3.1216 \(\int \frac{(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{10}} \, dx\)

Optimal. Leaf size=118 \[ \frac{16 \left (a+b x+c x^2\right )^{5/2}}{315 d^{10} \left (b^2-4 a c\right )^3 (b+2 c x)^5}+\frac{8 \left (a+b x+c x^2\right )^{5/2}}{63 d^{10} \left (b^2-4 a c\right )^2 (b+2 c x)^7}+\frac{2 \left (a+b x+c x^2\right )^{5/2}}{9 d^{10} \left (b^2-4 a c\right ) (b+2 c x)^9} \]

[Out]

(2*(a + b*x + c*x^2)^(5/2))/(9*(b^2 - 4*a*c)*d^10*(b + 2*c*x)^9) + (8*(a + b*x + c*x^2)^(5/2))/(63*(b^2 - 4*a*
c)^2*d^10*(b + 2*c*x)^7) + (16*(a + b*x + c*x^2)^(5/2))/(315*(b^2 - 4*a*c)^3*d^10*(b + 2*c*x)^5)

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Rubi [A]  time = 0.0572946, antiderivative size = 118, 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 = {693, 682} \[ \frac{16 \left (a+b x+c x^2\right )^{5/2}}{315 d^{10} \left (b^2-4 a c\right )^3 (b+2 c x)^5}+\frac{8 \left (a+b x+c x^2\right )^{5/2}}{63 d^{10} \left (b^2-4 a c\right )^2 (b+2 c x)^7}+\frac{2 \left (a+b x+c x^2\right )^{5/2}}{9 d^{10} \left (b^2-4 a c\right ) (b+2 c x)^9} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^10,x]

[Out]

(2*(a + b*x + c*x^2)^(5/2))/(9*(b^2 - 4*a*c)*d^10*(b + 2*c*x)^9) + (8*(a + b*x + c*x^2)^(5/2))/(63*(b^2 - 4*a*
c)^2*d^10*(b + 2*c*x)^7) + (16*(a + b*x + c*x^2)^(5/2))/(315*(b^2 - 4*a*c)^3*d^10*(b + 2*c*x)^5)

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{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{10}} \, dx &=\frac{2 \left (a+b x+c x^2\right )^{5/2}}{9 \left (b^2-4 a c\right ) d^{10} (b+2 c x)^9}+\frac{4 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^8} \, dx}{9 \left (b^2-4 a c\right ) d^2}\\ &=\frac{2 \left (a+b x+c x^2\right )^{5/2}}{9 \left (b^2-4 a c\right ) d^{10} (b+2 c x)^9}+\frac{8 \left (a+b x+c x^2\right )^{5/2}}{63 \left (b^2-4 a c\right )^2 d^{10} (b+2 c x)^7}+\frac{8 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^6} \, dx}{63 \left (b^2-4 a c\right )^2 d^4}\\ &=\frac{2 \left (a+b x+c x^2\right )^{5/2}}{9 \left (b^2-4 a c\right ) d^{10} (b+2 c x)^9}+\frac{8 \left (a+b x+c x^2\right )^{5/2}}{63 \left (b^2-4 a c\right )^2 d^{10} (b+2 c x)^7}+\frac{16 \left (a+b x+c x^2\right )^{5/2}}{315 \left (b^2-4 a c\right )^3 d^{10} (b+2 c x)^5}\\ \end{align*}

Mathematica [A]  time = 0.062807, size = 110, normalized size = 0.93 \[ \frac{2 (a+x (b+c x))^{5/2} \left (16 c^2 \left (35 a^2-20 a c x^2+8 c^2 x^4\right )+8 b^2 c \left (34 c x^2-45 a\right )+64 b c^2 x \left (4 c x^2-5 a\right )+144 b^3 c x+63 b^4\right )}{315 d^{10} \left (b^2-4 a c\right )^3 (b+2 c x)^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^10,x]

[Out]

