3.1224 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^3} \, dx\)

Optimal. Leaf size=147 \[ -\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{64 c^3 d^3}+\frac{5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{128 c^{7/2} d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2} \]

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])/(64*c^3*d^3) + (5*(a + b*x + c*x^2)^(3/2))/(48*c^2*d^3) - (a + b*x +
c*x^2)^(5/2)/(4*c*d^3*(b + 2*c*x)^2) + (5*(b^2 - 4*a*c)^(3/2)*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^
2 - 4*a*c]])/(128*c^(7/2)*d^3)

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Rubi [A]  time = 0.111583, antiderivative size = 147, 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, 685, 688, 205} \[ -\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{64 c^3 d^3}+\frac{5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{128 c^{7/2} d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^3,x]

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])/(64*c^3*d^3) + (5*(a + b*x + c*x^2)^(3/2))/(48*c^2*d^3) - (a + b*x +
c*x^2)^(5/2)/(4*c*d^3*(b + 2*c*x)^2) + (5*(b^2 - 4*a*c)^(3/2)*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^
2 - 4*a*c]])/(128*c^(7/2)*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 )^{5/2}}{(b d+2 c d x)^3} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac{5 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{b d+2 c d x} \, dx}{8 c d^2}\\ &=\frac{5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \frac{\sqrt{a+b x+c x^2}}{b d+2 c d x} \, dx}{32 c^2 d^2}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{64 c^3 d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac{\left (5 \left (b^2-4 a c\right )^2\right ) \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{128 c^3 d^2}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{64 c^3 d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac{\left (5 \left (b^2-4 a c\right )^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 )}{32 c^2 d^2}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}{64 c^3 d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac{5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{128 c^{7/2} d^3}\\ \end{align*}

Mathematica [C]  time = 0.0396225, size = 62, normalized size = 0.42 \[ \frac{2 (a+x (b+c x))^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{7 d^3 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^3,x]

[Out]

(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4
*a*c)^2*d^3)

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Maple [B]  time = 0.196, size = 840, normalized size = 5.7 \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)^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)^(7/2)+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)^(5/2)+5/12/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)*a-5/48/d^3
/c^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2+5/8/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^2-5/16/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)*a*b^2+5/128/d^3/c^3/(4*a*c
-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4-5/2/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^3+15/8/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^2*b^2-15/32/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))*a*b^4+5/128/d^3/c^4/(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^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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.40079, size = 1056, normalized size = 7.18 \begin{align*} \left [-\frac{15 \,{\left (b^{4} - 4 \, a b^{2} c + 4 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x\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 (32 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 15 \, b^{4} + 80 \, a b^{2} c - 48 \, a^{2} c^{2} - 8 \,{\left (b^{2} c^{2} - 28 \, a c^{3}\right )} x^{2} - 8 \,{\left (5 \, b^{3} c - 28 \, a b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{768 \,{\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}}, -\frac{15 \,{\left (b^{4} - 4 \, a b^{2} c + 4 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x\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 (32 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 15 \, b^{4} + 80 \, a b^{2} c - 48 \, a^{2} c^{2} - 8 \,{\left (b^{2} c^{2} - 28 \, a c^{3}\right )} x^{2} - 8 \,{\left (5 \, b^{3} c - 28 \, a b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{384 \,{\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\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)^3,x, algorithm="fricas")

[Out]

[-1/768*(15*(b^4 - 4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*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*(32*c^4*x^4 + 64*b*c^3*x^3 - 15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 8*(b^2*c^2 - 28*a*c^3)*x^2 - 8*(5*
b^3*c - 28*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3), -1/384*(15*(b^4 -
4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*sqrt((b^2 - 4*a*c)/c)*arctan(1/2*sqrt((b^2 -
4*a*c)/c)/sqrt(c*x^2 + b*x + a)) - 2*(32*c^4*x^4 + 64*b*c^3*x^3 - 15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 8*(b^2*c
^2 - 28*a*c^3)*x^2 - 8*(5*b^3*c - 28*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c
^3*d^3)]

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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^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac{b^{2} 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 + \int \frac{c^{2} x^{4} \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{2 a 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{2 a 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 + \int \frac{2 b c x^{3} \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)**(5/2)/(2*c*d*x+b*d)**3,x)

[Out]

(Integral(a**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) + Integral(b**2*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) + Integral(c**2*x**4*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(2*a*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(2*a*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) + Integral(2*b*c*x**3*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)^(5/2)/(2*c*d*x+b*d)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError