\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^5} \, dx\) [1226]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^5} \, dx=\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {15 \sqrt {b^2-4 a c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{512 c^{7/2} d^5} \]

[Out]

-5/64*(c*x^2+b*x+a)^(3/2)/c^2/d^5/(2*c*x+b)^2-1/8*(c*x^2+b*x+a)^(5/2)/c/d^5/(2*c*x+b)^4-15/512*arctan(2*c^(1/2
)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/c^(7/2)/d^5+15/256*(c*x^2+b*x+a)^(1/2)/c^3/d^5

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {698, 699, 702, 211} \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^5} \, dx=-\frac {15 \sqrt {b^2-4 a c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{512 c^{7/2} d^5}+\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4} \]

[In]

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

[Out]

(15*Sqrt[a + b*x + c*x^2])/(256*c^3*d^5) - (5*(a + b*x + c*x^2)^(3/2))/(64*c^2*d^5*(b + 2*c*x)^2) - (a + b*x +
c*x^2)^(5/2)/(8*c*d^5*(b + 2*c*x)^4) - (15*Sqrt[b^2 - 4*a*c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^
2 - 4*a*c]])/(512*c^(7/2)*d^5)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 702

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}+\frac {5 \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^3} \, dx}{16 c d^2} \\ & = -\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}+\frac {15 \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{128 c^2 d^4} \\ & = \frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {\left (15 \left (b^2-4 a c\right )\right ) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{512 c^3 d^4} \\ & = \frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {\left (15 \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{128 c^2 d^4} \\ & = \frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {15 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{512 c^{7/2} d^5} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^5} \, dx=\frac {2 (a+x (b+c x))^{7/2} \operatorname {Hypergeometric2F1}\left (3,\frac {7}{2},\frac {9}{2},\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{7 \left (b^2-4 a c\right )^3 d^5} \]

[In]

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

[Out]

(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4
*a*c)^3*d^5)

Maple [A] (verified)

Time = 3.80 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.19

method	result	size
pseudoelliptic	\(-\frac {\frac {15 \left (2 c x +b \right )^{4} \left (-\frac {b^{2}}{4}+a c \right ) \operatorname {arctanh}\left (\frac {2 c \sqrt {c \,x^{2}+b x +a}}{\sqrt {4 c^{2} a -b^{2} c}}\right )}{16}+\left (-4 c^{4} x^{4}+\left (-8 b \,x^{3}+\frac {9}{2} a \,x^{2}\right ) c^{3}+\left (-\frac {57}{8} b^{2} x^{2}+\frac {9}{2} a b x +a^{2}\right ) c^{2}+\frac {5 b^{2} \left (-5 b x +a \right ) c}{8}-\frac {15 b^{4}}{32}\right ) \sqrt {4 c^{2} a -b^{2} c}\, \sqrt {c \,x^{2}+b x +a}}{8 \sqrt {4 c^{2} a -b^{2} c}\, \left (2 c x +b \right )^{4} c^{3} d^{5}}\)	\(175\)
default	\(\frac {-\frac {c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4}}+\frac {3 c^{2} \left (-\frac {2 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {10 c^{2} \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{5}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 c}\right )}{4 a c -b^{2}}\right )}{4 a c -b^{2}}}{32 d^{5} c^{5}}\)	\(391\)
risch	\(\frac {\sqrt {c \,x^{2}+b x +a}}{32 c^{3} d^{5}}+\frac {-\frac {\left (12 a c -3 b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{c \sqrt {\frac {4 a c -b^{2}}{c}}}+\frac {\left (48 a^{2} c^{2}-24 a \,b^{2} c +3 b^{4}\right ) \left (-\frac {2 c \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {4 c^{2} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{8 c^{3}}+\frac {\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (-\frac {c \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4}}-\frac {3 c^{2} \left (-\frac {2 c \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {4 c^{2} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}\right )}{32 c^{5}}}{64 c^{3} d^{5}}\)	\(604\)