(2*(a + x*(b + c*x))^(5/2)*(63*b^4 + 144*b^3*c*x + 64*b*c^2*x*(-5*a + 4*c*x^2) + 8*b^2*c*(-45*a + 34*c*x^2) +
16*c^2*(35*a^2 - 20*a*c*x^2 + 8*c^2*x^4)))/(315*(b^2 - 4*a*c)^3*d^10*(b + 2*c*x)^9)

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Maple [A]  time = 0.046, size = 133, normalized size = 1.1 \begin{align*} -{\frac{256\,{c}^{4}{x}^{4}+512\,b{c}^{3}{x}^{3}-640\,a{c}^{3}{x}^{2}+544\,{b}^{2}{c}^{2}{x}^{2}-640\,ab{c}^{2}x+288\,{b}^{3}cx+1120\,{a}^{2}{c}^{2}-720\,ac{b}^{2}+126\,{b}^{4}}{315\, \left ( 2\,cx+b \right ) ^{9}{d}^{10} \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{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^10,x)

[Out]

-2/315*(128*c^4*x^4+256*b*c^3*x^3-320*a*c^3*x^2+272*b^2*c^2*x^2-320*a*b*c^2*x+144*b^3*c*x+560*a^2*c^2-360*a*b^
2*c+63*b^4)*(c*x^2+b*x+a)^(5/2)/(2*c*x+b)^9/d^10/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*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)^(3/2)/(2*c*d*x+b*d)^10,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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)^(3/2)/(2*c*d*x+b*d)^10,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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)**(3/2)/(2*c*d*x+b*d)**10,x)

[Out]

Timed out

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Giac [B]  time = 3.09737, size = 1881, normalized size = 15.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^10,x, algorithm="giac")

[Out]

1/630*(3360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^12*c^(13/2) + 20160*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*b*c
^6 + 54180*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^10*b^2*c^(11/2) + 5040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^10*a
*c^(13/2) + 86100*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*b^3*c^5 + 25200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*
a*b*c^6 + 90216*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*b^4*c^(9/2) + 53172*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
8*a*b^2*c^(11/2) + 7056*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*a^2*c^(13/2) + 66024*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^7*b^5*c^4 + 61488*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*a*b^3*c^5 + 28224*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^7*a^2*b*c^6 + 35028*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*b^6*c^(7/2) + 41832*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^6*a*b^4*c^(9/2) + 47880*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*a^2*b^2*c^(11/2) + 2016*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^6*a^3*c^(13/2) + 13860*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*b^7*c^3 + 16128*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^5*a*b^5*c^4 + 44856*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^2*b^3*c^5 + 6048*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^5*a^3*b*c^6 + 4176*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*b^8*c^(5/2) + 2484*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^4*a*b^6*c^(7/2) + 25416*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^2*b^4*c^(9/2)
+ 6984*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^3*b^2*c^(11/2) + 576*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^4
*c^(13/2) + 960*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^9*c^2 - 576*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^
7*c^3 + 9000*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*b^5*c^4 + 3888*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^
3*b^3*c^5 + 1152*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^4*b*c^6 + 162*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b
^10*c^(3/2) - 360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^8*c^(5/2) + 2016*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))^2*a^2*b^6*c^(7/2) + 936*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^3*b^4*c^(9/2) + 1044*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))^2*a^4*b^2*c^(11/2) - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^5*c^(13/2) + 18*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))*b^11*c - 72*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^9*c^2 + 288*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*a^2*b^7*c^3 + 468*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^4*b^3*c^5 - 144*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))*a^5*b*c^6 + b^12*sqrt(c) - 6*a*b^10*c^(3/2) + 24*a^2*b^8*c^(5/2) - 32*a^3*b^6*c^(7/2) + 96*a^4*b^4*c^
(9/2) - 60*a^5*b^2*c^(11/2) + 16*a^6*c^(13/2))/((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c + 2*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))*b*sqrt(c) + b^2 - 2*a*c)^9*c^3*d^10)