\(\int \frac {(e x)^{5/2} (c+d x^2)}{(a+b x^2)^{7/4}} \, dx\) [461]

Optimal result

Integrand size = 26, antiderivative size = 168 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=-\frac {2 (b c-a d) e (e x)^{3/2}}{3 b^2 \left (a+b x^2\right )^{3/4}}+\frac {d e (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b^2}-\frac {(4 b c-7 a d) e^{5/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{11/4}}+\frac {(4 b c-7 a d) e^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{11/4}} \] Output:

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

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.77 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {(e x)^{5/2} \left (\frac {2 b^{3/4} x^{3/2} \left (-4 b c+7 a d+3 b d x^2\right )}{\left (a+b x^2\right )^{3/4}}+3 (-4 b c+7 a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+3 (4 b c-7 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{12 b^{11/4} x^{5/2}} \] Input:

Integrate[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x]
 

Output:

((e*x)^(5/2)*((2*b^(3/4)*x^(3/2)*(-4*b*c + 7*a*d + 3*b*d*x^2))/(a + b*x^2) 
^(3/4) + 3*(-4*b*c + 7*a*d)*ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + 
3*(4*b*c - 7*a*d)*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(12*b^(11 
/4)*x^(5/2))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {362, 262, 266, 854, 27, 827, 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.

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 827

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

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]
 

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {5}{2}} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]

Input:

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)
 

Output:

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\text {Timed out} \] Input:

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 45.89 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate((e*x)**(5/2)*(d*x**2+c)/(b*x**2+a)**(7/4),x)
 

Output:

c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((7/4, 7/4), (11/4,), b*x**2*exp_polar 
(I*pi)/a)/(2*a**(7/4)*gamma(11/4)) + d*e**(5/2)*x**(11/2)*gamma(11/4)*hype 
r((7/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(15/4))
 

Maxima [F]

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

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(7/4), x)
 

Giac [F]

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

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(7/4), x)
 

Mupad [F(-1)]

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

int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x)
 

Output:

int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4), x)
 

Reduce [F]

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

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)
 

Output:

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