[In]

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

[Out]

-1/8*(15/16*(2*c*x+b)^4*(-1/4*b^2+a*c)*arctanh(2*c*(c*x^2+b*x+a)^(1/2)/(4*a*c^2-b^2*c)^(1/2))+(-4*c^4*x^4+(-8*
b*x^3+9/2*a*x^2)*c^3+(-57/8*b^2*x^2+9/2*a*b*x+a^2)*c^2+5/8*b^2*(-5*b*x+a)*c-15/32*b^4)*(4*a*c^2-b^2*c)^(1/2)*(
c*x^2+b*x+a)^(1/2))/(4*a*c^2-b^2*c)^(1/2)/(2*c*x+b)^4/c^3/d^5

Fricas [A] (verification not implemented)

none

Time = 1.03 (sec) , antiderivative size = 524, normalized size of antiderivative = 3.56 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^5} \, dx=\left [\frac {15 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (128 \, c^{4} x^{4} + 256 \, b c^{3} x^{3} + 15 \, b^{4} - 20 \, a b^{2} c - 32 \, a^{2} c^{2} + 12 \, {\left (19 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2} + 4 \, {\left (25 \, b^{3} c - 36 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1024 \, {\left (16 \, c^{7} d^{5} x^{4} + 32 \, b c^{6} d^{5} x^{3} + 24 \, b^{2} c^{5} d^{5} x^{2} + 8 \, b^{3} c^{4} d^{5} x + b^{4} c^{3} d^{5}\right )}}, \frac {15 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) + 2 \, {\left (128 \, c^{4} x^{4} + 256 \, b c^{3} x^{3} + 15 \, b^{4} - 20 \, a b^{2} c - 32 \, a^{2} c^{2} + 12 \, {\left (19 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2} + 4 \, {\left (25 \, b^{3} c - 36 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{512 \, {\left (16 \, c^{7} d^{5} x^{4} + 32 \, b c^{6} d^{5} x^{3} + 24 \, b^{2} c^{5} d^{5} x^{2} + 8 \, b^{3} c^{4} d^{5} x + b^{4} c^{3} d^{5}\right )}}\right ] \]

[In]

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

[Out]

[1/1024*(15*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*sqrt(-(b^2 - 4*a*c)/c)*log(-(4*c^2*
x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*c*sqrt(-(b^2 - 4*a*c)/c))/(4*c^2*x^2 + 4*b*c*x + b^2)) +
4*(128*c^4*x^4 + 256*b*c^3*x^3 + 15*b^4 - 20*a*b^2*c - 32*a^2*c^2 + 12*(19*b^2*c^2 - 12*a*c^3)*x^2 + 4*(25*b^
3*c - 36*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(16*c^7*d^5*x^4 + 32*b*c^6*d^5*x^3 + 24*b^2*c^5*d^5*x^2 + 8*b^3*c^
4*d^5*x + b^4*c^3*d^5), 1/512*(15*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*sqrt((b^2 - 4
*a*c)/c)*arctan(1/2*sqrt((b^2 - 4*a*c)/c)/sqrt(c*x^2 + b*x + a)) + 2*(128*c^4*x^4 + 256*b*c^3*x^3 + 15*b^4 - 2
0*a*b^2*c - 32*a^2*c^2 + 12*(19*b^2*c^2 - 12*a*c^3)*x^2 + 4*(25*b^3*c - 36*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/
(16*c^7*d^5*x^4 + 32*b*c^6*d^5*x^3 + 24*b^2*c^5*d^5*x^2 + 8*b^3*c^4*d^5*x + b^4*c^3*d^5)]

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^5} \, dx=\frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx}{d^{5}} \]

[In]

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

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*
x**4 + 32*c**5*x**5), x) + Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 +
80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b**5 + 10
*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(2*a*b*x*sqrt
(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5
), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3
+ 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b*
*3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x))/d**5

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^5} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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