\(\int \frac {(d+e x)^{13/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [192]

Optimal result

Integrand size = 37, antiderivative size = 186 \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {35 e^2 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{4 c^4 d^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}-\frac {35 e^2 \left (c d^2-a e^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{9/2} d^{9/2}} \] Output:

35/4*e^2*(-a*e^2+c*d^2)*(e*x+d)^(1/2)/c^4/d^4+35/12*e^2*(e*x+d)^(3/2)/c^3/ 
d^3-7/4*e*(e*x+d)^(5/2)/c^2/d^2/(c*d*x+a*e)-1/2*(e*x+d)^(7/2)/c/d/(c*d*x+a 
*e)^2-35/4*e^2*(-a*e^2+c*d^2)^(3/2)*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/ 
(-a*e^2+c*d^2)^(1/2))/c^(9/2)/d^(9/2)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.11 \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (105 a^3 e^6-35 a^2 c d e^4 (4 d-5 e x)+7 a c^2 d^2 e^2 \left (3 d^2-34 d e x+8 e^2 x^2\right )+c^3 d^3 \left (6 d^3+39 d^2 e x-80 d e^2 x^2-8 e^3 x^3\right )\right )}{12 c^4 d^4 (a e+c d x)^2}+\frac {35 \left (c d^2 e-a e^3\right )^2 \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 c^{9/2} d^{9/2} \sqrt {-c d^2+a e^2}} \] Input:

Integrate[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
 

Output:

-1/12*(Sqrt[d + e*x]*(105*a^3*e^6 - 35*a^2*c*d*e^4*(4*d - 5*e*x) + 7*a*c^2 
*d^2*e^2*(3*d^2 - 34*d*e*x + 8*e^2*x^2) + c^3*d^3*(6*d^3 + 39*d^2*e*x - 80 
*d*e^2*x^2 - 8*e^3*x^3)))/(c^4*d^4*(a*e + c*d*x)^2) + (35*(c*d^2*e - a*e^3 
)^2*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(4*c^( 
9/2)*d^(9/2)*Sqrt[-(c*d^2) + a*e^2])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1121, 51, 51, 60, 60, 73, 221}

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.

\(\displaystyle \int \frac {(d+e x)^{13/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3} \, dx\)

\(\Big \downarrow \) 1121

\(\displaystyle \int \frac {(d+e x)^{7/2}}{(a e+c d x)^3}dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {7 e \int \frac {(d+e x)^{5/2}}{(a e+c d x)^2}dx}{4 c d}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {7 e \left (\frac {5 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \sqrt {d+e x}}{c d}-\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} d^{3/2} \sqrt {c d^2-a e^2}}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{2 c d}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}\)

Input:

Int[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
 

Output:

-1/2*(d + e*x)^(7/2)/(c*d*(a*e + c*d*x)^2) + (7*e*(-((d + e*x)^(5/2)/(c*d* 
(a*e + c*d*x))) + (5*e*((2*(d + e*x)^(3/2))/(3*c*d) + ((d^2 - (a*e^2)/c)*( 
(2*Sqrt[d + e*x])/(c*d) - (2*(d^2 - (a*e^2)/c)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sq 
rt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(Sqrt[c]*d^(3/2)*Sqrt[c*d^2 - a*e^2]))) 
/d))/(2*c*d)))/(4*c*d)
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 1121

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int 
egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
 

Maple [A] (verified)

