3.1349 \(\int (b d+2 c d x)^{7/2} (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=321 \[ -\frac{5 d^{7/2} \left (b^2-4 a c\right )^{21/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{17556 c^4 \sqrt{a+b x+c x^2}}-\frac{5 d^3 \left (b^2-4 a c\right )^4 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{8778 c^3}-\frac{d \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{9/2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{9/2}}{114 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{9/2}}{19 c d} \]

[Out]

(-5*(b^2 - 4*a*c)^4*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(8778*c^3) - ((b^2 - 4*a*c)^3*d*(b*d + 2*c*
d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/(2926*c^3) + ((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2])/(
836*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(9/2)*(a + b*x + c*x^2)^(3/2))/(114*c^2*d) + ((b*d + 2*c*d*x)^(9/2
)*(a + b*x + c*x^2)^(5/2))/(19*c*d) - (5*(b^2 - 4*a*c)^(21/4)*d^(7/2)*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])/(17556*c^4*Sqrt[a + b*x + c*x^2
])

________________________________________________________________________________________

Rubi [A]  time = 0.295517, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {685, 692, 691, 689, 221} \[ -\frac{5 d^3 \left (b^2-4 a c\right )^4 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{8778 c^3}-\frac{5 d^{7/2} \left (b^2-4 a c\right )^{21/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{17556 c^4 \sqrt{a+b x+c x^2}}-\frac{d \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{9/2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{9/2}}{114 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{9/2}}{19 c d} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^(5/2),x]

[Out]

(-5*(b^2 - 4*a*c)^4*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(8778*c^3) - ((b^2 - 4*a*c)^3*d*(b*d + 2*c*
d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/(2926*c^3) + ((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2])/(
836*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(9/2)*(a + b*x + c*x^2)^(3/2))/(114*c^2*d) + ((b*d + 2*c*d*x)^(9/2
)*(a + b*x + c*x^2)^(5/2))/(19*c*d) - (5*(b^2 - 4*a*c)^(21/4)*d^(7/2)*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])/(17556*c^4*Sqrt[a + b*x + c*x^2
])

Rule 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && 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 (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2} \, dx}{38 c}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}+\frac{\left (b^2-4 a c\right )^2 \int (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2} \, dx}{76 c^2}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{\left (b^2-4 a c\right )^3 \int \frac{(b d+2 c d x)^{7/2}}{\sqrt{a+b x+c x^2}} \, dx}{1672 c^3}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{\left (5 \left (b^2-4 a c\right )^4 d^2\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx}{11704 c^3}\\ &=-\frac{5 \left (b^2-4 a c\right )^4 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{8778 c^3}-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{\left (5 \left (b^2-4 a c\right )^5 d^4\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{35112 c^3}\\ &=-\frac{5 \left (b^2-4 a c\right )^4 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{8778 c^3}-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{\left (5 \left (b^2-4 a c\right )^5 d^4 \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}{35112 c^3 \sqrt{a+b x+c x^2}}\\ &=-\frac{5 \left (b^2-4 a c\right )^4 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{8778 c^3}-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{\left (5 \left (b^2-4 a c\right )^5 d^3 \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 )}{17556 c^4 \sqrt{a+b x+c x^2}}\\ &=-\frac{5 \left (b^2-4 a c\right )^4 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{8778 c^3}-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{2926 c^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{836 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{3/2}}{114 c^2 d}+\frac{(b d+2 c d x)^{9/2} \left (a+b x+c x^2\right )^{5/2}}{19 c d}-\frac{5 \left (b^2-4 a c\right )^{21/4} d^{7/2} \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 )}{17556 c^4 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.410117, size = 161, normalized size = 0.5 \[ \frac{4 \sqrt{a+x (b+c x)} (d (b+2 c x))^{7/2} \left (3 (b+2 c x)^2 (a+x (b+c x))^3-2 c \left (a-\frac{b^2}{4 c}\right ) \left (\frac{\left (b^2-4 a c\right )^3 \, _2F_1\left (-\frac{5}{2},\frac{1}{4};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}+2 (a+x (b+c x))^3\right )\right )}{57 (b+2 c x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^(5/2),x]

