\(\int \frac {(a+b x^2)^{5/2}}{x^4 (c+d x)} \, dx\) [1134]

Optimal result

Integrand size = 22, antiderivative size = 237 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\frac {b^2 \sqrt {a+b x^2}}{d}-\frac {a^2 \sqrt {a+b x^2}}{3 c x^3}+\frac {a^2 d \sqrt {a+b x^2}}{2 c^2 x^2}-\frac {a \left (7 b c^2+3 a d^2\right ) \sqrt {a+b x^2}}{3 c^3 x}-\frac {b^{5/2} c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{d^2}-\frac {\left (b c^2+a d^2\right )^{5/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{c^4 d^2}+\frac {a^{3/2} d \left (5 b c^2+2 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 c^4} \] Output:

b^2*(b*x^2+a)^(1/2)/d-1/3*a^2*(b*x^2+a)^(1/2)/c/x^3+1/2*a^2*d*(b*x^2+a)^(1 
/2)/c^2/x^2-1/3*a*(3*a*d^2+7*b*c^2)*(b*x^2+a)^(1/2)/c^3/x-b^(5/2)*c*arctan 
h(b^(1/2)*x/(b*x^2+a)^(1/2))/d^2-(a*d^2+b*c^2)^(5/2)*arctanh((-b*c*x+a*d)/ 
(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/c^4/d^2+1/2*a^(3/2)*d*(2*a*d^2+5*b*c^ 
2)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/c^4
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\frac {-\frac {c d \sqrt {a+b x^2} \left (14 a b c^2 d x^2-6 b^2 c^3 x^3+a^2 d \left (2 c^2-3 c d x+6 d^2 x^2\right )\right )}{x^3}+12 \left (-b c^2-a d^2\right )^{5/2} \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )-6 a^{3/2} d^3 \left (5 b c^2+2 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+6 b^{5/2} c^5 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 c^4 d^2} \] Input:

Integrate[(a + b*x^2)^(5/2)/(x^4*(c + d*x)),x]
 

Output:

(-((c*d*Sqrt[a + b*x^2]*(14*a*b*c^2*d*x^2 - 6*b^2*c^3*x^3 + a^2*d*(2*c^2 - 
 3*c*d*x + 6*d^2*x^2)))/x^3) + 12*(-(b*c^2) - a*d^2)^(5/2)*ArcTan[(Sqrt[b] 
*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a*d^2]] - 6*a^(3/2)*d^3*(5 
*b*c^2 + 2*a*d^2)*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] + 6*b^(5/ 
2)*c^5*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(6*c^4*d^2)
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(589\) vs. \(2(237)=474\).

Time = 1.40 (sec) , antiderivative size = 589, normalized size of antiderivative = 2.49, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {617, 2009}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx\)

\(\Big \downarrow \) 617

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^{5/2} d^3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{c^4}+\frac {5 a^{3/2} b d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 c^2}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 d^4+20 a b c^2 d^2+8 b^2 c^4\right )}{8 c^3 d^2}+\frac {15 a^2 \sqrt {b} d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 c^3}-\frac {a^2 d^3 \sqrt {a+b x^2}}{c^4}+\frac {5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 c}-\frac {\left (a d^2+b c^2\right )^{5/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{c^4 d^2}+\frac {5 b^2 x \sqrt {a+b x^2}}{2 c}-\frac {a d^3 \left (a+b x^2\right )^{3/2}}{3 c^4}-\frac {d^2 \left (a+b x^2\right )^{5/2}}{c^3 x}+\frac {5 b d^2 x \left (a+b x^2\right )^{3/2}}{4 c^3}+\frac {15 a b d^2 x \sqrt {a+b x^2}}{8 c^3}+\frac {d \left (a+b x^2\right )^{5/2}}{2 c^2 x^2}-\frac {5 b d \left (a+b x^2\right )^{3/2}}{6 c^2}-\frac {5 a b d \sqrt {a+b x^2}}{2 c^2}+\frac {d \left (a+b x^2\right )^{3/2} \left (4 \left (a d^2+b c^2\right )-3 b c d x\right )}{12 c^4}+\frac {\sqrt {a+b x^2} \left (8 \left (a d^2+b c^2\right )^2-b c d x \left (7 a d^2+4 b c^2\right )\right )}{8 c^4 d}-\frac {5 b \left (a+b x^2\right )^{3/2}}{3 c x}-\frac {\left (a+b x^2\right )^{5/2}}{3 c x^3}\)

Input:

Int[(a + b*x^2)^(5/2)/(x^4*(c + d*x)),x]
 

Output:

(-5*a*b*d*Sqrt[a + b*x^2])/(2*c^2) - (a^2*d^3*Sqrt[a + b*x^2])/c^4 + (5*b^ 
2*x*Sqrt[a + b*x^2])/(2*c) + (15*a*b*d^2*x*Sqrt[a + b*x^2])/(8*c^3) + ((8* 
(b*c^2 + a*d^2)^2 - b*c*d*(4*b*c^2 + 7*a*d^2)*x)*Sqrt[a + b*x^2])/(8*c^4*d 
) - (5*b*d*(a + b*x^2)^(3/2))/(6*c^2) - (a*d^3*(a + b*x^2)^(3/2))/(3*c^4) 
- (5*b*(a + b*x^2)^(3/2))/(3*c*x) + (5*b*d^2*x*(a + b*x^2)^(3/2))/(4*c^3) 
+ (d*(4*(b*c^2 + a*d^2) - 3*b*c*d*x)*(a + b*x^2)^(3/2))/(12*c^4) - (a + b* 
x^2)^(5/2)/(3*c*x^3) + (d*(a + b*x^2)^(5/2))/(2*c^2*x^2) - (d^2*(a + b*x^2 
)^(5/2))/(c^3*x) + (5*a*b^(3/2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*c 
) + (15*a^2*Sqrt[b]*d^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*c^3) - (S 
qrt[b]*(8*b^2*c^4 + 20*a*b*c^2*d^2 + 15*a^2*d^4)*ArcTanh[(Sqrt[b]*x)/Sqrt[ 
a + b*x^2]])/(8*c^3*d^2) - ((b*c^2 + a*d^2)^(5/2)*ArcTanh[(a*d - b*c*x)/(S 
qrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(c^4*d^2) + (5*a^(3/2)*b*d*ArcTanh[S 
qrt[a + b*x^2]/Sqrt[a]])/(2*c^2) + (a^(5/2)*d^3*ArcTanh[Sqrt[a + b*x^2]/Sq 
rt[a]])/c^4
 

Defintions of rubi rules used

rule 617

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[(a + b*x^2)^p, x^m*(c + d*x)^n, x], x] /; FreeQ[{ 
a, b, c, d, p}, x] && ILtQ[n, 0] && IntegerQ[m] && IntegerQ[2*p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.34

method	result	size
risch	\(-\frac {a \sqrt {b \,x^{2}+a}\, \left (6 a \,d^{2} x^{2}+14 b \,c^{2} x^{2}-3 a d x c +2 a \,c^{2}\right )}{6 c^{3} x^{3}}+\frac {-\frac {2 \left (a^{3} d^{6}+3 a^{2} b \,c^{2} d^{4}+3 a \,b^{2} c^{4} d^{2}+b^{3} c^{6}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{3} c \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {2 b^{3} c^{3} \left (\frac {d \sqrt {b \,x^{2}+a}}{b}-\frac {c \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}\right )}{d^{2}}+\frac {d \,a^{\frac {3}{2}} \left (2 a \,d^{2}+5 b \,c^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{c}}{2 c^{3}}\)	\(317\)
default	\(\text {Expression too large to display}\)	\(1223\)

Input:

int((b*x^2+a)^(5/2)/x^4/(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

-1/6*a*(b*x^2+a)^(1/2)*(6*a*d^2*x^2+14*b*c^2*x^2-3*a*c*d*x+2*a*c^2)/c^3/x^ 
3+1/2/c^3*(-2/d^3*(a^3*d^6+3*a^2*b*c^2*d^4+3*a*b^2*c^4*d^2+b^3*c^6)/c/((a* 
d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b* 
c^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+ 
c/d))+2*b^3*c^3/d^2*(d*(b*x^2+a)^(1/2)/b-c*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b 
^(1/2))+d*a^(3/2)*(2*a*d^2+5*b*c^2)/c*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x 
))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((b*x^2+a)^(5/2)/x^4/(d*x+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{x^{4} \left (c + d x\right )}\, dx \] Input:

integrate((b*x**2+a)**(5/2)/x**4/(d*x+c),x)
 

Output:

Integral((a + b*x**2)**(5/2)/(x**4*(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{{\left (d x + c\right )} x^{4}} \,d x } \] Input:

integrate((b*x^2+a)^(5/2)/x^4/(d*x+c),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(5/2)/((d*x + c)*x^4), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((b*x^2+a)^(5/2)/x^4/(d*x+c),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^4\,\left (c+d\,x\right )} \,d x \] Input:

int((a + b*x^2)^(5/2)/(x^4*(c + d*x)),x)
 

Output:

int((a + b*x^2)^(5/2)/(x^4*(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{x^{4} \left (d x +c \right )}d x \] Input:

int((b*x^2+a)^(5/2)/x^4/(d*x+c),x)
 

Output:

int((b*x^2+a)^(5/2)/x^4/(d*x+c),x)