\(\int \frac {(c+d x^2)^3}{\sqrt {a+b x^2}} \, dx\) [88]

Optimal result

Integrand size = 21, antiderivative size = 161 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\frac {d \left (24 b^2 c^2-18 a b c d+5 a^2 d^2\right ) x \sqrt {a+b x^2}}{16 b^3}+\frac {d^2 (18 b c-5 a d) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {d^3 x^5 \sqrt {a+b x^2}}{6 b}+\frac {(2 b c-a d) \left (8 b^2 c^2-8 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \] Output:

1/16*d*(5*a^2*d^2-18*a*b*c*d+24*b^2*c^2)*x*(b*x^2+a)^(1/2)/b^3+1/24*d^2*(- 
5*a*d+18*b*c)*x^3*(b*x^2+a)^(1/2)/b^2+1/6*d^3*x^5*(b*x^2+a)^(1/2)/b+1/16*( 
-a*d+2*b*c)*(5*a^2*d^2-8*a*b*c*d+8*b^2*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1 
/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b} d x \sqrt {a+b x^2} \left (15 a^2 d^2-2 a b d \left (27 c+5 d x^2\right )+4 b^2 \left (18 c^2+9 c d x^2+2 d^2 x^4\right )\right )+\left (-48 b^3 c^3+72 a b^2 c^2 d-54 a^2 b c d^2+15 a^3 d^3\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{48 b^{7/2}} \] Input:

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

Output:

(Sqrt[b]*d*x*Sqrt[a + b*x^2]*(15*a^2*d^2 - 2*a*b*d*(27*c + 5*d*x^2) + 4*b^ 
2*(18*c^2 + 9*c*d*x^2 + 2*d^2*x^4)) + (-48*b^3*c^3 + 72*a*b^2*c^2*d - 54*a 
^2*b*c*d^2 + 15*a^3*d^3)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(48*b^(7/2))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {318, 403, 299, 224, 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.

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

\(\Big \downarrow \) 318

\(\displaystyle \frac {\int \frac {\left (d x^2+c\right ) \left (5 d (2 b c-a d) x^2+c (6 b c-a d)\right )}{\sqrt {b x^2+a}}dx}{6 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {\int \frac {d \left (44 b^2 c^2-44 a b d c+15 a^2 d^2\right ) x^2+c \left (24 b^2 c^2-14 a b d c+5 a^2 d^2\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {5 d x \sqrt {a+b x^2} \left (c+d x^2\right ) (2 b c-a d)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {3 (2 b c-a d) \left (5 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {d x \sqrt {a+b x^2} \left (15 a^2 d^2-44 a b c d+44 b^2 c^2\right )}{2 b}}{4 b}+\frac {5 d x \sqrt {a+b x^2} \left (c+d x^2\right ) (2 b c-a d)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {3 (2 b c-a d) \left (5 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {d x \sqrt {a+b x^2} \left (15 a^2 d^2-44 a b c d+44 b^2 c^2\right )}{2 b}}{4 b}+\frac {5 d x \sqrt {a+b x^2} \left (c+d x^2\right ) (2 b c-a d)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (2 b c-a d) \left (5 a^2 d^2-8 a b c d+8 b^2 c^2\right )}{2 b^{3/2}}+\frac {d x \sqrt {a+b x^2} \left (15 a^2 d^2-44 a b c d+44 b^2 c^2\right )}{2 b}}{4 b}+\frac {5 d x \sqrt {a+b x^2} \left (c+d x^2\right ) (2 b c-a d)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )^2}{6 b}\)

Input:

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

Output:

(d*x*Sqrt[a + b*x^2]*(c + d*x^2)^2)/(6*b) + ((5*d*(2*b*c - a*d)*x*Sqrt[a + 
 b*x^2]*(c + d*x^2))/(4*b) + ((d*(44*b^2*c^2 - 44*a*b*c*d + 15*a^2*d^2)*x* 
Sqrt[a + b*x^2])/(2*b) + (3*(2*b*c - a*d)*(8*b^2*c^2 - 8*a*b*c*d + 5*a^2*d 
^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/(4*b))/(6*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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(-\frac {5 \left (\left (a^{2} d^{2}-\frac {8}{5} a b c d +\frac {8}{5} b^{2} c^{2}\right ) \left (a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-d \sqrt {b \,x^{2}+a}\, x \left (\left (\frac {8}{15} d^{2} x^{4}+\frac {12}{5} c d \,x^{2}+\frac {24}{5} c^{2}\right ) b^{\frac {5}{2}}+a d \left (\left (-\frac {2 x^{2} d}{3}-\frac {18 c}{5}\right ) b^{\frac {3}{2}}+a d \sqrt {b}\right )\right )\right )}{16 b^{\frac {7}{2}}}\)	\(118\)
risch	\(\frac {x d \left (8 b^{2} d^{2} x^{4}-10 x^{2} a b \,d^{2}+36 x^{2} b^{2} c d +15 a^{2} d^{2}-54 a b c d +72 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}}{48 b^{3}}-\frac {\left (5 a^{3} d^{3}-18 a^{2} b c \,d^{2}+24 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\)	\(130\)
default	\(\frac {c^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+d^{3} \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+3 c \,d^{2} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+3 c^{2} d \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\)	\(227\)

Input:

int((d*x^2+c)^3/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-5/16*((a^2*d^2-8/5*a*b*c*d+8/5*b^2*c^2)*(a*d-2*b*c)*arctanh((b*x^2+a)^(1/ 
2)/x/b^(1/2))-d*(b*x^2+a)^(1/2)*x*((8/15*d^2*x^4+12/5*c*d*x^2+24/5*c^2)*b^ 
(5/2)+a*d*((-2/3*x^2*d-18/5*c)*b^(3/2)+a*d*b^(1/2))))/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.86 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (16 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{5} + 2 \, {\left (18 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{3} + 3 \, {\left (24 \, b^{3} c^{2} d - 18 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, -\frac {3 \, {\left (16 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d^{3} x^{5} + 2 \, {\left (18 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{3} + 3 \, {\left (24 \, b^{3} c^{2} d - 18 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \] Input:

integrate((d*x^2+c)^3/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/96*(3*(16*b^3*c^3 - 24*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt( 
b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*b^3*d^3*x^5 + 2* 
(18*b^3*c*d^2 - 5*a*b^2*d^3)*x^3 + 3*(24*b^3*c^2*d - 18*a*b^2*c*d^2 + 5*a^ 
2*b*d^3)*x)*sqrt(b*x^2 + a))/b^4, -1/48*(3*(16*b^3*c^3 - 24*a*b^2*c^2*d + 
18*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 
(8*b^3*d^3*x^5 + 2*(18*b^3*c*d^2 - 5*a*b^2*d^3)*x^3 + 3*(24*b^3*c^2*d - 18 
*a*b^2*c*d^2 + 5*a^2*b*d^3)*x)*sqrt(b*x^2 + a))/b^4]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {d^{3} x^{5}}{6 b} + \frac {x^{3} \left (- \frac {5 a d^{3}}{6 b} + 3 c d^{2}\right )}{4 b} + \frac {x \left (- \frac {3 a \left (- \frac {5 a d^{3}}{6 b} + 3 c d^{2}\right )}{4 b} + 3 c^{2} d\right )}{2 b}\right ) + \left (- \frac {a \left (- \frac {3 a \left (- \frac {5 a d^{3}}{6 b} + 3 c d^{2}\right )}{4 b} + 3 c^{2} d\right )}{2 b} + c^{3}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {c^{3} x + c^{2} d x^{3} + \frac {3 c d^{2} x^{5}}{5} + \frac {d^{3} x^{7}}{7}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((d*x**2+c)**3/(b*x**2+a)**(1/2),x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(d**3*x**5/(6*b) + x**3*(-5*a*d**3/(6*b) + 3*c 
*d**2)/(4*b) + x*(-3*a*(-5*a*d**3/(6*b) + 3*c*d**2)/(4*b) + 3*c**2*d)/(2*b 
)) + (-a*(-3*a*(-5*a*d**3/(6*b) + 3*c*d**2)/(4*b) + 3*c**2*d)/(2*b) + c**3 
)*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), ( 
x*log(x)/sqrt(b*x**2), True)), Ne(b, 0)), ((c**3*x + c**2*d*x**3 + 3*c*d** 
2*x**5/5 + d**3*x**7/7)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d^{3} x^{5}}{6 \, b} + \frac {3 \, \sqrt {b x^{2} + a} c d^{2} x^{3}}{4 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a d^{3} x^{3}}{24 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} c^{2} d x}{2 \, b} - \frac {9 \, \sqrt {b x^{2} + a} a c d^{2} x}{8 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a} a^{2} d^{3} x}{16 \, b^{3}} + \frac {c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {3 \, a c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {9 \, a^{2} c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {5 \, a^{3} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

