3.1328 \(\int (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2} \, dx\)

Optimal. Leaf size=180 \[ -\frac {d^{3/2} \left (b^2-4 a c\right )^{9/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{21 c^2 \sqrt {a+b x+c x^2}}-\frac {2 d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{21 c}+\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d} \]

[Out]

1/7*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(1/2)/c/d-2/21*(-4*a*c+b^2)*d*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c-
1/21*(-4*a*c+b^2)^(9/4)*d^(3/2)*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^2/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {d^{3/2} \left (b^2-4 a c\right )^{9/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{21 c^2 \sqrt {a+b x+c x^2}}-\frac {2 d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{21 c}+\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(21*c) + ((b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x +
c*x^2])/(7*c*d) - ((b^2 - 4*a*c)^(9/4)*d^(3/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[S
qrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(21*c^2*Sqrt[a + b*x + c*x^2])

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]

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 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 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 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])

Rubi steps

\begin {align*} \int (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2} \, dx &=\frac {(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{7 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx}{14 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}}{21 c}+\frac {(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{7 c d}-\frac {\left (\left (b^2-4 a c\right )^2 d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{42 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}}{21 c}+\frac {(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{7 c d}-\frac {\left (\left (b^2-4 a c\right )^2 d^2 \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}{42 c \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}}{21 c}+\frac {(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{7 c d}-\frac {\left (\left (b^2-4 a c\right )^2 d \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 )}{21 c^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}}{21 c}+\frac {(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{7 c d}-\frac {\left (b^2-4 a c\right )^{9/4} d^{3/2} \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 )}{21 c^2 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 110, normalized size = 0.61 \[ \frac {1}{14} d \sqrt {a+x (b+c x)} \sqrt {d (b+2 c x)} \left (\frac {\left (b^2-4 a c\right ) \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}}+8 (a+x (b+c x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*(8*(a + x*(b + c*x)) + ((b^2 - 4*a*c)*Hypergeometric2F1[-1/2, 1/4
, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/14

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} \sqrt {c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} \sqrt {c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 564, normalized size = 3.13 \[ -\frac {\sqrt {\left (2 c x +b \right ) d}\, \sqrt {c \,x^{2}+b x +a}\, \left (-48 c^{5} x^{5}-120 b \,c^{4} x^{4}-80 a \,c^{4} x^{3}-100 b^{2} c^{3} x^{3}-120 a b \,c^{3} x^{2}-30 b^{3} c^{2} x^{2}-32 a^{2} c^{3} x -44 a \,b^{2} c^{2} x -2 b^{4} c x -16 a^{2} b \,c^{2}+16 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, a^{2} c^{2} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-2 a \,b^{3} c -8 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, a \,b^{2} c \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )+\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b^{4} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right ) d}{42 \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/42*((2*c*x+b)*d)^(1/2)*(c*x^2+b*x+a)^(1/2)*d*(-48*x^5*c^5+16*((2*c*x+b+(-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)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*Ellip
ticF(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*a^2*c^2-8
*((2*c*x+b+(-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)*((-2*c*x-b+(-4*
a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2
)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*a*b^2*c+((2*c*x+b+(-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)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b
+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*b^4-120*x^4*b*c^4-80*x^3*a*
c^4-100*x^3*b^2*c^3-120*x^2*a*b*c^3-30*x^2*b^3*c^2-32*a^2*c^3*x-44*x*a*b^2*c^2-2*x*b^4*c-16*a^2*b*c^2-2*a*b^3*
c)/c^2/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} \sqrt {c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((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 = -0.01 \[ \int {\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\sqrt {c\,x^2+b\,x+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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