\(\int \frac {(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^{5/2}} \, dx\) [291]

Optimal result

Integrand size = 28, antiderivative size = 299 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt {a+b x+c x^2}}+\frac {1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {616 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{5 \sqrt {a+b x+c x^2}}-\frac {616 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{5 \sqrt {a+b x+c x^2}} \] Output:

-2/3*d*(2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2)-44/3*c*d^3*(2*c*d*x+b*d)^( 
7/2)/(c*x^2+b*x+a)^(1/2)+1232/15*c^2*d^5*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a) 
^(1/2)+616/5*c*(-4*a*c+b^2)^(7/4)*d^(13/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2)) 
^(1/2)*EllipticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/(c*x^2+ 
b*x+a)^(1/2)-616/5*c*(-4*a*c+b^2)^(7/4)*d^(13/2)*(-c*(c*x^2+b*x+a)/(-4*a*c 
+b^2))^(1/2)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/( 
c*x^2+b*x+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.60 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {4 d^5 (d (b+2 c x))^{3/2} \left (-41 b^4+156 b^3 c x+48 b c^2 x \left (-11 a+2 c x^2\right )+4 b^2 c \left (121 a+51 c x^2\right )+16 c^2 \left (-77 a^2-33 a c x^2+3 c^2 x^4\right )-616 c \left (b^2-4 a c\right ) (a+x (b+c x)) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{15 (a+x (b+c x))^{3/2}} \] Input:

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

Output:

(4*d^5*(d*(b + 2*c*x))^(3/2)*(-41*b^4 + 156*b^3*c*x + 48*b*c^2*x*(-11*a + 
2*c*x^2) + 4*b^2*c*(121*a + 51*c*x^2) + 16*c^2*(-77*a^2 - 33*a*c*x^2 + 3*c 
^2*x^4) - 616*c*(b^2 - 4*a*c)*(a + x*(b + c*x))*Sqrt[(c*(a + x*(b + c*x))) 
/(-b^2 + 4*a*c)]*Hypergeometric2F1[3/4, 5/2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a 
*c)]))/(15*(a + x*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1110, 1110, 1116, 1115, 1114, 836, 27, 762, 1389, 327}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1110

\(\displaystyle \frac {22}{3} c d^2 \int \frac {(b d+2 c x d)^{9/2}}{\left (c x^2+b x+a\right )^{3/2}}dx-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1110

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \int \frac {(b d+2 c x d)^{5/2}}{\sqrt {c x^2+b x+a}}dx-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {3}{5} d^2 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {3 d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{5 \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1114

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {b d+2 c x d}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 836

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{d \sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1389

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {\sqrt {\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}+1}}{\sqrt {1-\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {22}{3} c d^2 \left (14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d^{3/2} \left (b^2-4 a c\right )^{3/4} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

Input:

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

Output:

(-2*d*(b*d + 2*c*d*x)^(11/2))/(3*(a + b*x + c*x^2)^(3/2)) + (22*c*d^2*((-2 
*d*(b*d + 2*c*d*x)^(7/2))/Sqrt[a + b*x + c*x^2] + 14*c*d^2*((4*d*(b*d + 2* 
c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/5 + (6*(b^2 - 4*a*c)*d*Sqrt[-((c*(a + 
b*x + c*x^2))/(b^2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*EllipticE[ArcSi 
n[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1] - (b^2 - 4*a*c)^ 
(3/4)*d^(3/2)*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sq 
rt[d])], -1]))/(5*c*Sqrt[a + b*x + c*x^2]))))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[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 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 836

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

rule 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[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] && EqQ[2*c*d - b*e, 0] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1114

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symb 
ol] :> Simp[(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] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
 

rule 1115

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym 
bol] :> Simp[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] && EqQ[2*c* 
d - b*e, 0] && EqQ[m^2, 1/4]
 

rule 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   Int[(d 
 + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && 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])
 

rule 1389

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1327\) vs. \(2(249)=498\).

Time = 15.08 (sec) , antiderivative size = 1328, normalized size of antiderivative = 4.44

method	result	size
default	\(\text {Expression too large to display}\)	\(1328\)
elliptic	\(\text {Expression too large to display}\)	\(1360\)
risch	\(\text {Expression too large to display}\)	\(3986\)

Input:

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

Output:

-2/15*(d*(2*c*x+b))^(1/2)*(7392*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^ 
(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2 
)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*2^(1/2)*(1/(-4*a*c+b^2) 
^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*a^2*c^4*x^2-3696*(1/(- 
4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b 
^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)* 
EllipticE(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^ 
(1/2),2^(1/2))*a*b^2*c^3*x^2+462*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2) 
^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^ 
2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*2^(1/2)*(1/(-4*a*c+b^2 
)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*b^4*c^2*x^2-384*x^6*c 
^6+7392*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x 
+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^ 
(1/2))^(1/2)*EllipticE(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^ 
2)^(1/2)+b))^(1/2),2^(1/2))*a^2*b*c^3*x-3696*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+ 
(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c* 
x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*2^(1/2)*(1 
/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*a*b^3*c^2 
*x+462*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+ 
b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2...
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.21 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (924 \, \sqrt {2} {\left ({\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d^{6} x^{4} + 2 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{6} x^{3} + {\left (b^{4} c - 2 \, a b^{2} c^{2} - 8 \, a^{2} c^{3}\right )} d^{6} x^{2} + 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} d^{6} x + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} d^{6}\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) - {\left (192 \, c^{5} d^{6} x^{5} + 480 \, b c^{4} d^{6} x^{4} + 12 \, {\left (7 \, b^{2} c^{3} + 132 \, a c^{4}\right )} d^{6} x^{3} - 6 \, {\left (59 \, b^{3} c^{2} - 396 \, a b c^{3}\right )} d^{6} x^{2} - 4 \, {\left (40 \, b^{4} c - 143 \, a b^{2} c^{2} - 308 \, a^{2} c^{3}\right )} d^{6} x - {\left (5 \, b^{5} + 110 \, a b^{3} c - 616 \, a^{2} b c^{2}\right )} d^{6}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{15 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \] Input:

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

Output:

-2/15*(924*sqrt(2)*((b^2*c^3 - 4*a*c^4)*d^6*x^4 + 2*(b^3*c^2 - 4*a*b*c^3)* 
d^6*x^3 + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^6*x^2 + 2*(a*b^3*c - 4*a^2*b 
*c^2)*d^6*x + (a^2*b^2*c - 4*a^3*c^2)*d^6)*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)) - (192*c^5*d^6*x^5 + 480*b*c^4*d^6*x^4 + 12*(7*b^2*c^3 + 132*a*c^4) 
*d^6*x^3 - 6*(59*b^3*c^2 - 396*a*b*c^3)*d^6*x^2 - 4*(40*b^4*c - 143*a*b^2* 
c^2 - 308*a^2*c^3)*d^6*x - (5*b^5 + 110*a*b^3*c - 616*a^2*b*c^2)*d^6)*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)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:

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

Output:

(2*sqrt(d)*d**6*( - 14784*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*c**3 
 - 1232*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b**2*c**2 - 24640*sqrt 
(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b*c**3*x + 2068*sqrt(b + 2*c*x)*sq 
rt(a + b*x + c*x**2)*a*b**4*c + 4400*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2 
)*a*b**3*c**2*x - 15840*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2*c**3 
*x**2 - 10560*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b*c**4*x**3 - 179*s 
qrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**6 + 740*sqrt(b + 2*c*x)*sqrt(a + 
b*x + c*x**2)*b**5*c*x + 5160*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**4* 
c**2*x**2 + 5040*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**3*c**3*x**3 + 2 
400*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**2*c**4*x**4 + 960*sqrt(b + 2 
*c*x)*sqrt(a + b*x + c*x**2)*b*c**5*x**5 - 73920*int((sqrt(b + 2*c*x)*sqrt 
(a + b*x + c*x**2)*x**2)/(a**3*b + 2*a**3*c*x + 3*a**2*b**2*x + 9*a**2*b*c 
*x**2 + 6*a**2*c**2*x**3 + 3*a*b**3*x**2 + 12*a*b**2*c*x**3 + 15*a*b*c**2* 
x**4 + 6*a*c**3*x**5 + b**4*x**3 + 5*b**3*c*x**4 + 9*b**2*c**2*x**5 + 7*b* 
c**3*x**6 + 2*c**4*x**7),x)*a**5*c**5 + 55440*int((sqrt(b + 2*c*x)*sqrt(a 
+ b*x + c*x**2)*x**2)/(a**3*b + 2*a**3*c*x + 3*a**2*b**2*x + 9*a**2*b*c*x* 
*2 + 6*a**2*c**2*x**3 + 3*a*b**3*x**2 + 12*a*b**2*c*x**3 + 15*a*b*c**2*x** 
4 + 6*a*c**3*x**5 + b**4*x**3 + 5*b**3*c*x**4 + 9*b**2*c**2*x**5 + 7*b*c** 
3*x**6 + 2*c**4*x**7),x)*a**4*b**2*c**4 - 147840*int((sqrt(b + 2*c*x)*sqrt 
(a + b*x + c*x**2)*x**2)/(a**3*b + 2*a**3*c*x + 3*a**2*b**2*x + 9*a**2*...