\(\int \frac {(b d+2 c d x)^{15/2}}{(a+b x+c x^2)^{5/2}} \, dx\) [285]

Optimal result

Integrand size = 28, antiderivative size = 247 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {52 c d^3 (b d+2 c d x)^{9/2}}{3 \sqrt {a+b x+c x^2}}+\frac {1040}{7} c^2 \left (b^2-4 a c\right ) d^7 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {624}{7} c^2 d^5 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {520 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{7 \sqrt {a+b x+c x^2}} \] Output:

-2/3*d*(2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^(3/2)-52/3*c*d^3*(2*c*d*x+b*d)^( 
9/2)/(c*x^2+b*x+a)^(1/2)+1040/7*c^2*(-4*a*c+b^2)*d^7*(2*c*d*x+b*d)^(1/2)*( 
c*x^2+b*x+a)^(1/2)+624/7*c^2*d^5*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(1/2)+5 
20/7*c*(-4*a*c+b^2)^(9/4)*d^(15/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*E 
llipticF((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 = 10.38 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.02 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 d^7 \sqrt {d (b+2 c x)} \left (-7 b^6-266 b^5 c x+16 b^3 c^2 x \left (221 a+112 c x^2\right )+2 b^4 c \left (-91 a+219 c x^2\right )+32 b c^3 x \left (-273 a^2-104 a c x^2+36 c^2 x^4\right )+16 b^2 c^2 \left (156 a^2+117 a c x^2+116 c^2 x^4\right )-32 c^3 \left (195 a^3+273 a^2 c x^2+52 a c^2 x^4-12 c^3 x^6\right )+780 c \left (b^2-4 a c\right )^2 (a+x (b+c x)) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{21 (a+x (b+c x))^{3/2}} \] Input:

Integrate[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(2*d^7*Sqrt[d*(b + 2*c*x)]*(-7*b^6 - 266*b^5*c*x + 16*b^3*c^2*x*(221*a + 1 
12*c*x^2) + 2*b^4*c*(-91*a + 219*c*x^2) + 32*b*c^3*x*(-273*a^2 - 104*a*c*x 
^2 + 36*c^2*x^4) + 16*b^2*c^2*(156*a^2 + 117*a*c*x^2 + 116*c^2*x^4) - 32*c 
^3*(195*a^3 + 273*a^2*c*x^2 + 52*a*c^2*x^4 - 12*c^3*x^6) + 780*c*(b^2 - 4* 
a*c)^2*(a + x*(b + c*x))*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hyperg 
eometric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(21*(a + x*(b + 
c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1110, 1110, 1116, 1116, 1115, 1113, 762}

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.

\(\displaystyle \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle \frac {26}{3} c d^2 \int \frac {(b d+2 c x d)^{11/2}}{\left (c x^2+b x+a\right )^{3/2}}dx-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1110

\(\displaystyle \frac {26}{3} c d^2 \left (18 c d^2 \int \frac {(b d+2 c x d)^{7/2}}{\sqrt {c x^2+b x+a}}dx-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {26}{3} c d^2 \left (18 c d^2 \left (\frac {5}{7} d^2 \left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{3/2}}{\sqrt {c x^2+b x+a}}dx+\frac {4}{7} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {26}{3} c d^2 \left (18 c d^2 \left (\frac {5}{7} d^2 \left (b^2-4 a c\right ) \left (\frac {1}{3} d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )+\frac {4}{7} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {26}{3} c d^2 \left (18 c d^2 \left (\frac {5}{7} d^2 \left (b^2-4 a c\right ) \left (\frac {d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )+\frac {4}{7} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle \frac {26}{3} c d^2 \left (18 c d^2 \left (\frac {5}{7} d^2 \left (b^2-4 a c\right ) \left (\frac {2 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )+\frac {4}{7} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {26}{3} c d^2 \left (18 c d^2 \left (\frac {5}{7} d^2 \left (b^2-4 a c\right ) \left (\frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )+\frac {4}{7} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{13/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

Input:

Int[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*d*(b*d + 2*c*d*x)^(13/2))/(3*(a + b*x + c*x^2)^(3/2)) + (26*c*d^2*((-2 
*d*(b*d + 2*c*d*x)^(9/2))/Sqrt[a + b*x + c*x^2] + 18*c*d^2*((4*d*(b*d + 2* 
c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (5*(b^2 - 4*a*c)*d^2*((4*d*Sqrt[b* 
d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/3 + (2*(b^2 - 4*a*c)^(5/4)*d^(3/2)*Sqr 
t[-((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*c*Sqrt[a + b*x + c*x^2])))/7) 
))/3
 

Defintions of rubi rules used

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 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 1113

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_ 
Symbol] :> Simp[(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] && 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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1472\) vs. \(2(207)=414\).

Time = 15.68 (sec) , antiderivative size = 1473, normalized size of antiderivative = 5.96

method	result	size
default	\(\text {Expression too large to display}\)	\(1473\)
risch	\(\text {Expression too large to display}\)	\(1526\)
elliptic	\(\text {Expression too large to display}\)	\(1532\)

Input:

int((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/21*(-7*b^7+6240*(-4*a*c+b^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*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)*EllipticF(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^3*c^3-6240*a^3* 
b*c^3+8944*a*b^3*c^3*x^2-3744*a^2*b^2*c^3*x-3120*(-4*a*c+b^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*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)*E 
llipticF(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^2*c^2+390*(-4*a*c+b^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*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)*EllipticF(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^4 
*c+2688*b*c^6*x^6-3328*a*c^6*x^5+4864*b^2*c^5*x^5+5440*b^3*c^4*x^4-17472*a 
^2*c^5*x^3-94*b^5*c^2*x^2-12480*a^3*c^4*x-8320*a*b*c^5*x^4+416*a*b^2*c^4*x 
^3-26208*a^2*b*c^4*x^2+3172*a*b^4*c^2*x+768*c^7*x^7-3120*(-4*a*c+b^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*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)*EllipticF(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+6240*(-4*a*c+b^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*c*x+b)/(-4*a*c+b^2)^(1...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (206) = 412\).

Time = 0.11 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.72 \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (390 \, \sqrt {2} {\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{7} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{7} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d^{7} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d^{7} x + {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d^{7}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + {\left (384 \, c^{6} d^{7} x^{6} + 1152 \, b c^{5} d^{7} x^{5} + 64 \, {\left (29 \, b^{2} c^{4} - 26 \, a c^{5}\right )} d^{7} x^{4} + 256 \, {\left (7 \, b^{3} c^{3} - 13 \, a b c^{4}\right )} d^{7} x^{3} + 6 \, {\left (73 \, b^{4} c^{2} + 312 \, a b^{2} c^{3} - 1456 \, a^{2} c^{4}\right )} d^{7} x^{2} - 2 \, {\left (133 \, b^{5} c - 1768 \, a b^{3} c^{2} + 4368 \, a^{2} b c^{3}\right )} d^{7} x - {\left (7 \, b^{6} + 182 \, a b^{4} c - 2496 \, a^{2} b^{2} c^{2} + 6240 \, a^{3} c^{3}\right )} d^{7}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{21 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \] Input:

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

2/21*(390*sqrt(2)*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^7*x^4 + 2*(b^5*c 
 - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^7*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^ 
7*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^7*x + (a^2*b^4 - 8*a^3*b^ 
2*c + 16*a^4*c^2)*d^7)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 
0, 1/2*(2*c*x + b)/c) + (384*c^6*d^7*x^6 + 1152*b*c^5*d^7*x^5 + 64*(29*b^2 
*c^4 - 26*a*c^5)*d^7*x^4 + 256*(7*b^3*c^3 - 13*a*b*c^4)*d^7*x^3 + 6*(73*b^ 
4*c^2 + 312*a*b^2*c^3 - 1456*a^2*c^4)*d^7*x^2 - 2*(133*b^5*c - 1768*a*b^3* 
c^2 + 4368*a^2*b*c^3)*d^7*x - (7*b^6 + 182*a*b^4*c - 2496*a^2*b^2*c^2 + 62 
40*a^3*c^3)*d^7)*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. \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**(15/2)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {15}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((2*c*d*x + b*d)^(15/2)/(c*x^2 + b*x + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {15}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((2*c*d*x + b*d)^(15/2)/(c*x^2 + b*x + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{15/2}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:

int((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

(2*sqrt(d)*d**7*(29952*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*b**2*c* 
*3 - 29952*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*b*c**4*x - 29952*sq 
rt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*c**5*x**2 - 13312*sqrt(b + 2*c*x 
)*sqrt(a + b*x + c*x**2)*a**2*b**4*c**2 + 58240*sqrt(b + 2*c*x)*sqrt(a + b 
*x + c*x**2)*a**2*b**3*c**3*x + 54912*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x** 
2)*a**2*b**2*c**4*x**2 - 6656*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2* 
b*c**5*x**3 - 3328*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*c**6*x**4 + 
 1312*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**6*c - 21280*sqrt(b + 2*c 
*x)*sqrt(a + b*x + c*x**2)*a*b**5*c**2*x - 14880*sqrt(b + 2*c*x)*sqrt(a + 
b*x + c*x**2)*a*b**4*c**3*x**2 + 13568*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x* 
*2)*a*b**3*c**4*x**3 + 8704*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2* 
c**5*x**4 + 2304*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b*c**6*x**5 + 76 
8*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*c**7*x**6 + 21*sqrt(b + 2*c*x)* 
sqrt(a + b*x + c*x**2)*b**8 + 1968*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)* 
b**7*c*x - 144*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**6*c**2*x**2 - 537 
6*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**5*c**3*x**3 - 5568*sqrt(b + 2* 
c*x)*sqrt(a + b*x + c*x**2)*b**4*c**4*x**4 - 3456*sqrt(b + 2*c*x)*sqrt(a + 
 b*x + c*x**2)*b**3*c**5*x**5 - 1152*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2 
)*b**2*c**6*x**6 + 299520*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2 
)/(2*a**4*b*c + 4*a**4*c**2*x - 3*a**3*b**3 + 18*a**3*b*c**2*x**2 + 12*...