3.1230 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^9} \, dx\)

Optimal. Leaf size=197 \[ \frac{5 \sqrt{a+b x+c x^2}}{4096 c^3 d^9 \left (b^2-4 a c\right ) (b+2 c x)^2}+\frac{5 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{8192 c^{7/2} d^9 \left (b^2-4 a c\right )^{3/2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{5 \sqrt{a+b x+c x^2}}{2048 c^3 d^9 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8} \]

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(2048*c^3*d^9*(b + 2*c*x)^4) + (5*Sqrt[a + b*x + c*x^2])/(4096*c^3*(b^2 - 4*a*c)*d^
9*(b + 2*c*x)^2) - (5*(a + b*x + c*x^2)^(3/2))/(384*c^2*d^9*(b + 2*c*x)^6) - (a + b*x + c*x^2)^(5/2)/(16*c*d^9
*(b + 2*c*x)^8) + (5*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(8192*c^(7/2)*(b^2 - 4*a*c)^
(3/2)*d^9)

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Rubi [A]  time = 0.149922, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {684, 693, 688, 205} \[ \frac{5 \sqrt{a+b x+c x^2}}{4096 c^3 d^9 \left (b^2-4 a c\right ) (b+2 c x)^2}+\frac{5 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{8192 c^{7/2} d^9 \left (b^2-4 a c\right )^{3/2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{5 \sqrt{a+b x+c x^2}}{2048 c^3 d^9 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^9,x]

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(2048*c^3*d^9*(b + 2*c*x)^4) + (5*Sqrt[a + b*x + c*x^2])/(4096*c^3*(b^2 - 4*a*c)*d^
9*(b + 2*c*x)^2) - (5*(a + b*x + c*x^2)^(3/2))/(384*c^2*d^9*(b + 2*c*x)^6) - (a + b*x + c*x^2)^(5/2)/(16*c*d^9
*(b + 2*c*x)^8) + (5*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(8192*c^(7/2)*(b^2 - 4*a*c)^
(3/2)*d^9)

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 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 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)^9} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8}+\frac{5 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^7} \, dx}{32 c d^2}\\ &=-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8}+\frac{5 \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^5} \, dx}{256 c^2 d^4}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{2048 c^3 d^9 (b+2 c x)^4}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8}+\frac{5 \int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx}{4096 c^3 d^6}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{2048 c^3 d^9 (b+2 c x)^4}+\frac{5 \sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^9 (b+2 c x)^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8}+\frac{5 \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{8192 c^3 \left (b^2-4 a c\right ) d^8}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{2048 c^3 d^9 (b+2 c x)^4}+\frac{5 \sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^9 (b+2 c x)^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8}+\frac{5 \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 )}{2048 c^2 \left (b^2-4 a c\right ) d^8}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{2048 c^3 d^9 (b+2 c x)^4}+\frac{5 \sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^9 (b+2 c x)^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{384 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{16 c d^9 (b+2 c x)^8}+\frac{5 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{8192 c^{7/2} \left (b^2-4 a c\right )^{3/2} d^9}\\ \end{align*}

Mathematica [C]  time = 0.0399195, size = 62, normalized size = 0.31 \[ \frac{2 (a+x (b+c x))^{7/2} \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{7 d^9 \left (b^2-4 a c\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^9,x]

[Out]

(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4
*a*c)^5*d^9)

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Maple [B]  time = 0.225, size = 1020, normalized size = 5.2 \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)^9,x)

[Out]

-1/1024/d^9/c^8/(4*a*c-b^2)/(x+1/2*b/c)^8*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+1/1536/d^9/c^6/(4*a*c-b^2)
^2/(x+1/2*b/c)^6*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+1/1536/d^9/c^4/(4*a*c-b^2)^3/(x+1/2*b/c)^4*((x+1/2*
b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+1/256/d^9/c^2/(4*a*c-b^2)^4/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c
)^(7/2)-1/256/d^9/c/(4*a*c-b^2)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)-5/768/d^9/c/(4*a*c-b^2)^4*((x+1/2*
b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*a+5/3072/d^9/c^2/(4*a*c-b^2)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2
-5/512/d^9/c/(4*a*c-b^2)^4*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a^2+5/1024/d^9/c^2/(4*a*c-b^2)^4*(4*(x+1/2*
b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a*b^2-5/8192/d^9/c^3/(4*a*c-b^2)^4*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4+5
/128/d^9/c/(4*a*c-b^2)^4/((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^3-15/512/d^9/c^2/(4*a*c-b^2)^4/((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+15/2048/d^9/c^3/
(4*a*c-b^2)^4/((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^4-5/8192/d^9/c^4/(4*a*c-b^2)^4/((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)^9,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)^(5/2)/(2*c*d*x+b*d)^9,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)**(5/2)/(2*c*d*x+b*d)**9,x)

[Out]

Timed out

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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)^9,x, algorithm="giac")

[Out]

Exception raised: TypeError