\(\int \frac {1}{(b d+2 c d x)^{3/2} (a+b x+c x^2)^{3/2}} \, dx\) [1387]

Optimal result

Integrand size = 28, antiderivative size = 274 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}+\frac {12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\left (b^2-4 a c\right )^{5/4} d^{3/2} \sqrt {a+b x+c x^2}}-\frac {12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\left (b^2-4 a c\right )^{5/4} d^{3/2} \sqrt {a+b x+c x^2}} \]

[Out]

-2/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2)-24*c*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^2/d/(2*c*d*x+b
*d)^(1/2)+12*EllipticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)
/(-4*a*c+b^2)^(5/4)/d^(3/2)/(c*x^2+b*x+a)^(1/2)-12*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)
*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/(-4*a*c+b^2)^(5/4)/d^(3/2)/(c*x^2+b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {701, 707, 705, 704, 313, 227, 1213, 435} \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}+\frac {12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}} \]

[In]

Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

-2/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2]) - (24*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2
*d*Sqrt[b*d + 2*c*d*x]) + (12*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x
]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/((b^2 - 4*a*c)^(5/4)*d^(3/2)*Sqrt[a + b*x + c*x^2]) - (12*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])/((b^
2 - 4*a*c)^(5/4)*d^(3/2)*Sqrt[a + b*x + c*x^2])

Rule 227

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]

Rule 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 701

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*c*(d + e*x)^(m + 1
)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Dist[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*
c))), 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] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 704

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(4/e)*Sqrt[-c/(b^2 - 4*
a*c)], Subst[Int[x^2/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 705

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 707

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 1213

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {(6 c) \int \frac {1}{(b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c} \\ & = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}+\frac {(6 c) \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^2 d^2} \\ & = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}+\frac {\left (6 c \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac {\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}{\left (b^2-4 a c\right )^2 d^2 \sqrt {a+b x+c x^2}} \\ & = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}+\frac {\left (12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{\left (b^2-4 a c\right )^2 d^3 \sqrt {a+b x+c x^2}} \\ & = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {\left (12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {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 )}{\left (b^2-4 a c\right )^{3/2} d^2 \sqrt {a+b x+c x^2}}+\frac {\left (12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{\left (b^2-4 a c\right )^{3/2} d^2 \sqrt {a+b x+c x^2}} \\ & = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {12 \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 )}{\left (b^2-4 a c\right )^{5/4} d^{3/2} \sqrt {a+b x+c x^2}}+\frac {\left (12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}}{\sqrt {1-\frac {x^2}{\sqrt {b^2-4 a c} d}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{\left (b^2-4 a c\right )^{3/2} d^2 \sqrt {a+b x+c x^2}} \\ & = -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}-\frac {24 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}+\frac {12 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\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 )}{\left (b^2-4 a c\right )^{5/4} d^{3/2} \sqrt {a+b x+c x^2}}-\frac {12 \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 )}{\left (b^2-4 a c\right )^{5/4} d^{3/2} \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.35 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {8 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)}} \]

[In]

Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 4.38 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {2 \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (12 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a c -3 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{2}-12 c^{2} x^{2}-12 b c x -8 a c -b^{2}\right )}{d^{2} \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right )^{2}}\)	\(339\)
elliptic	\(\text {Expression too large to display}\)	\(1089\)

[In]

int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(12*((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)*EllipticE(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*c-3*((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)
*EllipticE(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^2-12*c^2*x^2-12*b*c*
x-8*a*c-b^2)/d^2/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(4*a*c-b^2)^2

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (6 \, \sqrt {2} {\left (2 \, c^{2} x^{3} + 3 \, b c x^{2} + a b + {\left (b^{2} + 2 \, a c\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (12 \, c^{2} x^{2} + 12 \, b c x + b^{2} + 8 \, a c\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{2 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{2} x^{3} + 3 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{2} x^{2} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d^{2} x + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d^{2}} \]

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

-2*(6*sqrt(2)*(2*c^2*x^3 + 3*b*c*x^2 + a*b + (b^2 + 2*a*c)*x)*sqrt(c^2*d)*weierstrassZeta((b^2 - 4*a*c)/c^2, 0
, weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c)) + (12*c^2*x^2 + 12*b*c*x + b^2 + 8*a*c)*sqrt(2
*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2
+ 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/((d*(b + 2*c*x))**(3/2)*(a + b*x + c*x**2)**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((2*c*d*x + b*d)^(3/2)*(c*x^2 + b*x + a)^(3/2)), x)

Giac [F]

\[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((2*c*d*x + b*d)^(3/2)*(c*x^2 + b*x + a)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

[In]

int(1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(3/2)),x)

[Out]

int(1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(3/2)), x)