\(\int \frac {\sqrt {e x} (a+b x^2)^2}{(c+d x^2)^{5/2}} \, dx\) [1087]

Optimal result

Integrand size = 28, antiderivative size = 403 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {(b c-a d)^2 (e x)^{3/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (3 b c+a d) (e x)^{3/2}}{2 c^2 d^2 e \sqrt {c+d x^2}}+\frac {\left (7 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt {e x} \sqrt {c+d x^2}}{2 c^2 d^{5/2} \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {\left (7 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt {e} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{2 c^{7/4} d^{11/4} \sqrt {c+d x^2}}+\frac {\left (7 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt {e} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{4 c^{7/4} d^{11/4} \sqrt {c+d x^2}} \] Output:

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

Mathematica [C] (verified)

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

Time = 11.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {\sqrt {e x} \left (-\left ((b c-a d) x \left (a d \left (5 c+3 d x^2\right )+b c \left (7 c+9 d x^2\right )\right )\right )+3 \left (7 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x \left (c+d x^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {c}{d x^2}\right )\right )}{6 c^2 d^2 \left (c+d x^2\right )^{3/2}} \] Input:

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

Output:

(Sqrt[e*x]*(-((b*c - a*d)*x*(a*d*(5*c + 3*d*x^2) + b*c*(7*c + 9*d*x^2))) + 
 3*(7*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*(c + d*x^2)*Hyp 
ergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(6*c^2*d^2*(c + d*x^2)^(3/2 
))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 398, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {366, 27, 362, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 366

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

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

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 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 366

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

rule 761

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

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.84

method	result	size
elliptic	\(\frac {\sqrt {e x \left (x^{2} d +c \right )}\, \sqrt {e x}\, \left (\frac {x \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d e \,x^{3}+c e x}}{3 c \,d^{4} \left (x^{2}+\frac {c}{d}\right )^{2}}+\frac {e \,x^{2} \left (a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}\right )}{2 d^{2} c^{2} \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {\left (\frac {b^{2} e}{d^{2}}-\frac {e \left (a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}\right )}{4 c^{2} d^{2}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{e x \sqrt {x^{2} d +c}}\)	\(337\)
default	\(\text {Expression too large to display}\)	\(1176\)

Input:

int((e*x)^(1/2)*(b*x^2+a)^2/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(e*x*(d*x^2+c))^(1/2)/e/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)*(1/3/c/d^4*x*(a^2*d^ 
2-2*a*b*c*d+b^2*c^2)*(d*e*x^3+c*e*x)^(1/2)/(x^2+c/d)^2+1/2/d^2*e*x^2/c^2*( 
a^2*d^2+2*a*b*c*d-3*b^2*c^2)/((x^2+c/d)*d*e*x)^(1/2)+(b^2*e/d^2-1/4/c^2/d^ 
2*e*(a^2*d^2+2*a*b*c*d-3*b^2*c^2))*(-c*d)^(1/2)/d*((x+(-c*d)^(1/2)/d)/(-c* 
d)^(1/2)*d)^(1/2)*(-2*(x-(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2)*(-d/(-c*d)^ 
(1/2)*x)^(1/2)/(d*e*x^3+c*e*x)^(1/2)*(-2*(-c*d)^(1/2)/d*EllipticE(((x+(-c* 
d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2),1/2*2^(1/2))+(-c*d)^(1/2)/d*EllipticF((( 
x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2),1/2*2^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {3 \, {\left (7 \, b^{2} c^{4} - 2 \, a b c^{3} d - a^{2} c^{2} d^{2} + {\left (7 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} - a^{2} d^{4}\right )} x^{4} + 2 \, {\left (7 \, b^{2} c^{3} d - 2 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d e} {\rm weierstrassZeta}\left (-\frac {4 \, c}{d}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right )\right ) + {\left (3 \, {\left (3 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} - a^{2} d^{4}\right )} x^{3} + {\left (7 \, b^{2} c^{3} d - 2 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{6 \, {\left (c^{2} d^{5} x^{4} + 2 \, c^{3} d^{4} x^{2} + c^{4} d^{3}\right )}} \] Input:

