### 3.1215 $$\int \frac{(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^9} \, dx$$

Optimal. Leaf size=207 $\frac{3 \sqrt{a+b x+c x^2}}{1024 c^2 d^9 \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{\sqrt{a+b x+c x^2}}{512 c^2 d^9 \left (b^2-4 a c\right ) (b+2 c x)^4}+\frac{3 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{2048 c^{5/2} d^9 \left (b^2-4 a c\right )^{5/2}}-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}$

[Out]

-Sqrt[a + b*x + c*x^2]/(128*c^2*d^9*(b + 2*c*x)^6) + Sqrt[a + b*x + c*x^2]/(512*c^2*(b^2 - 4*a*c)*d^9*(b + 2*c
*x)^4) + (3*Sqrt[a + b*x + c*x^2])/(1024*c^2*(b^2 - 4*a*c)^2*d^9*(b + 2*c*x)^2) - (a + b*x + c*x^2)^(3/2)/(16*
c*d^9*(b + 2*c*x)^8) + (3*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(2048*c^(5/2)*(b^2 - 4*
a*c)^(5/2)*d^9)

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Rubi [A]  time = 0.146302, antiderivative size = 207, 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} $\frac{3 \sqrt{a+b x+c x^2}}{1024 c^2 d^9 \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{\sqrt{a+b x+c x^2}}{512 c^2 d^9 \left (b^2-4 a c\right ) (b+2 c x)^4}+\frac{3 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{2048 c^{5/2} d^9 \left (b^2-4 a c\right )^{5/2}}-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^9,x]

[Out]

-Sqrt[a + b*x + c*x^2]/(128*c^2*d^9*(b + 2*c*x)^6) + Sqrt[a + b*x + c*x^2]/(512*c^2*(b^2 - 4*a*c)*d^9*(b + 2*c
*x)^4) + (3*Sqrt[a + b*x + c*x^2])/(1024*c^2*(b^2 - 4*a*c)^2*d^9*(b + 2*c*x)^2) - (a + b*x + c*x^2)^(3/2)/(16*
c*d^9*(b + 2*c*x)^8) + (3*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(2048*c^(5/2)*(b^2 - 4*
a*c)^(5/2)*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 )^{3/2}}{(b d+2 c d x)^9} \, dx &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}+\frac{3 \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^7} \, dx}{32 c d^2}\\ &=-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}+\frac{\int \frac{1}{(b d+2 c d x)^5 \sqrt{a+b x+c x^2}} \, dx}{256 c^2 d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{512 c^2 \left (b^2-4 a c\right ) d^9 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}+\frac{3 \int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx}{1024 c^2 \left (b^2-4 a c\right ) d^6}\\ &=-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{512 c^2 \left (b^2-4 a c\right ) d^9 (b+2 c x)^4}+\frac{3 \sqrt{a+b x+c x^2}}{1024 c^2 \left (b^2-4 a c\right )^2 d^9 (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}+\frac{3 \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{2048 c^2 \left (b^2-4 a c\right )^2 d^8}\\ &=-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{512 c^2 \left (b^2-4 a c\right ) d^9 (b+2 c x)^4}+\frac{3 \sqrt{a+b x+c x^2}}{1024 c^2 \left (b^2-4 a c\right )^2 d^9 (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}+\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 )}{512 c \left (b^2-4 a c\right )^2 d^8}\\ &=-\frac{\sqrt{a+b x+c x^2}}{128 c^2 d^9 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{512 c^2 \left (b^2-4 a c\right ) d^9 (b+2 c x)^4}+\frac{3 \sqrt{a+b x+c x^2}}{1024 c^2 \left (b^2-4 a c\right )^2 d^9 (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{16 c d^9 (b+2 c x)^8}+\frac{3 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{2048 c^{5/2} \left (b^2-4 a c\right )^{5/2} d^9}\\ \end{align*}

Mathematica [C]  time = 0.0297943, size = 62, normalized size = 0.3 $\frac{2 (a+x (b+c x))^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{5 d^9 \left (b^2-4 a c\right )^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^9,x]

[Out]

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

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Maple [B]  time = 0.238, size = 742, normalized size = 3.6 \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)^(3/2)/(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)^(5/2)+1/512/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)^(5/2)-1/512/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)^(5/2)-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)^
(5/2)+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)^(3/2)+3/512/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-3/2048/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)*b^2-3/128
/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^2+3/256/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*b^2-3/2048/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))*b^4

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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)^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)^(3/2)/(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)**(3/2)/(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)^(3/2)/(2*c*d*x+b*d)^9,x, algorithm="giac")

[Out]

Exception raised: TypeError