3.14.55 \(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{17/2}} \, dx\) [1355]

Optimal. Leaf size=258 \[ -\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{924 c^4 \left (b^2-4 a c\right )^{3/4} d^{17/2} \sqrt {a+b x+c x^2}} \]

[Out]

-1/66*(c*x^2+b*x+a)^(3/2)/c^2/d^3/(2*c*d*x+b*d)^(11/2)-1/15*(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b*d)^(15/2)-1/308
*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+b*d)^(7/2)+1/462*(c*x^2+b*x+a)^(1/2)/c^3/(-4*a*c+b^2)/d^7/(2*c*d*x+b*d)^
(3/2)+1/924*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/(-4*a*c+b^2)^(3/4)/d^(17/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {698, 707, 705, 703, 227} \begin {gather*} \frac {\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 )}{924 c^4 d^{17/2} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(17/2),x]

[Out]

-1/308*Sqrt[a + b*x + c*x^2]/(c^3*d^5*(b*d + 2*c*d*x)^(7/2)) + Sqrt[a + b*x + c*x^2]/(462*c^3*(b^2 - 4*a*c)*d^
7*(b*d + 2*c*d*x)^(3/2)) - (a + b*x + c*x^2)^(3/2)/(66*c^2*d^3*(b*d + 2*c*d*x)^(11/2)) - (a + b*x + c*x^2)^(5/
2)/(15*c*d*(b*d + 2*c*d*x)^(15/2)) + (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])/(924*c^4*(b^2 - 4*a*c)^(3/4)*d^(17/2)*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]

Rule 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx &=-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{13/2}} \, dx}{6 c d^2}\\ &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{9/2}} \, dx}{44 c^2 d^4}\\ &=-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx}{616 c^3 d^6}\\ &=-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{1848 c^3 \left (b^2-4 a c\right ) d^8}\\ &=-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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}{1848 c^3 \left (b^2-4 a c\right ) d^8 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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 )}{924 c^4 \left (b^2-4 a c\right ) d^9 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\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 )}{924 c^4 \left (b^2-4 a c\right )^{3/4} d^{17/2} \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 = 5.42, size = 109, normalized size = 0.42 \begin {gather*} -\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \, _2F_1\left (-\frac {15}{4},-\frac {5}{2};-\frac {11}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{480 c^3 d^9 (b+2 c x)^8 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(17/2),x]

[Out]

-1/480*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-15/4, -5/2, -11/4, (b + 2
*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^9*(b + 2*c*x)^8*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1430\) vs. \(2(218)=436\).
time = 0.80, size = 1431, normalized size = 5.55

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (16 a^{2} c^{2}-8 a c \,b^{2}+b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{61440 c^{11} d^{9} \left (x +\frac {b}{2 c}\right )^{8}}-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{5280 c^{9} d^{9} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {69 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{98560 c^{7} d^{9} \left (x +\frac {b}{2 c}\right )^{4}}-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{1848 c^{5} \left (4 a c -b^{2}\right ) d^{9} \left (x +\frac {b}{2 c}\right )^{2}}-\frac {\left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \EllipticF \left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{924 c^{3} \left (4 a c -b^{2}\right ) d^{8} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(684\)
default	\(\text {Expression too large to display}\)	\(1431\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(17/2),x,method=_RETURNVERBOSE)

[Out]

-1/9240*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(23712*a*b*c^6*x^5+22744*a*b^2*c^5*x^4+27584*a^2*b*c^5*x^3+596
8*a*b^3*c^4*x^3+13464*a^2*b^2*c^4*x^2-1128*a*b^4*c^3*x^2+9632*a^3*b*c^4*x-328*a^2*b^3*c^3*x-160*a*b^5*c^2*x-15
0*b^6*c^2*x^2+6984*b^2*c^6*x^6+3032*b^3*c^5*x^5-56*a^3*b^2*c^3+2240*((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))*(-4*a*c+b^2)^(1/2)*b*c^6
*x^6+3360*((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)*b^2*c^5*x^5-20*a^2*b^4*c^2+2800*((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)*b^3*c^4*x^4+1400*((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)*b^4*c^3*x^3+420*((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)*b^5*c^2*x^2+70*((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)*b^6*c*x+1280*c^8*x^8+2464*a^4*c^4-10*a*b^6*c-948*b^4*c^
4*x^4-976*b^5*c^3*x^3-10*b^7*c*x+5120*b*c^7*x^7+7904*a*c^7*x^6+13792*a^2*c^6*x^4+9632*a^3*c^5*x^2+5*((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)*b^7+640*((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)*c^7*x^7)/d^9/(2*c^2*x^3+3*b*c*x^2+2*a*
c*x+b^2*x+a*b)/(2*c*x+b)^7/(4*a*c-b^2)/c^4

