2*x + b*c^4*d^2), -1/512*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2 + 2*(b^4*c - 8*a*b^2*c^2 + 16*a^2
*c^3)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(32*c^5*x
^4 + 64*b*c^4*x^3 - 15*b^4*c + 100*a*b^2*c^2 - 128*a^2*c^3 + 12*(b^2*c^3 + 12*a*c^4)*x^2 - 4*(5*b^3*c^2 - 36*a
*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(2*c^5*d^2*x + b*c^4*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + Integral(b**2*x**2*sqrt(a + b*x + c
*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2
*x**2), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + Integral(2*a*c*x**2*
sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**2 +
4*b*c*x + 4*c**2*x**2), x))/d**2

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out