3.14.71 \(\int \frac {1}{(b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}} \, dx\) [1371]

Optimal. Leaf size=237 \[ \frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{c \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {a+b x+c x^2}}+\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{c \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {707, 705, 704, 313, 227, 1213, 435} \begin {gather*} \frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{c d^{3/2} \sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\text {ArcSin}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{c d^{3/2} \sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]) - (2*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])/(c*(b^2 - 4*a*c)^(1/4)*d^(3/2)*S
qrt[a + b*x + c*x^2]) + (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])/(c*(b^2 - 4*a*c)^(1/4)*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 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*} \int \frac {1}{(b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}} \, dx &=\frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}-\frac {\int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) d^2}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}-\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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 ) d^2 \sqrt {a+b x+c x^2}}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}-\frac {\left (2 \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 )}{c \left (b^2-4 a c\right ) d^3 \sqrt {a+b x+c x^2}}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}+\frac {\left (2 \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 )}{c \sqrt {b^2-4 a c} d^2 \sqrt {a+b x+c x^2}}-\frac {\left (2 \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 )}{c \sqrt {b^2-4 a c} d^2 \sqrt {a+b x+c x^2}}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}+\frac {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 )}{c \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {a+b x+c x^2}}-\frac {\left (2 \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 )}{c \sqrt {b^2-4 a c} d^2 \sqrt {a+b x+c x^2}}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x}}-\frac {2 \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 )}{c \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {a+b x+c x^2}}+\frac {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 )}{c \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 89, normalized size = 0.38 \begin {gather*} -\frac {2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c d \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.81, size = 336, normalized size = 1.42

method	result	size
default	\(\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (4 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticE \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a c -\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticE \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{2}-4 c^{2} x^{2}-4 b c x -4 a c \right )}{c \,d^{2} \left (2 c^{2} x^{3}+3 c \,x^{2} b +2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right )}\)	\(336\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {2 \left (2 c^{2} d \,x^{2}+2 x b c d +2 a c d \right )}{d^{2} \left (4 a c -b^{2}\right ) c \sqrt {\left (x +\frac {b}{2 c}\right ) \left (2 c^{2} d \,x^{2}+2 x b c d +2 a c d \right )}}+\frac {2 b \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \EllipticF \left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\left (4 a c -b^{2}\right ) d \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}+\frac {4 c \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \left (\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}\right ) \EllipticE \left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )-\frac {b \EllipticF \left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{2 c}\right )}{\left (4 a c -b^{2}\right ) d \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(1038\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(4*((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-((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)*Elli
pticE(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^2-4*c^2*x^2-4*b*c*x-4*a*c
)/c/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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.40, size = 125, normalized size = 0.53 \begin {gather*} \frac {2 \, {\left (\sqrt {2} \sqrt {c^{2} d} {\left (2 \, c x + b\right )} {\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 ) + 2 \, \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c\right )}}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} x + {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sqrt(2)*sqrt(c^2*d)*(2*c*x + b)*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)) + 2*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)*c)/(2*(b^2*c^2 - 4*a*c^3)*d^2*x + (b^3*
c - 4*a*b*c^2)*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}} \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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