integrate((e*x)^(1/2)*(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

-1/6*(3*(7*b^2*c^4 - 2*a*b*c^3*d - a^2*c^2*d^2 + (7*b^2*c^2*d^2 - 2*a*b*c* 
d^3 - a^2*d^4)*x^4 + 2*(7*b^2*c^3*d - 2*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt 
(d*e)*weierstrassZeta(-4*c/d, 0, weierstrassPInverse(-4*c/d, 0, x)) + (3*( 
3*b^2*c^2*d^2 - 2*a*b*c*d^3 - a^2*d^4)*x^3 + (7*b^2*c^3*d - 2*a*b*c^2*d^2 
- 5*a^2*c*d^3)*x)*sqrt(d*x^2 + c)*sqrt(e*x))/(c^2*d^5*x^4 + 2*c^3*d^4*x^2 
+ c^4*d^3)
 

Sympy [F]

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

integrate((e*x)**(1/2)*(b*x**2+a)**2/(d*x**2+c)**(5/2),x)
 

Output:

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

Maxima [F]

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

integrate((e*x)^(1/2)*(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x)^(1/2)*(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-4 \sqrt {x}\, \sqrt {d \,x^{2}+c}\, a b d x +14 \sqrt {x}\, \sqrt {d \,x^{2}+c}\, b^{2} c x +6 \sqrt {x}\, \sqrt {d \,x^{2}+c}\, b^{2} d \,x^{3}+3 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) a^{2} c^{2} d^{2}+6 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) a^{2} c \,d^{3} x^{2}+3 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) a^{2} d^{4} x^{4}+6 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) a b \,c^{3} d +12 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) a b \,c^{2} d^{2} x^{2}+6 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) a b c \,d^{3} x^{4}-21 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) b^{2} c^{4}-42 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) b^{2} c^{3} d \,x^{2}-21 \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{2}+c}}{d^{3} x^{6}+3 c \,d^{2} x^{4}+3 c^{2} d \,x^{2}+c^{3}}d x \right ) b^{2} c^{2} d^{2} x^{4}\right )}{3 d^{2} \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right )} \] Input:

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

Output:

(sqrt(e)*( - 4*sqrt(x)*sqrt(c + d*x**2)*a*b*d*x + 14*sqrt(x)*sqrt(c + d*x* 
*2)*b**2*c*x + 6*sqrt(x)*sqrt(c + d*x**2)*b**2*d*x**3 + 3*int((sqrt(x)*sqr 
t(c + d*x**2))/(c**3 + 3*c**2*d*x**2 + 3*c*d**2*x**4 + d**3*x**6),x)*a**2* 
c**2*d**2 + 6*int((sqrt(x)*sqrt(c + d*x**2))/(c**3 + 3*c**2*d*x**2 + 3*c*d 
**2*x**4 + d**3*x**6),x)*a**2*c*d**3*x**2 + 3*int((sqrt(x)*sqrt(c + d*x**2 
))/(c**3 + 3*c**2*d*x**2 + 3*c*d**2*x**4 + d**3*x**6),x)*a**2*d**4*x**4 + 
6*int((sqrt(x)*sqrt(c + d*x**2))/(c**3 + 3*c**2*d*x**2 + 3*c*d**2*x**4 + d 
**3*x**6),x)*a*b*c**3*d + 12*int((sqrt(x)*sqrt(c + d*x**2))/(c**3 + 3*c**2 
*d*x**2 + 3*c*d**2*x**4 + d**3*x**6),x)*a*b*c**2*d**2*x**2 + 6*int((sqrt(x 
)*sqrt(c + d*x**2))/(c**3 + 3*c**2*d*x**2 + 3*c*d**2*x**4 + d**3*x**6),x)* 
a*b*c*d**3*x**4 - 21*int((sqrt(x)*sqrt(c + d*x**2))/(c**3 + 3*c**2*d*x**2 
+ 3*c*d**2*x**4 + d**3*x**6),x)*b**2*c**4 - 42*int((sqrt(x)*sqrt(c + d*x** 
2))/(c**3 + 3*c**2*d*x**2 + 3*c*d**2*x**4 + d**3*x**6),x)*b**2*c**3*d*x**2 
 - 21*int((sqrt(x)*sqrt(c + d*x**2))/(c**3 + 3*c**2*d*x**2 + 3*c*d**2*x**4 
 + d**3*x**6),x)*b**2*c**2*d**2*x**4))/(3*d**2*(c**2 + 2*c*d*x**2 + d**2*x 
**4))