\(\int \frac {(c+d x) (a x^2+b x^3)^{5/2}}{(e x)^{11/2}} \, dx\) [324]

Optimal result

Integrand size = 28, antiderivative size = 214 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\frac {11 a^2 (8 b c-a d) \sqrt {a x^2+b x^3}}{64 b e^5 \sqrt {e x}}+\frac {13 a (8 b c-a d) \sqrt {e x} \sqrt {a x^2+b x^3}}{96 e^6}+\frac {b (8 b c-a d) (e x)^{3/2} \sqrt {a x^2+b x^3}}{24 e^7}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}+\frac {5 a^3 (8 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{64 b^{3/2} e^{11/2}} \] Output:

11/64*a^2*(-a*d+8*b*c)*(b*x^3+a*x^2)^(1/2)/b/e^5/(e*x)^(1/2)+13/96*a*(-a*d 
+8*b*c)*(e*x)^(1/2)*(b*x^3+a*x^2)^(1/2)/e^6+1/24*b*(-a*d+8*b*c)*(e*x)^(3/2 
)*(b*x^3+a*x^2)^(1/2)/e^7+1/4*d*e*(b*x^3+a*x^2)^(7/2)/b/(e*x)^(13/2)+5/64* 
a^3*(-a*d+8*b*c)*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b*x^3+a*x^2)^(1/2))/ 
b^(3/2)/e^(11/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.72 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\frac {\sqrt {x} \sqrt {x^2 (a+b x)} \left (\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (15 a^3 d+16 b^3 x^2 (4 c+3 d x)+8 a b^2 x (26 c+17 d x)+2 a^2 b (132 c+59 d x)\right )+15 a^3 (-8 b c+a d) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{192 b^{3/2} e^4 (e x)^{3/2} \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/(e*x)^(11/2),x]
 

Output:

(Sqrt[x]*Sqrt[x^2*(a + b*x)]*(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(15*a^3*d + 16 
*b^3*x^2*(4*c + 3*d*x) + 8*a*b^2*x*(26*c + 17*d*x) + 2*a^2*b*(132*c + 59*d 
*x)) + 15*a^3*(-8*b*c + a*d)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]))/(19 
2*b^(3/2)*e^4*(e*x)^(3/2)*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1945, 1927, 1927, 1927, 1937, 1935, 219}

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.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2} (c+d x)}{(e x)^{11/2}} \, dx\)

\(\Big \downarrow \) 1945

\(\displaystyle \frac {(8 b c-a d) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{(e x)^{11/2}}dx}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(8 b c-a d) \left (\frac {5 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{(e x)^{7/2}}dx}{6 e^2}+\frac {\left (a x^2+b x^3\right )^{5/2}}{3 e (e x)^{9/2}}\right )}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(8 b c-a d) \left (\frac {5 a \left (\frac {3 a \int \frac {\sqrt {b x^3+a x^2}}{(e x)^{3/2}}dx}{4 e^2}+\frac {\left (a x^2+b x^3\right )^{3/2}}{2 e (e x)^{5/2}}\right )}{6 e^2}+\frac {\left (a x^2+b x^3\right )^{5/2}}{3 e (e x)^{9/2}}\right )}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(8 b c-a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \int \frac {\sqrt {e x}}{\sqrt {b x^3+a x^2}}dx}{2 e^2}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{4 e^2}+\frac {\left (a x^2+b x^3\right )^{3/2}}{2 e (e x)^{5/2}}\right )}{6 e^2}+\frac {\left (a x^2+b x^3\right )^{5/2}}{3 e (e x)^{9/2}}\right )}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

\(\Big \downarrow \) 1937

\(\displaystyle \frac {(8 b c-a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \sqrt {e x} \int \frac {\sqrt {x}}{\sqrt {b x^3+a x^2}}dx}{2 e^2 \sqrt {x}}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{4 e^2}+\frac {\left (a x^2+b x^3\right )^{3/2}}{2 e (e x)^{5/2}}\right )}{6 e^2}+\frac {\left (a x^2+b x^3\right )^{5/2}}{3 e (e x)^{9/2}}\right )}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

\(\Big \downarrow \) 1935

\(\displaystyle \frac {(8 b c-a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \sqrt {e x} \int \frac {1}{1-\frac {b x^3}{b x^3+a x^2}}d\frac {x^{3/2}}{\sqrt {b x^3+a x^2}}}{e^2 \sqrt {x}}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{4 e^2}+\frac {\left (a x^2+b x^3\right )^{3/2}}{2 e (e x)^{5/2}}\right )}{6 e^2}+\frac {\left (a x^2+b x^3\right )^{5/2}}{3 e (e x)^{9/2}}\right )}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(8 b c-a d) \left (\frac {5 a \left (\frac {3 a \left (\frac {a \sqrt {e x} \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {b} e^2 \sqrt {x}}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{4 e^2}+\frac {\left (a x^2+b x^3\right )^{3/2}}{2 e (e x)^{5/2}}\right )}{6 e^2}+\frac {\left (a x^2+b x^3\right )^{5/2}}{3 e (e x)^{9/2}}\right )}{8 b}+\frac {d e \left (a x^2+b x^3\right )^{7/2}}{4 b (e x)^{13/2}}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/(e*x)^(11/2),x]
 

