\(\int \frac {(e x)^{7/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [471]

Optimal result

Integrand size = 26, antiderivative size = 210 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 a (b c-a d) e^3 \sqrt {e x}}{5 b^3 \left (a+b x^2\right )^{5/4}}-\frac {2 (6 b c-11 a d) e^3 \sqrt {e x}}{5 b^3 \sqrt [4]{a+b x^2}}+\frac {d e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}{2 b^3}+\frac {(4 b c-9 a d) e^{7/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}}+\frac {(4 b c-9 a d) e^{7/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{13/4}} \] Output:

2/5*a*(-a*d+b*c)*e^3*(e*x)^(1/2)/b^3/(b*x^2+a)^(5/4)-2/5*(-11*a*d+6*b*c)*e 
^3*(e*x)^(1/2)/b^3/(b*x^2+a)^(1/4)+1/2*d*e^3*(e*x)^(1/2)*(b*x^2+a)^(3/4)/b 
^3+1/4*(-9*a*d+4*b*c)*e^(7/2)*arctan(b^(1/4)*(e*x)^(1/2)/e^(1/2)/(b*x^2+a) 
^(1/4))/b^(13/4)+1/4*(-9*a*d+4*b*c)*e^(7/2)*arctanh(b^(1/4)*(e*x)^(1/2)/e^ 
(1/2)/(b*x^2+a)^(1/4))/b^(13/4)
 

Mathematica [A] (verified)

Time = 1.88 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.71 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {(e x)^{7/2} \left (\frac {2 \sqrt [4]{b} \sqrt {x} \left (45 a^2 d+b^2 x^2 \left (-24 c+5 d x^2\right )+a b \left (-20 c+54 d x^2\right )\right )}{\left (a+b x^2\right )^{5/4}}+5 (4 b c-9 a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+5 (4 b c-9 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{20 b^{13/4} x^{7/2}} \] Input:

Integrate[((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]
 

Output:

((e*x)^(7/2)*((2*b^(1/4)*Sqrt[x]*(45*a^2*d + b^2*x^2*(-24*c + 5*d*x^2) + a 
*b*(-20*c + 54*d*x^2)))/(a + b*x^2)^(5/4) + 5*(4*b*c - 9*a*d)*ArcTan[(b^(1 
/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + 5*(4*b*c - 9*a*d)*ArcTanh[(b^(1/4)*Sqrt[ 
x])/(a + b*x^2)^(1/4)]))/(20*b^(13/4)*x^(7/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {362, 250, 252, 266, 770, 756, 218, 221}

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.

\(\displaystyle \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 250

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

\(\Big \downarrow \) 252

\(\displaystyle \frac {2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {e (e x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e x} \sqrt [4]{b x^2+a}}dx}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\right )}{5 a b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {e (e x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a e^2 \left (\frac {2 e \int \frac {1}{\sqrt [4]{b x^2+a}}d\sqrt {e x}}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\right )}{5 a b}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {e (e x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a e^2 \left (\frac {2 e \int \frac {1}{1-b x^2}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\right )}{5 a b}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {e (e x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a e^2 \left (\frac {2 e \left (\frac {1}{2} e \int \frac {1}{e-\sqrt {b} e x}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}+\frac {1}{2} e \int \frac {1}{\sqrt {b} x e+e}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}\right )}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\right )}{5 a b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {e (e x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a e^2 \left (\frac {2 e \left (\frac {1}{2} e \int \frac {1}{e-\sqrt {b} e x}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}+\frac {\sqrt {e} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\right )}{5 a b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (e x)^{9/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {e (e x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a e^2 \left (\frac {2 e \left (\frac {\sqrt {e} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\right )}{5 a b}\)

Input:

Int[((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]
 

Output:

(2*(b*c - a*d)*(e*x)^(9/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - ((4*b*c - 9*a*d) 
*((e*(e*x)^(5/2))/(2*b*(a + b*x^2)^(1/4)) - (5*a*e^2*((-2*e*Sqrt[e*x])/(b* 
(a + b*x^2)^(1/4)) + (2*e*((Sqrt[e]*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a 
 + b*x^2)^(1/4))])/(2*b^(1/4)) + (Sqrt[e]*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqr 
t[e]*(a + b*x^2)^(1/4))])/(2*b^(1/4))))/b))/(4*b)))/(5*a*b)
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

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

rule 250

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*(( 
c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( 
b*(2*m - 3)))   Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, 
 c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {7}{2}} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

Input:

int((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Output:

int((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 904, normalized size of antiderivative = 4.30 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx =\text {Too large to display} \] Input:

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

Output:

1/40*(4*(5*b^2*d*e^3*x^4 - 6*(4*b^2*c - 9*a*b*d)*e^3*x^2 - 5*(4*a*b*c - 9* 
a^2*d)*e^3)*(b*x^2 + a)^(3/4)*sqrt(e*x) + 5*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b 
^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b* 
c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 9* 
a*d)*sqrt(e*x)*e^3 + (b^4*x^2 + a*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 
7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4)) 
/(b*x^2 + a)) - 5*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*((256*b^4*c^4 - 2304*a 
*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14 
/b^13)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 9*a*d)*sqrt(e*x)*e^3 - (b^4* 
x^2 + a*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 116 
64*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4))/(b*x^2 + a)) - 5*(-I*b^5* 
x^4 - 2*I*a*b^4*x^2 - I*a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a 
^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4)*log(-( 
(b*x^2 + a)^(3/4)*(4*b*c - 9*a*d)*sqrt(e*x)*e^3 + (I*b^4*x^2 + I*a*b^3)*(( 
256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 
+ 6561*a^4*d^4)*e^14/b^13)^(1/4))/(b*x^2 + a)) - 5*(I*b^5*x^4 + 2*I*a*b^4* 
x^2 + I*a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 
 11664*a^3*b*c*d^3 + 6561*a^4*d^4)*e^14/b^13)^(1/4)*log(-((b*x^2 + a)^(3/4 
)*(4*b*c - 9*a*d)*sqrt(e*x)*e^3 + (-I*b^4*x^2 - I*a*b^3)*((256*b^4*c^4 - 2 
304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \] Input:

integrate((e*x)**(7/2)*(d*x**2+c)/(b*x**2+a)**(9/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*(e*x)^(7/2)/(b*x^2 + a)^(9/4), x)
 

Giac [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*(e*x)^(7/2)/(b*x^2 + a)^(9/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {{\left (e\,x\right )}^{7/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \] Input:

int(((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x)
 

Output:

int(((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(9/4), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\sqrt {e}\, e^{3} \left (\left (\int \frac {\sqrt {x}\, x^{5}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d +\left (\int \frac {\sqrt {x}\, x^{3}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c \right ) \] Input:

int((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Output:

sqrt(e)*e**3*(int((sqrt(x)*x**5)/((a + b*x**2)**(1/4)*a**2 + 2*(a + b*x**2 
)**(1/4)*a*b*x**2 + (a + b*x**2)**(1/4)*b**2*x**4),x)*d + int((sqrt(x)*x** 
3)/((a + b*x**2)**(1/4)*a**2 + 2*(a + b*x**2)**(1/4)*a*b*x**2 + (a + b*x** 
2)**(1/4)*b**2*x**4),x)*c)