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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((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(17/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.88, size = 521, normalized size = 2.02 \begin {gather*} \frac {5 \, \sqrt {2} {\left (256 \, c^{8} x^{8} + 1024 \, b c^{7} x^{7} + 1792 \, b^{2} c^{6} x^{6} + 1792 \, b^{3} c^{5} x^{5} + 1120 \, b^{4} c^{4} x^{4} + 448 \, b^{5} c^{3} x^{3} + 112 \, b^{6} c^{2} x^{2} + 16 \, b^{7} c x + b^{8}\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 (640 \, c^{8} x^{6} + 1920 \, b c^{7} x^{5} - 5 \, b^{6} c^{2} - 10 \, a b^{4} c^{3} - 28 \, a^{2} b^{2} c^{4} + 1232 \, a^{3} c^{5} + 12 \, {\left (131 \, b^{2} c^{6} + 276 \, a c^{7}\right )} x^{4} - 8 \, {\left (7 \, b^{3} c^{5} - 828 \, a b c^{6}\right )} x^{3} - 2 \, {\left (209 \, b^{4} c^{4} - 1588 \, a b^{2} c^{5} - 1792 \, a^{2} c^{6}\right )} x^{2} - 2 \, {\left (35 \, b^{5} c^{3} + 68 \, a b^{3} c^{4} - 1792 \, a^{2} b c^{5}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{9240 \, {\left (256 \, {\left (b^{2} c^{13} - 4 \, a c^{14}\right )} d^{9} x^{8} + 1024 \, {\left (b^{3} c^{12} - 4 \, a b c^{13}\right )} d^{9} x^{7} + 1792 \, {\left (b^{4} c^{11} - 4 \, a b^{2} c^{12}\right )} d^{9} x^{6} + 1792 \, {\left (b^{5} c^{10} - 4 \, a b^{3} c^{11}\right )} d^{9} x^{5} + 1120 \, {\left (b^{6} c^{9} - 4 \, a b^{4} c^{10}\right )} d^{9} x^{4} + 448 \, {\left (b^{7} c^{8} - 4 \, a b^{5} c^{9}\right )} d^{9} x^{3} + 112 \, {\left (b^{8} c^{7} - 4 \, a b^{6} c^{8}\right )} d^{9} x^{2} + 16 \, {\left (b^{9} c^{6} - 4 \, a b^{7} c^{7}\right )} d^{9} x + {\left (b^{10} c^{5} - 4 \, a b^{8} c^{6}\right )} d^{9}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="fricas")

[Out]

1/9240*(5*sqrt(2)*(256*c^8*x^8 + 1024*b*c^7*x^7 + 1792*b^2*c^6*x^6 + 1792*b^3*c^5*x^5 + 1120*b^4*c^4*x^4 + 448
*b^5*c^3*x^3 + 112*b^6*c^2*x^2 + 16*b^7*c*x + b^8)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(
2*c*x + b)/c) + 2*(640*c^8*x^6 + 1920*b*c^7*x^5 - 5*b^6*c^2 - 10*a*b^4*c^3 - 28*a^2*b^2*c^4 + 1232*a^3*c^5 + 1
2*(131*b^2*c^6 + 276*a*c^7)*x^4 - 8*(7*b^3*c^5 - 828*a*b*c^6)*x^3 - 2*(209*b^4*c^4 - 1588*a*b^2*c^5 - 1792*a^2
*c^6)*x^2 - 2*(35*b^5*c^3 + 68*a*b^3*c^4 - 1792*a^2*b*c^5)*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(256*
(b^2*c^13 - 4*a*c^14)*d^9*x^8 + 1024*(b^3*c^12 - 4*a*b*c^13)*d^9*x^7 + 1792*(b^4*c^11 - 4*a*b^2*c^12)*d^9*x^6
+ 1792*(b^5*c^10 - 4*a*b^3*c^11)*d^9*x^5 + 1120*(b^6*c^9 - 4*a*b^4*c^10)*d^9*x^4 + 448*(b^7*c^8 - 4*a*b^5*c^9)
*d^9*x^3 + 112*(b^8*c^7 - 4*a*b^6*c^8)*d^9*x^2 + 16*(b^9*c^6 - 4*a*b^7*c^7)*d^9*x + (b^10*c^5 - 4*a*b^8*c^6)*d
^9)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(17/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5985 deep

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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((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(17/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{17/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(17/2),x)

[Out]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(17/2), x)

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