\(\int \frac {(e x)^{11/2} (c+d x)}{(a x^2+b x^3)^{5/2}} \, dx\) [349]

Optimal result

Integrand size = 28, antiderivative size = 120 \[ \int \frac {(e x)^{11/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 (b c-a d) e (e x)^{9/2}}{3 a b \left (a x^2+b x^3\right )^{3/2}}-\frac {2 d e^4 (e x)^{3/2}}{b^2 \sqrt {a x^2+b x^3}}+\frac {2 d e^{11/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{b^{5/2}} \] Output:

2/3*(-a*d+b*c)*e*(e*x)^(9/2)/a/b/(b*x^3+a*x^2)^(3/2)-2*d*e^4*(e*x)^(3/2)/b 
^2/(b*x^3+a*x^2)^(1/2)+2*d*e^(11/2)*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b 
*x^3+a*x^2)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.90 \[ \int \frac {(e x)^{11/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 e^5 x^{5/2} \sqrt {e x} \left (\sqrt {b} \sqrt {x} \left (3 a^2 d-b^2 c x+4 a b d x\right )+3 a d (a+b x)^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{3 a b^{5/2} \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(-2*e^5*x^(5/2)*Sqrt[e*x]*(Sqrt[b]*Sqrt[x]*(3*a^2*d - b^2*c*x + 4*a*b*d*x) 
 + 3*a*d*(a + b*x)^(3/2)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]))/(3*a*b^ 
(5/2)*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1943, 1928, 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.

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

\(\Big \downarrow \) 1943

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

\(\Big \downarrow \) 1928

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

\(\Big \downarrow \) 1937

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

\(\Big \downarrow \) 1935

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*(b*c - a*d)*e*(e*x)^(9/2))/(3*a*b*(a*x^2 + b*x^3)^(3/2)) + (d*e^2*((-2* 
e^2*(e*x)^(3/2))/(b*Sqrt[a*x^2 + b*x^3]) + (2*e^3*Sqrt[e*x]*ArcTanh[(Sqrt[ 
b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(b^(3/2)*Sqrt[x])))/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 1928

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

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 1943

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(246\) vs. \(2(98)=196\).

Time = 0.40 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {x^{5} \left (b x +a \right ) \left (-3 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a \,b^{2} d e \,x^{2}-6 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{2} b d e x +8 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a b d x -2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, b^{2} c x -3 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{3} d e +6 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a^{2} d \right ) \sqrt {e x}\, e^{5}}{3 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \sqrt {b e}\, a \sqrt {e x \left (b x +a \right )}\, b^{2}}\)	\(247\)

Input:

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

Output:

-1/3*x^5*(b*x+a)*(-3*ln(1/2*(2*b*e*x+2*(e*x*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e 
)/(b*e)^(1/2))*a*b^2*d*e*x^2-6*ln(1/2*(2*b*e*x+2*(e*x*(b*x+a))^(1/2)*(b*e) 
^(1/2)+a*e)/(b*e)^(1/2))*a^2*b*d*e*x+8*(e*x*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b 
*d*x-2*(e*x*(b*x+a))^(1/2)*(b*e)^(1/2)*b^2*c*x-3*ln(1/2*(2*b*e*x+2*(e*x*(b 
*x+a))^(1/2)*(b*e)^(1/2)+a*e)/(b*e)^(1/2))*a^3*d*e+6*(e*x*(b*x+a))^(1/2)*( 
b*e)^(1/2)*a^2*d)*(e*x)^(1/2)*e^5/(b*x^3+a*x^2)^(5/2)/(b*e)^(1/2)/a/(e*x*( 
b*x+a))^(1/2)/b^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.80 \[ \int \frac {(e x)^{11/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a b^{2} d e^{5} x^{3} + 2 \, a^{2} b d e^{5} x^{2} + a^{3} d e^{5} x\right )} \sqrt {\frac {e}{b}} \log \left (\frac {2 \, b e x^{2} + a e x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {e x} b \sqrt {\frac {e}{b}}}{x}\right ) - 2 \, {\left (3 \, a^{2} d e^{5} - {\left (b^{2} c - 4 \, a b d\right )} e^{5} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{3 \, {\left (a b^{4} x^{3} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2} x\right )}}, -\frac {2 \, {\left (3 \, {\left (a b^{2} d e^{5} x^{3} + 2 \, a^{2} b d e^{5} x^{2} + a^{3} d e^{5} x\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {e x} b \sqrt {-\frac {e}{b}}}{b e x^{2} + a e x}\right ) + {\left (3 \, a^{2} d e^{5} - {\left (b^{2} c - 4 \, a b d\right )} e^{5} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}\right )}}{3 \, {\left (a b^{4} x^{3} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2} x\right )}}\right ] \] Input:

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

Output:

[1/3*(3*(a*b^2*d*e^5*x^3 + 2*a^2*b*d*e^5*x^2 + a^3*d*e^5*x)*sqrt(e/b)*log( 
(2*b*e*x^2 + a*e*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*sqrt(e/b))/x) - 2*( 
3*a^2*d*e^5 - (b^2*c - 4*a*b*d)*e^5*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(a*b 
^4*x^3 + 2*a^2*b^3*x^2 + a^3*b^2*x), -2/3*(3*(a*b^2*d*e^5*x^3 + 2*a^2*b*d* 
e^5*x^2 + a^3*d*e^5*x)*sqrt(-e/b)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*s 
qrt(-e/b)/(b*e*x^2 + a*e*x)) + (3*a^2*d*e^5 - (b^2*c - 4*a*b*d)*e^5*x)*sqr 
t(b*x^3 + a*x^2)*sqrt(e*x))/(a*b^4*x^3 + 2*a^2*b^3*x^2 + a^3*b^2*x)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.26 \[ \int \frac {(e x)^{11/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {d e^{5} {\left | e \right |} \log \left (e^{2} {\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {b e} b^{2}} - \frac {2 \, d e^{5} {\left | e \right |} \log \left ({\left | -\sqrt {b e} \sqrt {e x} + \sqrt {b e^{2} x + a e^{2}} \right |}\right )}{\sqrt {b e} b^{2} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (\frac {3 \, a d e^{7} {\left | e \right |}}{b^{2} \mathrm {sgn}\left (x\right )} - \frac {{\left (b^{3} c e^{9} {\left | e \right |} \mathrm {sgn}\left (x\right ) - 4 \, a b^{2} d e^{9} {\left | e \right |} \mathrm {sgn}\left (x\right )\right )} x}{a b^{3} e^{2}}\right )} \sqrt {e x}}{3 \, {\left (b e^{2} x + a e^{2}\right )}^{\frac {3}{2}}} \] Input:

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

Output:

d*e^5*abs(e)*log(e^2*abs(a))*sgn(x)/(sqrt(b*e)*b^2) - 2*d*e^5*abs(e)*log(a 
bs(-sqrt(b*e)*sqrt(e*x) + sqrt(b*e^2*x + a*e^2)))/(sqrt(b*e)*b^2*sgn(x)) - 
 2/3*(3*a*d*e^7*abs(e)/(b^2*sgn(x)) - (b^3*c*e^9*abs(e)*sgn(x) - 4*a*b^2*d 
*e^9*abs(e)*sgn(x))*x/(a*b^3*e^2))*sqrt(e*x)/(b*e^2*x + a*e^2)^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.22 \[ \int \frac {(e x)^{11/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {e}\, e^{5} \left (3 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} d +3 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a b d x +\sqrt {b}\, \sqrt {b x +a}\, a b c +\sqrt {b}\, \sqrt {b x +a}\, b^{2} c x -3 \sqrt {x}\, a^{2} b d -4 \sqrt {x}\, a \,b^{2} d x +\sqrt {x}\, b^{3} c x \right )}{3 \sqrt {b x +a}\, a \,b^{3} \left (b x +a \right )} \] Input:

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

Output:

(2*sqrt(e)*e**5*(3*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt 
(b))/sqrt(a))*a**2*d + 3*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x 
)*sqrt(b))/sqrt(a))*a*b*d*x + sqrt(b)*sqrt(a + b*x)*a*b*c + sqrt(b)*sqrt(a 
 + b*x)*b**2*c*x - 3*sqrt(x)*a**2*b*d - 4*sqrt(x)*a*b**2*d*x + sqrt(x)*b** 
3*c*x))/(3*sqrt(a + b*x)*a*b**3*(a + b*x))