integrate((d*x^2+c)^3/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

1/6*sqrt(b*x^2 + a)*d^3*x^5/b + 3/4*sqrt(b*x^2 + a)*c*d^2*x^3/b - 5/24*sqr 
t(b*x^2 + a)*a*d^3*x^3/b^2 + 3/2*sqrt(b*x^2 + a)*c^2*d*x/b - 9/8*sqrt(b*x^ 
2 + a)*a*c*d^2*x/b^2 + 5/16*sqrt(b*x^2 + a)*a^2*d^3*x/b^3 + c^3*arcsinh(b* 
x/sqrt(a*b))/sqrt(b) - 3/2*a*c^2*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 9/8*a^ 
2*c*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 5/16*a^3*d^3*arcsinh(b*x/sqrt(a*b 
))/b^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, d^{3} x^{2}}{b} + \frac {18 \, b^{4} c d^{2} - 5 \, a b^{3} d^{3}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (24 \, b^{4} c^{2} d - 18 \, a b^{3} c d^{2} + 5 \, a^{2} b^{2} d^{3}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (16 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

integrate((d*x^2+c)^3/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

1/48*(2*(4*d^3*x^2/b + (18*b^4*c*d^2 - 5*a*b^3*d^3)/b^5)*x^2 + 3*(24*b^4*c 
^2*d - 18*a*b^3*c*d^2 + 5*a^2*b^2*d^3)/b^5)*sqrt(b*x^2 + a)*x - 1/16*(16*b 
^3*c^3 - 24*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b)*x + 
 sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

int((c + d*x^2)^3/(a + b*x^2)^(1/2),x)
 

Output:

int((c + d*x^2)^3/(a + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.49 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt {a+b x^2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3} x -54 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2} x -10 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{3} x^{3}+72 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} d x +36 \sqrt {b \,x^{2}+a}\, b^{3} c \,d^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} d^{3} x^{5}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d^{3}+54 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c \,d^{2}-72 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c^{2} d +48 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c^{3}}{48 b^{4}} \] Input:

int((d*x^2+c)^3/(b*x^2+a)^(1/2),x)
 

Output:

(15*sqrt(a + b*x**2)*a**2*b*d**3*x - 54*sqrt(a + b*x**2)*a*b**2*c*d**2*x - 
 10*sqrt(a + b*x**2)*a*b**2*d**3*x**3 + 72*sqrt(a + b*x**2)*b**3*c**2*d*x 
+ 36*sqrt(a + b*x**2)*b**3*c*d**2*x**3 + 8*sqrt(a + b*x**2)*b**3*d**3*x**5 
 - 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*d**3 + 54*s 
qrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*c*d**2 - 72*sqrt 
(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c**2*d + 48*sqrt(b) 
*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**3*c**3)/(48*b**4)