3.14.63 \(\int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx\) [1363]

Optimal. Leaf size=174 \[ \frac {20}{21} \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {10 \left (b^2-4 a c\right )^{9/4} d^{7/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{21 c \sqrt {a+b x+c x^2}} \]

[Out]

4/7*d*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(1/2)+20/21*(-4*a*c+b^2)*d^3*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)+1
0/21*(-4*a*c+b^2)^(9/4)*d^(7/2)*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/(c*x^2+b*x+a)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {706, 705, 703, 227} \begin {gather*} \frac {10 d^{7/2} \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 .\text {ArcSin}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{21 c \sqrt {a+b x+c x^2}}+\frac {20}{21} d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}+\frac {4}{7} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(20*(b^2 - 4*a*c)*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/21 + (4*d*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x
+ c*x^2])/7 + (10*(b^2 - 4*a*c)^(9/4)*d^(7/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sq
rt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(21*c*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 703

Int[1/(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[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 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 706

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 \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx &=\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {1}{7} \left (5 \left (b^2-4 a c\right ) d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx\\ &=\frac {20}{21} \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {1}{21} \left (5 \left (b^2-4 a c\right )^2 d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx\\ &=\frac {20}{21} \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {\left (5 \left (b^2-4 a c\right )^2 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}{21 \sqrt {a+b x+c x^2}}\\ &=\frac {20}{21} \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {\left (10 \left (b^2-4 a c\right )^2 d^3 \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 )}{21 c \sqrt {a+b x+c x^2}}\\ &=\frac {20}{21} \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {10 \left (b^2-4 a c\right )^{9/4} 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 )}{21 c \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 = 10.14, size = 166, normalized size = 0.95 \begin {gather*} \frac {2 d^3 \sqrt {d (b+2 c x)} \left (8 c \left (-5 a^2 c+2 a \left (b^2-b c x-c^2 x^2\right )+x \left (2 b^3+5 b^2 c x+6 b c^2 x^2+3 c^3 x^3\right )\right )+5 \left (b^2-4 a c\right )^2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{21 c \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^3*Sqrt[d*(b + 2*c*x)]*(8*c*(-5*a^2*c + 2*a*(b^2 - b*c*x - c^2*x^2) + x*(2*b^3 + 5*b^2*c*x + 6*b*c^2*x^2 +
3*c^3*x^3)) + 5*(b^2 - 4*a*c)^2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[1/4, 1/2, 5/4, (
b + 2*c*x)^2/(b^2 - 4*a*c)]))/(21*c*Sqrt[a + x*(b + c*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(566\) vs. \(2(146)=292\).
time = 0.77, size = 567, normalized size = 3.26 Too large to display

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/21*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^3*(96*c^5*x^5+80*(-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*c^2-4
0*(-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^2*c+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)*E
llipticF(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^4+240*b*c^4*x^4-64*a*c
^4*x^3+256*b^2*c^3*x^3-96*a*b*c^3*x^2+144*b^3*c^2*x^2-160*a^2*c^3*x+32*b^2*c^2*a*x+32*c*b^4*x-80*a^2*b*c^2+32*
a*b^3*c)/c/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/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 = 0.46, size = 131, normalized size = 0.75 \begin {gather*} \frac {5 \, \sqrt {2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c^{2} d} d^{3} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 16 \, {\left (3 \, c^{4} d^{3} x^{2} + 3 \, b c^{3} d^{3} x + {\left (2 \, b^{2} c^{2} - 5 \, a c^{3}\right )} d^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{21 \, c^{2}} \end {gather*}

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)^(1/2),x, algorithm="fricas")

[Out]

1/21*(5*sqrt(2)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c^2*d)*d^3*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(
2*c*x + b)/c) + 16*(3*c^4*d^3*x^2 + 3*b*c^3*d^3*x + (2*b^2*c^2 - 5*a*c^3)*d^3)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2
+ b*x + a))/c^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {7}{2}}}{\sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

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)**(1/2),x)

[Out]

Integral((d*(b + 2*c*x))**(7/2)/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((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(1/2),x)

[Out]

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(1/2), x)

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