\(\int \frac {c+d x^2}{(e x)^{9/2} (a+b x^2)^{7/4}} \, dx\) [470]

Optimal result

Integrand size = 26, antiderivative size = 181 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{7/4}} \, dx=-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (10 b c-7 a d)}{21 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {4 (10 b c-7 a d) \sqrt [4]{a+b x^2}}{21 a^3 e^3 (e x)^{3/2}}-\frac {8 b^{3/2} (10 b c-7 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 a^{7/2} e^6 \left (a+b x^2\right )^{3/4}} \] Output:

-2/7*c/a/e/(e*x)^(7/2)/(b*x^2+a)^(3/4)-2/21*(-7*a*d+10*b*c)/a^2/e^3/(e*x)^ 
(3/2)/(b*x^2+a)^(3/4)+4/21*(-7*a*d+10*b*c)*(b*x^2+a)^(1/4)/a^3/e^3/(e*x)^( 
3/2)-8/21*b^(3/2)*(-7*a*d+10*b*c)*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*InverseJac 
obiAM(1/2*arccot(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(7/2)/e^6/(b*x^2+a)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.45 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{7/4}} \, dx=-\frac {2 \sqrt {e x} \left (3 a c+(-10 b c+7 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {7}{4},\frac {1}{4},-\frac {b x^2}{a}\right )\right )}{21 a^2 e^5 x^4 \left (a+b x^2\right )^{3/4}} \] Input:

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

Output:

(-2*Sqrt[e*x]*(3*a*c + (-10*b*c + 7*a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*Hyperge 
ometric2F1[-3/4, 7/4, 1/4, -((b*x^2)/a)]))/(21*a^2*e^5*x^4*(a + b*x^2)^(3/ 
4))
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {359, 253, 264, 266, 768, 858, 807, 229}

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.

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

\(\Big \downarrow \) 359

\(\displaystyle -\frac {(10 b c-7 a d) \int \frac {1}{(e x)^{5/2} \left (b x^2+a\right )^{7/4}}dx}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \int \frac {1}{(e x)^{5/2} \left (b x^2+a\right )^{3/4}}dx}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \left (-\frac {2 b \int \frac {1}{\sqrt {e x} \left (b x^2+a\right )^{3/4}}dx}{3 a e^2}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \left (-\frac {4 b \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {e x}}{3 a e^3}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \left (-\frac {4 b (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (e x)^{3/2}}d\sqrt {e x}}{3 a e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \left (\frac {4 b (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {e x} \left (\frac {a x^2 e^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {e x}}}{3 a e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \left (\frac {2 b (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x e^3}{b}+1\right )^{3/4}}d(e x)}{3 a e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle -\frac {(10 b c-7 a d) \left (\frac {2 \left (\frac {4 b^{3/2} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} e^2 x}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} e^4 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}}\right )}{a}+\frac {2}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{3/4}}\)

Input:

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

Output:

(-2*c)/(7*a*e*(e*x)^(7/2)*(a + b*x^2)^(3/4)) - ((10*b*c - 7*a*d)*(2/(3*a*e 
*(e*x)^(3/2)*(a + b*x^2)^(3/4)) + (2*((-2*(a + b*x^2)^(1/4))/(3*a*e*(e*x)^ 
(3/2)) + (4*b^(3/2)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcTan[(Sq 
rt[a]*e^2*x)/Sqrt[b]]/2, 2])/(3*a^(3/2)*e^4*(a + b*x^2)^(3/4))))/a))/(7*a* 
e^2)
 

Defintions of rubi rules used

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 253

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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 359

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

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)*sqrt(e*x)/(b^2*e^5*x^9 + 2*a*b*e^5* 
x^7 + a^2*e^5*x^5), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.43 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{7/4}} \, dx=\frac {2 \sqrt {e}\, \left (-28 a b d \,x^{4}+32 b^{2} c \,x^{4}-7 a^{2} d \,x^{2}+8 a b c \,x^{2}-3 a^{2} c \right )}{21 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{3} e^{5} x^{3}} \] Input:

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

Output:

(2*sqrt(e)*(a + b*x**2)**(3/4)*( - 3*a**2*c - 7*a**2*d*x**2 + 8*a*b*c*x**2 
 - 28*a*b*d*x**4 + 32*b**2*c*x**4))/(21*sqrt(x)*sqrt(a + b*x**2)*a**3*e**5 
*x**3*(a + b*x**2))