Time = 15.89 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {2 e^{2} \left (-c d x e +9 a \,e^{2}-10 c \,d^{2}\right ) \sqrt {e x +d}}{3 d^{4} c^{4}}+\frac {\left (2 a^{2} e^{4}-4 a c \,d^{2} e^{2}+2 c^{2} d^{4}\right ) e^{2} \left (\frac {-\frac {13 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (-\frac {11 a \,e^{2}}{8}+\frac {11 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {35 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{d^{4} c^{4}}\)	\(180\)
pseudoelliptic	\(\frac {\frac {35 e^{2} \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{4}-\frac {35 \sqrt {e x +d}\, \left (\frac {2 d^{3} \left (-\frac {4}{3} e^{3} x^{3}-\frac {40}{3} d \,e^{2} x^{2}+\frac {13}{2} d^{2} e x +d^{3}\right ) c^{3}}{35}+\frac {e^{2} a \,d^{2} \left (\frac {8}{3} e^{2} x^{2}-\frac {34}{3} d e x +d^{2}\right ) c^{2}}{5}-\frac {4 \left (-\frac {5 e x}{4}+d \right ) e^{4} a^{2} d c}{3}+e^{6} a^{3}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}{4}}{d^{4} c^{4} \left (c d x +a e \right )^{2} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\)	\(208\)
derivativedivides	\(2 e^{2} \left (-\frac {-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a \,e^{2} \sqrt {e x +d}-3 c \,d^{2} \sqrt {e x +d}}{c^{4} d^{4}}+\frac {\frac {\left (-\frac {13}{8} d \,e^{4} a^{2} c +\frac {13}{4} d^{3} e^{2} a \,c^{2}-\frac {13}{8} d^{5} c^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {11}{8} e^{6} a^{3}+\frac {33}{8} d^{2} e^{4} a^{2} c -\frac {33}{8} d^{4} e^{2} a \,c^{2}+\frac {11}{8} d^{6} c^{3}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {35 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{4} d^{4}}\right )\)	\(245\)
default	\(2 e^{2} \left (-\frac {-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a \,e^{2} \sqrt {e x +d}-3 c \,d^{2} \sqrt {e x +d}}{c^{4} d^{4}}+\frac {\frac {\left (-\frac {13}{8} d \,e^{4} a^{2} c +\frac {13}{4} d^{3} e^{2} a \,c^{2}-\frac {13}{8} d^{5} c^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {11}{8} e^{6} a^{3}+\frac {33}{8} d^{2} e^{4} a^{2} c -\frac {33}{8} d^{4} e^{2} a \,c^{2}+\frac {11}{8} d^{6} c^{3}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {35 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{4} d^{4}}\right )\)	\(245\)

Input:

int((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^3,x,method=_RETURNVER 
BOSE)
 

Output:

-2/3*e^2*(-c*d*e*x+9*a*e^2-10*c*d^2)*(e*x+d)^(1/2)/d^4/c^4+1/d^4/c^4*(2*a^ 
2*e^4-4*a*c*d^2*e^2+2*c^2*d^4)*e^2*((-13/8*c*d*(e*x+d)^(3/2)+(-11/8*a*e^2+ 
11/8*c*d^2)*(e*x+d)^(1/2))/(c*d*(e*x+d)+a*e^2-c*d^2)^2+35/8/(c*d*(a*e^2-c* 
d^2))^(1/2)*arctan(c*d*(e*x+d)^(1/2)/(c*d*(a*e^2-c*d^2))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (154) = 308\).

Time = 0.12 (sec) , antiderivative size = 638, normalized size of antiderivative = 3.43 \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [-\frac {105 \, {\left (a^{2} c d^{2} e^{4} - a^{3} e^{6} + {\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) - 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{3} - 6 \, c^{3} d^{6} - 21 \, a c^{2} d^{4} e^{2} + 140 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 8 \, {\left (10 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} - {\left (39 \, c^{3} d^{5} e - 238 \, a c^{2} d^{3} e^{3} + 175 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (c^{6} d^{6} x^{2} + 2 \, a c^{5} d^{5} e x + a^{2} c^{4} d^{4} e^{2}\right )}}, -\frac {105 \, {\left (a^{2} c d^{2} e^{4} - a^{3} e^{6} + {\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (8 \, c^{3} d^{3} e^{3} x^{3} - 6 \, c^{3} d^{6} - 21 \, a c^{2} d^{4} e^{2} + 140 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 8 \, {\left (10 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} - {\left (39 \, c^{3} d^{5} e - 238 \, a c^{2} d^{3} e^{3} + 175 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{12 \, {\left (c^{6} d^{6} x^{2} + 2 \, a c^{5} d^{5} e x + a^{2} c^{4} d^{4} e^{2}\right )}}\right ] \] Input:

integrate((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= 
"fricas")
 

Output:

[-1/24*(105*(a^2*c*d^2*e^4 - a^3*e^6 + (c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 + 
 2*(a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e 
*x + 2*c*d^2 - a*e^2 + 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c 
*d*x + a*e)) - 2*(8*c^3*d^3*e^3*x^3 - 6*c^3*d^6 - 21*a*c^2*d^4*e^2 + 140*a 
^2*c*d^2*e^4 - 105*a^3*e^6 + 8*(10*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 - (3 
9*c^3*d^5*e - 238*a*c^2*d^3*e^3 + 175*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^6* 
d^6*x^2 + 2*a*c^5*d^5*e*x + a^2*c^4*d^4*e^2), -1/12*(105*(a^2*c*d^2*e^4 - 
a^3*e^6 + (c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 + 2*(a*c^2*d^3*e^3 - a^2*c*d*e 
^5)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c*d*sqrt(-(c*d^2 
 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (8*c^3*d^3*e^3*x^3 - 6*c^3*d^6 - 21*a* 
c^2*d^4*e^2 + 140*a^2*c*d^2*e^4 - 105*a^3*e^6 + 8*(10*c^3*d^4*e^2 - 7*a*c^ 
2*d^2*e^4)*x^2 - (39*c^3*d^5*e - 238*a*c^2*d^3*e^3 + 175*a^2*c*d*e^5)*x)*s 
qrt(e*x + d))/(c^6*d^6*x^2 + 2*a*c^5*d^5*e*x + a^2*c^4*d^4*e^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(13/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= 
"maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {35 \, {\left (c^{2} d^{4} e^{2} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} c^{4} d^{4}} - \frac {13 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{3} d^{5} e^{2} - 11 \, \sqrt {e x + d} c^{3} d^{6} e^{2} - 26 \, {\left (e x + d\right )}^{\frac {3}{2}} a c^{2} d^{3} e^{4} + 33 \, \sqrt {e x + d} a c^{2} d^{4} e^{4} + 13 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} c d e^{6} - 33 \, \sqrt {e x + d} a^{2} c d^{2} e^{6} + 11 \, \sqrt {e x + d} a^{3} e^{8}}{4 \, {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{4} d^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{6} d^{6} e^{2} + 9 \, \sqrt {e x + d} c^{6} d^{7} e^{2} - 9 \, \sqrt {e x + d} a c^{5} d^{5} e^{4}\right )}}{3 \, c^{9} d^{9}} \] Input:

integrate((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= 
"giac")
 

Output:

35/4*(c^2*d^4*e^2 - 2*a*c*d^2*e^4 + a^2*e^6)*arctan(sqrt(e*x + d)*c*d/sqrt 
(-c^2*d^3 + a*c*d*e^2))/(sqrt(-c^2*d^3 + a*c*d*e^2)*c^4*d^4) - 1/4*(13*(e* 
x + d)^(3/2)*c^3*d^5*e^2 - 11*sqrt(e*x + d)*c^3*d^6*e^2 - 26*(e*x + d)^(3/ 
2)*a*c^2*d^3*e^4 + 33*sqrt(e*x + d)*a*c^2*d^4*e^4 + 13*(e*x + d)^(3/2)*a^2 
*c*d*e^6 - 33*sqrt(e*x + d)*a^2*c*d^2*e^6 + 11*sqrt(e*x + d)*a^3*e^8)/(((e 
*x + d)*c*d - c*d^2 + a*e^2)^2*c^4*d^4) + 2/3*((e*x + d)^(3/2)*c^6*d^6*e^2 
 + 9*sqrt(e*x + d)*c^6*d^7*e^2 - 9*sqrt(e*x + d)*a*c^5*d^5*e^4)/(c^9*d^9)
 

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.72 \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2\,e^2\,{\left (d+e\,x\right )}^{3/2}}{3\,c^3\,d^3}-\frac {{\left (d+e\,x\right )}^{3/2}\,\left (\frac {13\,a^2\,c\,d\,e^6}{4}-\frac {13\,a\,c^2\,d^3\,e^4}{2}+\frac {13\,c^3\,d^5\,e^2}{4}\right )+\sqrt {d+e\,x}\,\left (\frac {11\,a^3\,e^8}{4}-\frac {33\,a^2\,c\,d^2\,e^6}{4}+\frac {33\,a\,c^2\,d^4\,e^4}{4}-\frac {11\,c^3\,d^6\,e^2}{4}\right )}{c^6\,d^8-\left (2\,c^6\,d^7-2\,a\,c^5\,d^5\,e^2\right )\,\left (d+e\,x\right )+c^6\,d^6\,{\left (d+e\,x\right )}^2-2\,a\,c^5\,d^6\,e^2+a^2\,c^4\,d^4\,e^4}+\frac {2\,e^2\,\left (3\,c^3\,d^4-3\,a\,c^2\,d^2\,e^2\right )\,\sqrt {d+e\,x}}{c^6\,d^6}+\frac {35\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,{\left (a\,e^2-c\,d^2\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^6-2\,a\,c\,d^2\,e^4+c^2\,d^4\,e^2}\right )\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}{4\,c^{9/2}\,d^{9/2}} \] Input:

int((d + e*x)^(13/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)
 

