3.1338 \(\int (b d+2 c d x)^{3/2} (a+b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=227 \[ \frac {d^{3/2} \left (b^2-4 a c\right )^{13/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{154 c^3 \sqrt {a+b x+c x^2}}+\frac {d \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{77 c^2}-\frac {3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{154 c^2 d}+\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d} \]

[Out]

1/11*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(3/2)/c/d-3/154*(-4*a*c+b^2)*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(1/2)/c^
2/d+1/77*(-4*a*c+b^2)^2*d*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c^2+1/154*(-4*a*c+b^2)^(13/4)*d^(3/2)*Ellipt
icF((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^3/(c*x^2+b*x+a)^
(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {d^{3/2} \left (b^2-4 a c\right )^{13/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{154 c^3 \sqrt {a+b x+c x^2}}+\frac {d \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{77 c^2}-\frac {3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{154 c^2 d}+\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(77*c^2) - (3*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2
)*Sqrt[a + b*x + c*x^2])/(154*c^2*d) + ((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(3/2))/(11*c*d) + ((b^2 - 4*a*
c)^(13/4)*d^(3/2)*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])/(154*c^3*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} \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2} \, dx}{22 c}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{154 c^2 d}+\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx}{308 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{77 c^2}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{154 c^2 d}+\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac {\left (\left (b^2-4 a c\right )^3 d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{308 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{77 c^2}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{154 c^2 d}+\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac {\left (\left (b^2-4 a c\right )^3 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}{308 c^2 \sqrt {a+b x+c x^2}}\\ &=\frac {\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{77 c^2}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{154 c^2 d}+\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac {\left (\left (b^2-4 a c\right )^3 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 )}{154 c^3 \sqrt {a+b x+c x^2}}\\ &=\frac {\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{77 c^2}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{154 c^2 d}+\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac {\left (b^2-4 a c\right )^{13/4} 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 )}{154 c^3 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 117, normalized size = 0.52 \[ \frac {2}{11} d \sqrt {a+x (b+c x)} \sqrt {d (b+2 c x)} \left (2 (a+x (b+c x))^2-\frac {\left (b^2-4 a c\right )^2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{16 c^2 \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*(2*(a + x*(b + c*x))^2 - ((b^2 - 4*a*c)^2*Hypergeometric2F1[-3/
2, 1/4, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(16*c^2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/11

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (2 \, c^{2} d x^{3} + 3 \, b c d x^{2} + a b d + {\left (b^{2} + 2 \, a c\right )} d x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.16, size = 796, normalized size = 3.51 \[ -\frac {\sqrt {\left (2 c x +b \right ) d}\, \sqrt {c \,x^{2}+b x +a}\, \left (-224 c^{7} x^{7}-784 b \,c^{6} x^{6}-640 a \,c^{6} x^{5}-1016 b^{2} c^{5} x^{5}-1600 a b \,c^{5} x^{4}-580 b^{3} c^{4} x^{4}-544 a^{2} c^{5} x^{3}-1328 a \,b^{2} c^{4} x^{3}-124 b^{4} c^{3} x^{3}-816 a^{2} b \,c^{4} x^{2}-392 a \,b^{3} c^{3} x^{2}+2 b^{5} c^{2} x^{2}-128 a^{3} c^{4} x -312 a^{2} b^{2} c^{3} x -20 a \,b^{4} c^{2} x +2 b^{6} c x -64 a^{3} b \,c^{3}+64 \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^{3} c^{3} \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 )-20 a^{2} b^{3} c^{2}-48 \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} b^{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^{5} c +12 \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^{4} 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^{6} \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}{308 \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c^{3}} \]

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)^(3/2),x)

[Out]

-1/308*((2*c*x+b)*d)^(1/2)*(c*x^2+b*x+a)^(1/2)*d*(-224*c^7*x^7-784*b*c^6*x^6-640*a*c^6*x^5-1016*b^2*c^5*x^5+64
*((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^3*c^3-48*((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^2*b^2*c^2+12*((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^4*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^6-1600*a*b*c^5*x^4-580*b^3*c^4*x^4-
544*a^2*c^5*x^3-1328*a*b^2*c^4*x^3-124*b^4*c^3*x^3-816*a^2*b*c^4*x^2-392*a*b^3*c^3*x^2+2*b^5*c^2*x^2-128*a^3*c
^4*x-312*a^2*b^2*c^3*x-20*a*b^4*c^2*x+2*b^6*c*x-64*a^3*b*c^3-20*a^2*b^3*c^2+2*a*b^5*c)/c^3/(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 \[ \int {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \]

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)^(3/2),x)

[Out]

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

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

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)**(3/2),x)

[Out]

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

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