\(\int \frac {x (c+d x)}{(a x^2+b x^3)^{5/2}} \, dx\) [298]

Optimal result

Integrand size = 22, antiderivative size = 211 \[ \int \frac {x (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 b^2 (b c-a d) x^3}{3 a^4 \left (a x^2+b x^3\right )^{3/2}}-\frac {2 b^2 (4 b c-3 a d) x}{a^5 \sqrt {a x^2+b x^3}}-\frac {c \sqrt {a x^2+b x^3}}{3 a^3 x^4}+\frac {(17 b c-6 a d) \sqrt {a x^2+b x^3}}{12 a^4 x^3}-\frac {b (41 b c-22 a d) \sqrt {a x^2+b x^3}}{8 a^5 x^2}+\frac {35 b^2 (3 b c-2 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{8 a^{11/2}} \] Output:

-2/3*b^2*(-a*d+b*c)*x^3/a^4/(b*x^3+a*x^2)^(3/2)-2*b^2*(-3*a*d+4*b*c)*x/a^5 
/(b*x^3+a*x^2)^(1/2)-1/3*c*(b*x^3+a*x^2)^(1/2)/a^3/x^4+1/12*(-6*a*d+17*b*c 
)*(b*x^3+a*x^2)^(1/2)/a^4/x^3-1/8*b*(-22*a*d+41*b*c)*(b*x^3+a*x^2)^(1/2)/a 
^5/x^2+35/8*b^2*(-2*a*d+3*b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(1 
1/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.68 \[ \int \frac {x (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {a} \left (-315 b^4 c x^4+210 a b^3 x^3 (-2 c+d x)-4 a^4 (2 c+3 d x)+6 a^3 b x (3 c+7 d x)+7 a^2 b^2 x^2 (-9 c+40 d x)\right )+105 b^2 (3 b c-2 a d) x^3 (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{11/2} \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(Sqrt[a]*(-315*b^4*c*x^4 + 210*a*b^3*x^3*(-2*c + d*x) - 4*a^4*(2*c + 3*d*x 
) + 6*a^3*b*x*(3*c + 7*d*x) + 7*a^2*b^2*x^2*(-9*c + 40*d*x)) + 105*b^2*(3* 
b*c - 2*a*d)*x^3*(a + b*x)^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(24*a^(11 
/2)*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1944, 1929, 1912, 1931, 1931, 1914, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1912

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 1912

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

rule 1914

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

rule 1929

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

rule 1931

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

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.21

method	result	size
pseudoelliptic	\(\frac {2 \left (d \,x^{2}-c x \right ) b^{2}-\frac {4 a \left (-6 d x +c \right ) b}{3}+\frac {16 a^{2} d}{3}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\)	\(45\)
risch	\(-\frac {\left (b x +a \right ) \left (-66 a b d \,x^{2}+123 b^{2} c \,x^{2}+12 a^{2} d x -34 a b c x +8 a^{2} c \right )}{24 a^{5} x^{2} \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{2} \left (-\frac {2 \left (70 a d -105 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \left (-48 a d +64 b c \right )}{\sqrt {b x +a}}+\frac {32 a \left (a d -b c \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}\right ) \sqrt {b x +a}\, x}{16 a^{5} \sqrt {x^{2} \left (b x +a \right )}}\)	\(153\)
default	\(-\frac {x^{2} \left (b x +a \right ) \left (210 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} d \,x^{3}-315 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} c \,x^{3}+12 a^{\frac {9}{2}} d x -42 a^{\frac {7}{2}} b d \,x^{2}-280 a^{\frac {5}{2}} b^{2} d \,x^{3}-210 a^{\frac {3}{2}} b^{3} d \,x^{4}+8 a^{\frac {9}{2}} c -18 a^{\frac {7}{2}} b c x +63 a^{\frac {5}{2}} b^{2} c \,x^{2}+420 a^{\frac {3}{2}} b^{3} c \,x^{3}+315 \sqrt {a}\, b^{4} c \,x^{4}\right )}{24 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{\frac {11}{2}}}\)	\(176\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.28 \[ \int \frac {x (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [-\frac {105 \, {\left ({\left (3 \, b^{5} c - 2 \, a b^{4} d\right )} x^{6} + 2 \, {\left (3 \, a b^{4} c - 2 \, a^{2} b^{3} d\right )} x^{5} + {\left (3 \, a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, a^{5} c + 105 \, {\left (3 \, a b^{4} c - 2 \, a^{2} b^{3} d\right )} x^{4} + 140 \, {\left (3 \, a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} x^{3} + 21 \, {\left (3 \, a^{3} b^{2} c - 2 \, a^{4} b d\right )} x^{2} - 6 \, {\left (3 \, a^{4} b c - 2 \, a^{5} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}}, -\frac {105 \, {\left ({\left (3 \, b^{5} c - 2 \, a b^{4} d\right )} x^{6} + 2 \, {\left (3 \, a b^{4} c - 2 \, a^{2} b^{3} d\right )} x^{5} + {\left (3 \, a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, a^{5} c + 105 \, {\left (3 \, a b^{4} c - 2 \, a^{2} b^{3} d\right )} x^{4} + 140 \, {\left (3 \, a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} x^{3} + 21 \, {\left (3 \, a^{3} b^{2} c - 2 \, a^{4} b d\right )} x^{2} - 6 \, {\left (3 \, a^{4} b c - 2 \, a^{5} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}}\right ] \] Input:

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

Output:

[-1/48*(105*((3*b^5*c - 2*a*b^4*d)*x^6 + 2*(3*a*b^4*c - 2*a^2*b^3*d)*x^5 + 
 (3*a^2*b^3*c - 2*a^3*b^2*d)*x^4)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^ 
3 + a*x^2)*sqrt(a))/x^2) + 2*(8*a^5*c + 105*(3*a*b^4*c - 2*a^2*b^3*d)*x^4 
+ 140*(3*a^2*b^3*c - 2*a^3*b^2*d)*x^3 + 21*(3*a^3*b^2*c - 2*a^4*b*d)*x^2 - 
 6*(3*a^4*b*c - 2*a^5*d)*x)*sqrt(b*x^3 + a*x^2))/(a^6*b^2*x^6 + 2*a^7*b*x^ 
5 + a^8*x^4), -1/24*(105*((3*b^5*c - 2*a*b^4*d)*x^6 + 2*(3*a*b^4*c - 2*a^2 
*b^3*d)*x^5 + (3*a^2*b^3*c - 2*a^3*b^2*d)*x^4)*sqrt(-a)*arctan(sqrt(b*x^3 
+ a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (8*a^5*c + 105*(3*a*b^4*c - 2*a^2*b^3*d 
)*x^4 + 140*(3*a^2*b^3*c - 2*a^3*b^2*d)*x^3 + 21*(3*a^3*b^2*c - 2*a^4*b*d) 
*x^2 - 6*(3*a^4*b*c - 2*a^5*d)*x)*sqrt(b*x^3 + a*x^2))/(a^6*b^2*x^6 + 2*a^ 
7*b*x^5 + a^8*x^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.99 \[ \int \frac {x (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {35 \, {\left (3 \, b^{3} c - 2 \, a b^{2} d\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (x\right )} - \frac {315 \, {\left (b x + a\right )}^{4} b^{3} c - 840 \, {\left (b x + a\right )}^{3} a b^{3} c + 693 \, {\left (b x + a\right )}^{2} a^{2} b^{3} c - 144 \, {\left (b x + a\right )} a^{3} b^{3} c - 16 \, a^{4} b^{3} c - 210 \, {\left (b x + a\right )}^{4} a b^{2} d + 560 \, {\left (b x + a\right )}^{3} a^{2} b^{2} d - 462 \, {\left (b x + a\right )}^{2} a^{3} b^{2} d + 96 \, {\left (b x + a\right )} a^{4} b^{2} d + 16 \, a^{5} b^{2} d}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

-35/8*(3*b^3*c - 2*a*b^2*d)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5*s 
gn(x)) - 1/24*(315*(b*x + a)^4*b^3*c - 840*(b*x + a)^3*a*b^3*c + 693*(b*x 
+ a)^2*a^2*b^3*c - 144*(b*x + a)*a^3*b^3*c - 16*a^4*b^3*c - 210*(b*x + a)^ 
4*a*b^2*d + 560*(b*x + a)^3*a^2*b^2*d - 462*(b*x + a)^2*a^3*b^2*d + 96*(b* 
x + a)*a^4*b^2*d + 16*a^5*b^2*d)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)^3*a^ 
5*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.64 \[ \int \frac {x (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {210 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a^{2} b^{2} d \,x^{3}-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} c \,x^{3}+210 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} d \,x^{4}-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} c \,x^{4}-210 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a^{2} b^{2} d \,x^{3}+315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} c \,x^{3}-210 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} d \,x^{4}+315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} c \,x^{4}-16 a^{5} c -24 a^{5} d x +36 a^{4} b c x +84 a^{4} b d \,x^{2}-126 a^{3} b^{2} c \,x^{2}+560 a^{3} b^{2} d \,x^{3}-840 a^{2} b^{3} c \,x^{3}+420 a^{2} b^{3} d \,x^{4}-630 a \,b^{4} c \,x^{4}}{48 \sqrt {b x +a}\, a^{6} x^{3} \left (b x +a \right )} \] Input:

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

Output:

(210*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a**2*b**2*d*x**3 - 
 315*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**3*c*x**3 + 21 
0*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**3*d*x**4 - 315*s 
qrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**4*c*x**4 - 210*sqrt(a 
)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a**2*b**2*d*x**3 + 315*sqrt(a 
)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**3*c*x**3 - 210*sqrt(a)*s 
qrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**3*d*x**4 + 315*sqrt(a)*sqrt 
(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**4*c*x**4 - 16*a**5*c - 24*a**5*d 
*x + 36*a**4*b*c*x + 84*a**4*b*d*x**2 - 126*a**3*b**2*c*x**2 + 560*a**3*b* 
*2*d*x**3 - 840*a**2*b**3*c*x**3 + 420*a**2*b**3*d*x**4 - 630*a*b**4*c*x** 
4)/(48*sqrt(a + b*x)*a**6*x**3*(a + b*x))