\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{7/2}} \, dx\) [1360]

Optimal result

Integrand size = 28, antiderivative size = 310 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {698, 699, 705, 704, 313, 227, 1213, 435} \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}}+\frac {3 \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}} \]

[In]

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

[Out]

(3*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(20*c^3*d^5) - (a + b*x + c*x^2)^(3/2)/(2*c^2*d^3*Sqrt[b*d + 2
*c*d*x]) - (a + b*x + c*x^2)^(5/2)/(5*c*d*(b*d + 2*c*d*x)^(5/2)) - (3*(b^2 - 4*a*c)^(7/4)*Sqrt[-((c*(a + b*x +
c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(20*c^4*d^(
7/2)*Sqrt[a + b*x + c*x^2]) + (3*(b^2 - 4*a*c)^(7/4)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[Ar
cSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(20*c^4*d^(7/2)*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 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && 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 699

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 704

Int[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[x^2/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 1213

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + e*(x^2/d)]/Sqrt
[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac {\int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{3/2}} \, dx}{2 c d^2} \\ & = -\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac {3 \int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx}{4 c^2 d^4} \\ & = \frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{40 c^3 d^4} \\ & = \frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac {\left (3 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac {\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}{40 c^3 d^4 \sqrt {a+b x+c x^2}} \\ & = \frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac {\left (3 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{20 c^4 d^5 \sqrt {a+b x+c x^2}} \\ & = \frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac {\left (3 \left (b^2-4 a c\right )^{3/2} \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 )}{20 c^4 d^4 \sqrt {a+b x+c x^2}}-\frac {\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{20 c^4 d^4 \sqrt {a+b x+c x^2}} \\ & = \frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac {3 \left (b^2-4 a c\right )^{7/4} \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}}-\frac {\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}}{\sqrt {1-\frac {x^2}{\sqrt {b^2-4 a c} d}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{20 c^4 d^4 \sqrt {a+b x+c x^2}} \\ & = \frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\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 )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}}+\frac {3 \left (b^2-4 a c\right )^{7/4} \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \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 = 10.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{4},-\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{160 c^3 d (d (b+2 c x))^{5/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]

[In]

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

[Out]

-1/160*((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -5/4, -1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]
)/(c^3*d*(d*(b + 2*c*x))^(5/2)*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. \(1294\) vs. \(2(260)=520\).

Time = 6.43 (sec) , antiderivative size = 1295, normalized size of antiderivative = 4.18

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(1295\)
default	\(\text {Expression too large to display}\)	\(1362\)
risch	\(\text {Expression too large to display}\)	\(3978\)

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),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)*(-1/640*(16*a^2*c^2-8*a*b^2*c+b^4)/c
^6/d^4*(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)^3-3/40*(2*c^2*d*x^2+2*b*c*d*x+2*a*c
*d)*(4*a*c-b^2)/c^4/d^4/((x+1/2/c*b)*(2*c^2*d*x^2+2*b*c*d*x+2*a*c*d))^(1/2)+1/40/d^4/c^2*x*(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)+1/80*b/c^3/d^4*(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)+2*
(1/32*b*(6*a*c-b^2)/c^3/d^3+3/80/c^3*b*(4*a*c-b^2)/d^3-1/40/d^3/c^2*a*b-1/80*b/c^3/d^4*(a*c*d+1/2*b^2*d))*(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)*EllipticF(((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))+2*(3/8/c*a/d^3+3/40*(4*a*c-b^2)/c^2/d^3-1/40/d^4/c^2*(3*a*
c*d+3/2*b^2*d)-3/80*b^2/c^2/d^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)*((-1/2*(b+(-4*a*c+b^2)^(
1/2))/c+1/2/c*b)*EllipticE(((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))-1/2/c*b*EllipticF(((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 = 274, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\frac {3 \, \sqrt {2} {\left (b^{5} - 4 \, a b^{3} c + 8 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + 12 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + 6 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (4 \, c^{5} x^{4} + 8 \, b c^{4} x^{3} + 3 \, b^{4} c - 10 \, a b^{2} c^{2} - 4 \, a^{2} c^{3} + 6 \, {\left (3 \, b^{2} c^{3} - 8 \, a c^{4}\right )} x^{2} + 2 \, {\left (7 \, b^{3} c^{2} - 24 \, a b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{20 \, {\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}} \]

[In]

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

[Out]

1/20*(3*sqrt(2)*(b^5 - 4*a*b^3*c + 8*(b^2*c^3 - 4*a*c^4)*x^3 + 12*(b^3*c^2 - 4*a*b*c^3)*x^2 + 6*(b^4*c - 4*a*b
^2*c^2)*x)*sqrt(c^2*d)*weierstrassZeta((b^2 - 4*a*c)/c^2, 0, weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*
c*x + b)/c)) + (4*c^5*x^4 + 8*b*c^4*x^3 + 3*b^4*c - 10*a*b^2*c^2 - 4*a^2*c^3 + 6*(3*b^2*c^3 - 8*a*c^4)*x^2 + 2
*(7*b^3*c^2 - 24*a*b*c^3)*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*
b^2*c^5*d^4*x + b^3*c^4*d^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{7/2}} \,d x \]

[In]

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

[Out]

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