3.1232 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{11}} \, dx\)

Optimal. Leaf size=239 \[ \frac{3 \sqrt{a+b x+c x^2}}{8192 c^3 d^{11} \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{\sqrt{a+b x+c x^2}}{4096 c^3 d^{11} \left (b^2-4 a c\right ) (b+2 c x)^4}+\frac{3 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{16384 c^{7/2} d^{11} \left (b^2-4 a c\right )^{5/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}} \]

[Out]

-Sqrt[a + b*x + c*x^2]/(1024*c^3*d^11*(b + 2*c*x)^6) + Sqrt[a + b*x + c*x^2]/(4096*c^3*(b^2 - 4*a*c)*d^11*(b +
2*c*x)^4) + (3*Sqrt[a + b*x + c*x^2])/(8192*c^3*(b^2 - 4*a*c)^2*d^11*(b + 2*c*x)^2) - (a + b*x + c*x^2)^(3/2)
/(128*c^2*d^11*(b + 2*c*x)^8) - (a + b*x + c*x^2)^(5/2)/(20*c*d^11*(b + 2*c*x)^10) + (3*ArcTan[(2*Sqrt[c]*Sqrt
[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(16384*c^(7/2)*(b^2 - 4*a*c)^(5/2)*d^11)

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Rubi [A]  time = 0.184909, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {684, 693, 688, 205} \[ \frac{3 \sqrt{a+b x+c x^2}}{8192 c^3 d^{11} \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{\sqrt{a+b x+c x^2}}{4096 c^3 d^{11} \left (b^2-4 a c\right ) (b+2 c x)^4}+\frac{3 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{16384 c^{7/2} d^{11} \left (b^2-4 a c\right )^{5/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^11,x]

[Out]

-Sqrt[a + b*x + c*x^2]/(1024*c^3*d^11*(b + 2*c*x)^6) + Sqrt[a + b*x + c*x^2]/(4096*c^3*(b^2 - 4*a*c)*d^11*(b +
2*c*x)^4) + (3*Sqrt[a + b*x + c*x^2])/(8192*c^3*(b^2 - 4*a*c)^2*d^11*(b + 2*c*x)^2) - (a + b*x + c*x^2)^(3/2)
/(128*c^2*d^11*(b + 2*c*x)^8) - (a + b*x + c*x^2)^(5/2)/(20*c*d^11*(b + 2*c*x)^10) + (3*ArcTan[(2*Sqrt[c]*Sqrt
[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(16384*c^(7/2)*(b^2 - 4*a*c)^(5/2)*d^11)

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)^{11}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{\int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^9} \, dx}{8 c d^2}\\ &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{3 \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^7} \, dx}{256 c^2 d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{\int \frac{1}{(b d+2 c d x)^5 \sqrt{a+b x+c x^2}} \, dx}{2048 c^3 d^6}\\ &=-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^{11} (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{3 \int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx}{8192 c^3 \left (b^2-4 a c\right ) d^8}\\ &=-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^{11} (b+2 c x)^4}+\frac{3 \sqrt{a+b x+c x^2}}{8192 c^3 \left (b^2-4 a c\right )^2 d^{11} (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{3 \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{16384 c^3 \left (b^2-4 a c\right )^2 d^{10}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^{11} (b+2 c x)^4}+\frac{3 \sqrt{a+b x+c x^2}}{8192 c^3 \left (b^2-4 a c\right )^2 d^{11} (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{3 \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 )}{4096 c^2 \left (b^2-4 a c\right )^2 d^{10}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{1024 c^3 d^{11} (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{4096 c^3 \left (b^2-4 a c\right ) d^{11} (b+2 c x)^4}+\frac{3 \sqrt{a+b x+c x^2}}{8192 c^3 \left (b^2-4 a c\right )^2 d^{11} (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{128 c^2 d^{11} (b+2 c x)^8}-\frac{\left (a+b x+c x^2\right )^{5/2}}{20 c d^{11} (b+2 c x)^{10}}+\frac{3 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{16384 c^{7/2} \left (b^2-4 a c\right )^{5/2} d^{11}}\\ \end{align*}

Mathematica [C]  time = 0.0372217, size = 62, normalized size = 0.26 \[ \frac{2 (a+x (b+c x))^{7/2} \, _2F_1\left (\frac{7}{2},6;\frac{9}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{7 d^{11} \left (b^2-4 a c\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^11,x]

[Out]

(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[7/2, 6, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4
*a*c)^6*d^11)

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Maple [B]  time = 0.262, size = 1080, normalized size = 4.5 \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)^11,x)

[Out]

-1/5120/d^11/c^10/(4*a*c-b^2)/(x+1/2*b/c)^10*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+3/10240/d^11/c^8/(4*a*c
-b^2)^2/(x+1/2*b/c)^8*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)-1/5120/d^11/c^6/(4*a*c-b^2)^3/(x+1/2*b/c)^6*((
x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)-1/5120/d^11/c^4/(4*a*c-b^2)^4/(x+1/2*b/c)^4*((x+1/2*b/c)^2*c+1/4*(4*a*
c-b^2)/c)^(7/2)-3/2560/d^11/c^2/(4*a*c-b^2)^5/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+3/2560/d
^11/c/(4*a*c-b^2)^5*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+1/512/d^11/c/(4*a*c-b^2)^5*((x+1/2*b/c)^2*c+1/4*
(4*a*c-b^2)/c)^(3/2)*a-1/2048/d^11/c^2/(4*a*c-b^2)^5*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2+3/1024/d^11
/c/(4*a*c-b^2)^5*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a^2-3/2048/d^11/c^2/(4*a*c-b^2)^5*(4*(x+1/2*b/c)^2*c+
(4*a*c-b^2)/c)^(1/2)*a*b^2+3/16384/d^11/c^3/(4*a*c-b^2)^5*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4-3/256/d^
11/c/(4*a*c-b^2)^5/((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+9/1024/d^11/c^2/(4*a*c-b^2)^5/((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^2-9/4096/d^11/c^3/(4*a*
c-b^2)^5/((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+3/16384/d^11/c^4/(4*a*c-b^2)^5/((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)^11,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)^11,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)**11,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)^11,x, algorithm="giac")

[Out]

Exception raised: TypeError