Output:

(2*e^2*(d + e*x)^(3/2))/(3*c^3*d^3) - ((d + e*x)^(3/2)*((13*c^3*d^5*e^2)/4 
 - (13*a*c^2*d^3*e^4)/2 + (13*a^2*c*d*e^6)/4) + (d + e*x)^(1/2)*((11*a^3*e 
^8)/4 - (11*c^3*d^6*e^2)/4 + (33*a*c^2*d^4*e^4)/4 - (33*a^2*c*d^2*e^6)/4)) 
/(c^6*d^8 - (2*c^6*d^7 - 2*a*c^5*d^5*e^2)*(d + e*x) + c^6*d^6*(d + e*x)^2 
- 2*a*c^5*d^6*e^2 + a^2*c^4*d^4*e^4) + (2*e^2*(3*c^3*d^4 - 3*a*c^2*d^2*e^2 
)*(d + e*x)^(1/2))/(c^6*d^6) + (35*e^2*atan((c^(1/2)*d^(1/2)*e^2*(a*e^2 - 
c*d^2)^(3/2)*(d + e*x)^(1/2))/(a^2*e^6 + c^2*d^4*e^2 - 2*a*c*d^2*e^4))*(a* 
e^2 - c*d^2)^(3/2))/(4*c^(9/2)*d^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.19 \[ \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{3} e^{6}-105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} c \,d^{2} e^{4}+210 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} c d \,e^{5} x -210 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a \,c^{2} d^{3} e^{3} x +105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a \,c^{2} d^{2} e^{4} x^{2}-105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{3} d^{4} e^{2} x^{2}-105 \sqrt {e x +d}\, a^{3} c d \,e^{6}+140 \sqrt {e x +d}\, a^{2} c^{2} d^{3} e^{4}-175 \sqrt {e x +d}\, a^{2} c^{2} d^{2} e^{5} x -21 \sqrt {e x +d}\, a \,c^{3} d^{5} e^{2}+238 \sqrt {e x +d}\, a \,c^{3} d^{4} e^{3} x -56 \sqrt {e x +d}\, a \,c^{3} d^{3} e^{4} x^{2}-6 \sqrt {e x +d}\, c^{4} d^{7}-39 \sqrt {e x +d}\, c^{4} d^{6} e x +80 \sqrt {e x +d}\, c^{4} d^{5} e^{2} x^{2}+8 \sqrt {e x +d}\, c^{4} d^{4} e^{3} x^{3}}{12 c^{5} d^{5} \left (c^{2} d^{2} x^{2}+2 a c d e x +a^{2} e^{2}\right )} \] Input:

int((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)
 

Output:

(105*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt( 
d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**3*e**6 - 105*sqrt(d)*sqrt(c)*sqrt(a* 
e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d 
**2)))*a**2*c*d**2*e**4 + 210*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan(( 
sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**2*c*d*e**5* 
x - 210*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sq 
rt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a*c**2*d**3*e**3*x + 105*sqrt(d)*sqr 
t(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt( 
a*e**2 - c*d**2)))*a*c**2*d**2*e**4*x**2 - 105*sqrt(d)*sqrt(c)*sqrt(a*e**2 
 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2) 
))*c**3*d**4*e**2*x**2 - 105*sqrt(d + e*x)*a**3*c*d*e**6 + 140*sqrt(d + e* 
x)*a**2*c**2*d**3*e**4 - 175*sqrt(d + e*x)*a**2*c**2*d**2*e**5*x - 21*sqrt 
(d + e*x)*a*c**3*d**5*e**2 + 238*sqrt(d + e*x)*a*c**3*d**4*e**3*x - 56*sqr 
t(d + e*x)*a*c**3*d**3*e**4*x**2 - 6*sqrt(d + e*x)*c**4*d**7 - 39*sqrt(d + 
 e*x)*c**4*d**6*e*x + 80*sqrt(d + e*x)*c**4*d**5*e**2*x**2 + 8*sqrt(d + e* 
x)*c**4*d**4*e**3*x**3)/(12*c**5*d**5*(a**2*e**2 + 2*a*c*d*e*x + c**2*d**2 
*x**2))