Output:

(d*e*(a*x^2 + b*x^3)^(7/2))/(4*b*(e*x)^(13/2)) + ((8*b*c - a*d)*((a*x^2 + 
b*x^3)^(5/2)/(3*e*(e*x)^(9/2)) + (5*a*((a*x^2 + b*x^3)^(3/2)/(2*e*(e*x)^(5 
/2)) + (3*a*(Sqrt[a*x^2 + b*x^3]/(e*Sqrt[e*x]) + (a*Sqrt[e*x]*ArcTanh[(Sqr 
t[b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(Sqrt[b]*e^2*Sqrt[x])))/(4*e^2)))/(6*e 
^2)))/(8*b)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1935

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp 
[-2/(n - j)   Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], 
 x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
 

rule 1937

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
:> Simp[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a*x^j + b 
*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2] && 
NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]
 

rule 1945

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* 
p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1))   Int[(e* 
x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, 
x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p 
*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\left (48 b^{3} d \,x^{3}+136 a \,b^{2} d \,x^{2}+64 b^{3} c \,x^{2}+118 a^{2} b d x +208 a \,b^{2} c x +15 a^{3} d +264 c \,a^{2} b \right ) \sqrt {x^{2} \left (b x +a \right )}}{192 b \,e^{5} \sqrt {e x}}-\frac {5 a^{3} \left (a d -8 b c \right ) \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right ) \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x \left (b x +a \right )}}{128 b \sqrt {b e}\, e^{5} x \left (b x +a \right ) \sqrt {e x}}\)	\(174\)
default	\(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (96 b^{3} d \,x^{3} \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+272 a \,b^{2} d \,x^{2} \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+128 b^{3} c \,x^{2} \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}-15 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{4} d e +120 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{3} b c e +236 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a^{2} b d x +416 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a \,b^{2} c x +30 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a^{3} d +528 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a^{2} b c \right )}{384 x^{4} \left (b x +a \right )^{2} e^{5} b \sqrt {e x}\, \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}}\)	\(298\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/(e*x)^(11/2),x,method=_RETURNVERBOSE)
 

Output:

1/192/b*(48*b^3*d*x^3+136*a*b^2*d*x^2+64*b^3*c*x^2+118*a^2*b*d*x+208*a*b^2 
*c*x+15*a^3*d+264*a^2*b*c)/e^5*(x^2*(b*x+a))^(1/2)/(e*x)^(1/2)-5/128*a^3/b 
*(a*d-8*b*c)*ln((1/2*a*e+b*e*x)/(b*e)^(1/2)+(b*e*x^2+a*e*x)^(1/2))/(b*e)^( 
1/2)/e^5*(x^2*(b*x+a))^(1/2)/x/(b*x+a)*(e*x*(b*x+a))^(1/2)/(e*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.50 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\left [-\frac {15 \, {\left (8 \, a^{3} b c - a^{4} d\right )} \sqrt {b e} x \log \left (\frac {2 \, b e x^{2} + a e x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b e} \sqrt {e x}}{x}\right ) - 2 \, {\left (48 \, b^{4} d x^{3} + 264 \, a^{2} b^{2} c + 15 \, a^{3} b d + 8 \, {\left (8 \, b^{4} c + 17 \, a b^{3} d\right )} x^{2} + 2 \, {\left (104 \, a b^{3} c + 59 \, a^{2} b^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{384 \, b^{2} e^{6} x}, -\frac {15 \, {\left (8 \, a^{3} b c - a^{4} d\right )} \sqrt {-b e} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b e} \sqrt {e x}}{b e x^{2} + a e x}\right ) - {\left (48 \, b^{4} d x^{3} + 264 \, a^{2} b^{2} c + 15 \, a^{3} b d + 8 \, {\left (8 \, b^{4} c + 17 \, a b^{3} d\right )} x^{2} + 2 \, {\left (104 \, a b^{3} c + 59 \, a^{2} b^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{192 \, b^{2} e^{6} x}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/(e*x)^(11/2),x, algorithm="fricas")
 

Output:

[-1/384*(15*(8*a^3*b*c - a^4*d)*sqrt(b*e)*x*log((2*b*e*x^2 + a*e*x - 2*sqr 
t(b*x^3 + a*x^2)*sqrt(b*e)*sqrt(e*x))/x) - 2*(48*b^4*d*x^3 + 264*a^2*b^2*c 
 + 15*a^3*b*d + 8*(8*b^4*c + 17*a*b^3*d)*x^2 + 2*(104*a*b^3*c + 59*a^2*b^2 
*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(b^2*e^6*x), -1/192*(15*(8*a^3*b*c - 
 a^4*d)*sqrt(-b*e)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-b*e)*sqrt(e*x)/(b*e* 
x^2 + a*e*x)) - (48*b^4*d*x^3 + 264*a^2*b^2*c + 15*a^3*b*d + 8*(8*b^4*c + 
17*a*b^3*d)*x^2 + 2*(104*a*b^3*c + 59*a^2*b^2*d)*x)*sqrt(b*x^3 + a*x^2)*sq 
rt(e*x))/(b^2*e^6*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/(e*x)**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}}{\left (e x\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/(e*x)^(11/2),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(5/2)*(d*x + c)/(e*x)^(11/2), x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\frac {{\left (\sqrt {{\left (b x + a\right )} b e - a b e} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )} d \mathrm {sgn}\left (x\right )}{b^{2} e} + \frac {8 \, b^{3} c e \mathrm {sgn}\left (x\right ) - a b^{2} d e \mathrm {sgn}\left (x\right )}{b^{4} e^{2}}\right )} + \frac {5 \, {\left (8 \, a b^{3} c e \mathrm {sgn}\left (x\right ) - a^{2} b^{2} d e \mathrm {sgn}\left (x\right )\right )}}{b^{4} e^{2}}\right )} + \frac {15 \, {\left (8 \, a^{2} b^{3} c e \mathrm {sgn}\left (x\right ) - a^{3} b^{2} d e \mathrm {sgn}\left (x\right )\right )}}{b^{4} e^{2}}\right )} \sqrt {b x + a} - \frac {15 \, {\left (8 \, a^{3} b c \mathrm {sgn}\left (x\right ) - a^{4} d \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {{\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b}\right )} b}{192 \, e^{5} {\left | b \right |}} + \frac {5 \, {\left (8 \, a^{3} b c \log \left (\sqrt {b e} \sqrt {a}\right ) - a^{4} d \log \left (\sqrt {b e} \sqrt {a}\right )\right )} \mathrm {sgn}\left (x\right )}{64 \, \sqrt {b e} e^{5} {\left | b \right |}} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/(e*x)^(11/2),x, algorithm="giac")
 

Output:

1/192*(sqrt((b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)* 
d*sgn(x)/(b^2*e) + (8*b^3*c*e*sgn(x) - a*b^2*d*e*sgn(x))/(b^4*e^2)) + 5*(8 
*a*b^3*c*e*sgn(x) - a^2*b^2*d*e*sgn(x))/(b^4*e^2)) + 15*(8*a^2*b^3*c*e*sgn 
(x) - a^3*b^2*d*e*sgn(x))/(b^4*e^2))*sqrt(b*x + a) - 15*(8*a^3*b*c*sgn(x) 
- a^4*d*sgn(x))*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt((b*x + a)*b*e - a* 
b*e)))/(sqrt(b*e)*b))*b/(e^5*abs(b)) + 5/64*(8*a^3*b*c*log(sqrt(b*e)*sqrt( 
a)) - a^4*d*log(sqrt(b*e)*sqrt(a)))*sgn(x)/(sqrt(b*e)*e^5*abs(b))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}\,\left (c+d\,x\right )}{{\left (e\,x\right )}^{11/2}} \,d x \] Input:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/(e*x)^(11/2),x)
 

Output:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/(e*x)^(11/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{(e x)^{11/2}} \, dx=\frac {\sqrt {e}\, \left (15 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b d +264 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} c +118 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} d x +208 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c x +136 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} d \,x^{2}+64 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c \,x^{2}+48 \sqrt {x}\, \sqrt {b x +a}\, b^{4} d \,x^{3}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{4} d +120 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} b c \right )}{192 b^{2} e^{6}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/(e*x)^(11/2),x)
 

Output:

(sqrt(e)*(15*sqrt(x)*sqrt(a + b*x)*a**3*b*d + 264*sqrt(x)*sqrt(a + b*x)*a* 
*2*b**2*c + 118*sqrt(x)*sqrt(a + b*x)*a**2*b**2*d*x + 208*sqrt(x)*sqrt(a + 
 b*x)*a*b**3*c*x + 136*sqrt(x)*sqrt(a + b*x)*a*b**3*d*x**2 + 64*sqrt(x)*sq 
rt(a + b*x)*b**4*c*x**2 + 48*sqrt(x)*sqrt(a + b*x)*b**4*d*x**3 - 15*sqrt(b 
)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**4*d + 120*sqrt(b)*log( 
(sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**3*b*c))/(192*b**2*e**6)