[Out]

(4*(d*(b + 2*c*x))^(7/2)*Sqrt[a + x*(b + c*x)]*(3*(b + 2*c*x)^2*(a + x*(b + c*x))^3 - 2*(a - b^2/(4*c))*c*(2*(
a + x*(b + c*x))^3 + ((b^2 - 4*a*c)^3*Hypergeometric2F1[-5/2, 1/4, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64*c^3*
Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]))))/(57*(b + 2*c*x)^3)

________________________________________________________________________________________

Maple [B]  time = 0.26, size = 1344, normalized size = 4.2 \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)^(7/2)*(c*x^2+b*x+a)^(5/2),x)

[Out]

1/35112*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^3*(-5120*a^5*b*c^5+5888*a^4*b^3*c^4+1256*a^3*b^5*c^3-180*a^2
*b^7*c^2+10*a*b^9*c+325248*x^10*b*c^10+5120*(-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)*Ellipt
icF(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^5*c^5-6400*(-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^4*b^2*c^4+59136*x^11*c^11-800*(-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^6*c^2+100*(-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^8*c+216832*x^9*a*c^10+758912*x^9*b^2*c^9+975744*x^8*b^3*c^8+277760*x^7*a^
2*c^9+749168*x^7*b^4*c^7+345352*x^6*b^5*c^6+126208*x^5*a^3*c^8+89168*x^5*b^6*c^5+10036*x^4*b^7*c^4-4096*x^3*a^
4*c^7-4*x^3*b^8*c^3+10*x^2*b^9*c^2-10240*x*a^5*c^6+10*x*b^10*c+3200*(-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*b
^4*c^3+1812608*x^7*a*b^2*c^8+305336*x^4*a*b^5*c^5+319616*x^3*a^3*b^2*c^6+1790656*x^6*a*b^3*c^7+1363584*x^5*a^2
*b^2*c^7+1002096*x^5*a*b^4*c^6+9728*x*a^4*b^2*c^5+972160*x^6*a^2*b*c^8-192*x^2*a*b^7*c^3+315520*x^4*a^3*b*c^7+
978560*x^4*a^2*b^3*c^6+1248*x*a^2*b^6*c^3+36112*x*a^3*b^4*c^4-6144*x^2*a^4*b*c^6+163904*x^2*a^3*b^3*c^5-180*x*
a*b^8*c^2+369424*x^3*a^2*b^4*c^5+40208*x^3*a*b^6*c^4+61656*x^2*a^2*b^5*c^4+975744*x^8*a*b*c^9-5*(-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))*b^10)/c^4/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, c d x + b d\right )}^{\frac{7}{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)^(7/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(7/2)*(c*x^2 + b*x + a)^(5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (8 \, c^{5} d^{3} x^{7} + 28 \, b c^{4} d^{3} x^{6} + 2 \,{\left (19 \, b^{2} c^{3} + 8 \, a c^{4}\right )} d^{3} x^{5} + a^{2} b^{3} d^{3} + 5 \,{\left (5 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} d^{3} x^{4} + 4 \,{\left (2 \, b^{4} c + 9 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} d^{3} x^{3} +{\left (b^{5} + 14 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} d^{3} x^{2} + 2 \,{\left (a b^{4} + 3 \, a^{2} b^{2} c\right )} d^{3} x\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(7/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((8*c^5*d^3*x^7 + 28*b*c^4*d^3*x^6 + 2*(19*b^2*c^3 + 8*a*c^4)*d^3*x^5 + a^2*b^3*d^3 + 5*(5*b^3*c^2 + 8
*a*b*c^3)*d^3*x^4 + 4*(2*b^4*c + 9*a*b^2*c^2 + 2*a^2*c^3)*d^3*x^3 + (b^5 + 14*a*b^3*c + 12*a^2*b*c^2)*d^3*x^2
+ 2*(a*b^4 + 3*a^2*b^2*c)*d^3*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a), x)

________________________________________________________________________________________

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)**(7/2)*(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, c d x + b d\right )}^{\frac{7}{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)^(7/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(7/2)*(c*x^2 + b*x + a)^(5/2), x)