3.14.91 \(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^{5/2}} \, dx\) [1391]

Optimal. Leaf size=165 \[ -\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}+\frac {40 c \sqrt [4]{b^2-4 a c} d^{7/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{3 \sqrt {a+b x+c x^2}} \]

[Out]

-2/3*d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(3/2)-20/3*c*d^3*(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2)+40/3*c*(-4*a
*c+b^2)^(1/4)*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*x^2+b*x+a)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 165, 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 = {700, 705, 703, 227} \begin {gather*} \frac {40 c d^{7/2} \sqrt [4]{b^2-4 a c} \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 )}{3 \sqrt {a+b x+c x^2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (20*c*d^3*Sqrt[b*d + 2*c*d*x])/(3*Sqrt[a + b*x + c*
x^2]) + (40*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*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])/(3*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 700

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*(d + e*x)^(m - 1)*(
(a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] - Dist[d*e*((m - 1)/(b*(p + 1))), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && IntegerQ[2*p]

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]

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac {1}{3} \left (10 c d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}+\frac {1}{3} \left (20 c^2 d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}+\frac {\left (20 c^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}{3 \sqrt {a+b x+c x^2}}\\ &=-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}+\frac {\left (40 c 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 )}{3 \sqrt {a+b x+c x^2}}\\ &=-\frac {2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {20 c d^3 \sqrt {b d+2 c d x}}{3 \sqrt {a+b x+c x^2}}+\frac {40 c \sqrt [4]{b^2-4 a c} 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 )}{3 \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 = 10.09, size = 122, normalized size = 0.74 \begin {gather*} -\frac {2 d^3 \sqrt {d (b+2 c x)} \left (b^2+14 b c x+2 c \left (5 a+7 c x^2\right )-20 c (a+x (b+c x)) \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 )}{3 (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^3*Sqrt[d*(b + 2*c*x)]*(b^2 + 14*b*c*x + 2*c*(5*a + 7*c*x^2) - 20*c*(a + x*(b + c*x))*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)]))/(3*(a + x*(b + c*x))^(
3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(478\) vs. \(2(137)=274\).
time = 0.90, size = 479, normalized size = 2.90

method	result	size
default	\(\frac {2 \left (10 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) c^{2} x^{2}+10 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b c x +10 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, a c -28 c^{3} x^{3}-42 b \,c^{2} x^{2}-20 a \,c^{2} x -16 b^{2} c x -10 a b c -b^{3}\right ) d^{3} \sqrt {d \left (2 c x +b \right )}}{3 \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(479\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (\frac {2 \left (4 a c -b^{2}\right ) d^{3} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{3 c^{2} \left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right )^{2}}-\frac {28 \left (2 c^{2} d x +b c d \right ) d^{3}}{3 \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +b c d \right )}}+\frac {40 c^{2} d^{4} \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \EllipticF \left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(544\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(10*(-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))*c^2*x^2+10*(-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))*b*c*x+10*((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))*(-4*a*c+b^2)^(1/2)*a*c-28*c^3*x^3-42*b*c^2*x^2-20*a*c^2*x-16*b^2*c*x-10*a*b*c-b^3)*d^3*(d*(2*c*x+b))
^(1/2)/(2*c*x+b)/(c*x^2+b*x+a)^(3/2)

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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((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.80, size = 185, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (10 \, \sqrt {2} {\left (c^{2} d^{3} x^{4} + 2 \, b c d^{3} x^{3} + 2 \, a b d^{3} x + {\left (b^{2} + 2 \, a c\right )} d^{3} x^{2} + a^{2} d^{3}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - {\left (14 \, c^{2} d^{3} x^{2} + 14 \, b c d^{3} x + {\left (b^{2} + 10 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \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)^(5/2),x, algorithm="fricas")

[Out]

2/3*(10*sqrt(2)*(c^2*d^3*x^4 + 2*b*c*d^3*x^3 + 2*a*b*d^3*x + (b^2 + 2*a*c)*d^3*x^2 + a^2*d^3)*sqrt(c^2*d)*weie
rstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) - (14*c^2*d^3*x^2 + 14*b*c*d^3*x + (b^2 + 10*a*c)*d^3
)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**(5/2),x)

[Out]

Timed out

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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)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(5/2), 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}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,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)^(5/2),x)

[Out]

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2), x)

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