### 3.1389 $$\int \frac{(b d+2 c d x)^{15/2}}{(a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=247 $\frac{520 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{7 \sqrt{a+b x+c x^2}}+\frac{1040}{7} c^2 d^7 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}+\frac{624}{7} c^2 d^5 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

[Out]

(-2*d*(b*d + 2*c*d*x)^(13/2))/(3*(a + b*x + c*x^2)^(3/2)) - (52*c*d^3*(b*d + 2*c*d*x)^(9/2))/(3*Sqrt[a + b*x +
c*x^2]) + (1040*c^2*(b^2 - 4*a*c)*d^7*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/7 + (624*c^2*d^5*(b*d + 2*c*
d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (520*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 -
4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(7*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.202617, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.179, Rules used = {686, 692, 691, 689, 221} $\frac{1040}{7} c^2 d^7 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}+\frac{520 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{7 \sqrt{a+b x+c x^2}}+\frac{624}{7} c^2 d^5 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(13/2))/(3*(a + b*x + c*x^2)^(3/2)) - (52*c*d^3*(b*d + 2*c*d*x)^(9/2))/(3*Sqrt[a + b*x +
c*x^2]) + (1040*c^2*(b^2 - 4*a*c)*d^7*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/7 + (624*c^2*d^5*(b*d + 2*c*
d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (520*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 -
4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(7*Sqrt[a + b*x + c*x^2])

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 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] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rule 691

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]/Sqrt[a + b*x + c*x^2], Int[(d + e*x)^m/Sqrt[-((a*c)/(b^2 - 4*a*c)) - (b*c*x)/(b^2 - 4*a
*c) - (c^2*x^2)/(b^2 - 4*a*c)], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && EqQ[m^2, 1/4]

Rule 689

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 -
4*a*c))])/e, Subst[Int[1/Sqrt[Simp[1 - (b^2*x^4)/(d^2*(b^2 - 4*a*c)), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac{1}{3} \left (26 c d^2\right ) \int \frac{(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}+\left (156 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{7/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}+\frac{624}{7} c^2 d^5 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{1}{7} \left (780 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1040}{7} c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{624}{7} c^2 d^5 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{1}{7} \left (260 c^2 \left (b^2-4 a c\right )^2 d^8\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1040}{7} c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{624}{7} c^2 d^5 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{\left (260 c^2 \left (b^2-4 a c\right )^2 d^8 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{-\frac{a c}{b^2-4 a c}-\frac{b c x}{b^2-4 a c}-\frac{c^2 x^2}{b^2-4 a c}}} \, dx}{7 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1040}{7} c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{624}{7} c^2 d^5 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{\left (520 c \left (b^2-4 a c\right )^2 d^7 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{7 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1040}{7} c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{624}{7} c^2 d^5 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{520 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{7 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.324416, size = 252, normalized size = 1.02 $\frac{2 d^7 \sqrt{d (b+2 c x)} \left (16 b^2 c^2 \left (156 a^2+117 a c x^2+116 c^2 x^4\right )+32 b c^3 x \left (-273 a^2-104 a c x^2+36 c^2 x^4\right )-32 c^3 \left (273 a^2 c x^2+195 a^3+52 a c^2 x^4-12 c^3 x^6\right )+16 b^3 c^2 x \left (221 a+112 c x^2\right )+780 c \left (b^2-4 a c\right )^2 (a+x (b+c x)) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+2 b^4 c \left (219 c x^2-91 a\right )-266 b^5 c x-7 b^6\right )}{21 (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(2*d^7*Sqrt[d*(b + 2*c*x)]*(-7*b^6 - 266*b^5*c*x + 16*b^3*c^2*x*(221*a + 112*c*x^2) + 2*b^4*c*(-91*a + 219*c*x
^2) + 32*b*c^3*x*(-273*a^2 - 104*a*c*x^2 + 36*c^2*x^4) + 16*b^2*c^2*(156*a^2 + 117*a*c*x^2 + 116*c^2*x^4) - 32
*c^3*(195*a^3 + 273*a^2*c*x^2 + 52*a*c^2*x^4 - 12*c^3*x^6) + 780*c*(b^2 - 4*a*c)^2*(a + x*(b + c*x))*Sqrt[(c*(
a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(21*(a + x*(
b + c*x))^(3/2))

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Maple [B]  time = 0.313, size = 1473, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/21*(6240*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^2*a^2*c^4*
(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/
2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)+390*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-
4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^2*b^4*c^2*(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+
b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/
2)+390*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x*b^5*c*(-4*a*c+
b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-
2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)-8320*x^4*a*b*c^5+416*x^3*a*b^2*c^4-26208*x^2*a^2*b*c^4+894
4*x^2*a*b^3*c^3-3744*x*a^2*b^2*c^3+3172*x*a*b^4*c^2+6240*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b
^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x*a^2*b*c^3*(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1
/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)-3120
*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x*a*b^3*c^2*(-4*a*c+b^
2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*
c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)-3120*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2
)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^2*a*b^2*c^3*(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1
/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)-7*b^
7-6240*a^3*b*c^3+2496*a^2*b^3*c^2-182*a*b^5*c+390*(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2
)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*
EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a*b^4*c+6240*(-4*a*c+b^
2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*
c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/
2))^(1/2)*2^(1/2),2^(1/2))*a^3*c^3+2688*x^6*b*c^6-3120*(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*
c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(
1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a^2*b^2*c^2-3328*x
^5*a*c^6+4864*x^5*b^2*c^5+5440*x^4*b^3*c^4-17472*x^3*a^2*c^5+2668*x^3*b^4*c^3-94*x^2*b^5*c^2-12480*x*a^3*c^4-2
80*x*b^6*c+768*x^7*c^7)*d^7*(d*(2*c*x+b))^(1/2)/(2*c*x+b)/(c*x^2+b*x+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{15}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(15/2)/(c*x^2 + b*x + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (128 \, c^{7} d^{7} x^{7} + 448 \, b c^{6} d^{7} x^{6} + 672 \, b^{2} c^{5} d^{7} x^{5} + 560 \, b^{3} c^{4} d^{7} x^{4} + 280 \, b^{4} c^{3} d^{7} x^{3} + 84 \, b^{5} c^{2} d^{7} x^{2} + 14 \, b^{6} c d^{7} x + b^{7} d^{7}\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((128*c^7*d^7*x^7 + 448*b*c^6*d^7*x^6 + 672*b^2*c^5*d^7*x^5 + 560*b^3*c^4*d^7*x^4 + 280*b^4*c^3*d^7*x^
3 + 84*b^5*c^2*d^7*x^2 + 14*b^6*c*d^7*x + b^7*d^7)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(c^3*x^6 + 3*b*c^
2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)

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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((2*c*d*x+b*d)**(15/2)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{15}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(15/2)/(c*x^2 + b*x + a)^(5/2), x)