\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{13/2}} \, dx\) [1354]

Optimal result

Integrand size = 28, antiderivative size = 211 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {5 \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{308 c^4 d^{13/2} \sqrt {a+b x+c x^2}} \]

[Out]

-5/154*(c*x^2+b*x+a)^(3/2)/c^2/d^3/(2*c*d*x+b*d)^(7/2)-1/11*(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b*d)^(11/2)-5/308
*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+b*d)^(3/2)+5/308*(-4*a*c+b^2)^(1/4)*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/d^(13/2)/(c*x^2+b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {698, 705, 703, 227} \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\frac {5 \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{308 c^4 d^{13/2} \sqrt {a+b x+c x^2}}-\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}} \]

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x]

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(308*c^3*d^5*(b*d + 2*c*d*x)^(3/2)) - (5*(a + b*x + c*x^2)^(3/2))/(154*c^2*d^3*(b*d
+ 2*c*d*x)^(7/2)) - (a + b*x + c*x^2)^(5/2)/(11*c*d*(b*d + 2*c*d*x)^(11/2)) + (5*(b^2 - 4*a*c)^(1/4)*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])
/(308*c^4*d^(13/2)*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 698

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 + 1))), x] - Dist[b*(p/(d*e*(m + 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] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && 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*} \text {integral}& = -\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {5 \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{9/2}} \, dx}{22 c d^2} \\ & = -\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {15 \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx}{308 c^2 d^4} \\ & = -\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {5 \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{616 c^3 d^6} \\ & = -\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {\left (5 \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}{616 c^3 d^6 \sqrt {a+b x+c x^2}} \\ & = -\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {\left (5 \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 )}{308 c^4 d^7 \sqrt {a+b x+c x^2}} \\ & = -\frac {5 \sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {5 \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 .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{308 c^4 d^{13/2} \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.52 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {5}{2},-\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{352 c^3 d^7 (b+2 c x)^6 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x]

[Out]

-1/352*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-11/4, -5/2, -7/4, (b + 2*
c*x)^2/(b^2 - 4*a*c)])/(c^3*d^7*(b + 2*c*x)^6*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(605\) vs. \(2(177)=354\).

Time = 5.23 (sec) , antiderivative size = 606, normalized size of antiderivative = 2.87

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{11264 c^{9} d^{7} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {3 \left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{2464 c^{7} d^{7} \left (x +\frac {b}{2 c}\right )^{4}}-\frac {37 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{4928 d^{7} c^{5} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {5 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{308 d^{6} c^{3} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(606\)
default	\(\text {Expression too large to display}\)	\(1035\)

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x,method=_RETURNVERBOSE)

[Out]

(d*(2*c*x+b)*(c*x^2+b*x+a))^(1/2)/(d*(2*c*x+b))^(1/2)/(c*x^2+b*x+a)^(1/2)*(-1/11264*(16*a^2*c^2-8*a*b^2*c+b^4)
/c^9/d^7*(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x+1/2/c*b)^6-3/2464*(4*a*c-b^2)/c^7/d^7*(2*c
^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x+1/2/c*b)^4-37/4928/d^7/c^5*(2*c^2*d*x^3+3*b*c*d*x^2+2*a
*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x+1/2/c*b)^2+5/308/d^6/c^3*(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/
2))/c)*((x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)*(
(x+1/2/c*b)/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c+1/2/c*b))^(1/2)*((x-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+(-4*a*c
+b^2)^(1/2))/c-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)/(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)*E
llipticF(((x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)
,((-1/2*(b+(-4*a*c+b^2)^(1/2))/c-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c+1/2/c*b))^(1/2)
))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\frac {5 \, \sqrt {2} {\left (64 \, c^{6} x^{6} + 192 \, b c^{5} x^{5} + 240 \, b^{2} c^{4} x^{4} + 160 \, b^{3} c^{3} x^{3} + 60 \, b^{4} c^{2} x^{2} + 12 \, b^{5} c x + b^{6}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - 2 \, {\left (148 \, c^{6} x^{4} + 296 \, b c^{5} x^{3} + 5 \, b^{4} c^{2} + 10 \, a b^{2} c^{3} + 28 \, a^{2} c^{4} + 6 \, {\left (33 \, b^{2} c^{4} + 16 \, a c^{5}\right )} x^{2} + 2 \, {\left (25 \, b^{3} c^{3} + 48 \, a b c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{616 \, {\left (64 \, c^{11} d^{7} x^{6} + 192 \, b c^{10} d^{7} x^{5} + 240 \, b^{2} c^{9} d^{7} x^{4} + 160 \, b^{3} c^{8} d^{7} x^{3} + 60 \, b^{4} c^{7} d^{7} x^{2} + 12 \, b^{5} c^{6} d^{7} x + b^{6} c^{5} d^{7}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="fricas")

[Out]

1/616*(5*sqrt(2)*(64*c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x
+ b^6)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) - 2*(148*c^6*x^4 + 296*b*c^5*
x^3 + 5*b^4*c^2 + 10*a*b^2*c^3 + 28*a^2*c^4 + 6*(33*b^2*c^4 + 16*a*c^5)*x^2 + 2*(25*b^3*c^3 + 48*a*b*c^4)*x)*s
qrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(64*c^11*d^7*x^6 + 192*b*c^10*d^7*x^5 + 240*b^2*c^9*d^7*x^4 + 160*b^
3*c^8*d^7*x^3 + 60*b^4*c^7*d^7*x^2 + 12*b^5*c^6*d^7*x + b^6*c^5*d^7)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(13/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {13}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(13/2), x)

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {13}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(13/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{13/2}} \,d x \]

[In]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x)

[Out]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2), x)