3.3.9 \(\int \sqrt {f x} (d+e x^2) (a+b x^2+c x^4)^{3/2} \, dx\) [209]

3.3.9.1 Optimal result

Integrand size = 31, antiderivative size = 299 \[ \int \sqrt {f x} \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {2 a d (f x)^{3/2} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {3}{2},-\frac {3}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {2 a e (f x)^{7/2} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {7}{4},-\frac {3}{2},-\frac {3}{2},\frac {11}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{7 f^3 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \]

output

2/3*a*d*(f*x)^(3/2)*AppellF1(3/4,-3/2,-3/2,7/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1 
/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(c*x^4+b*x^2+a)^(1/2)/f/(1+2*c*x^2/( 
b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)+2/7* 
a*e*(f*x)^(7/2)*AppellF1(7/4,-3/2,-3/2,11/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2) 
),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(c*x^4+b*x^2+a)^(1/2)/f^3/(1+2*c*x^2/(b 
-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)
 

3.3.9.2 Mathematica [A] (verified)

Time = 11.94 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.64 \[ \int \sqrt {f x} \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {2 x \sqrt {f x} \left (7 \left (a+b x^2+c x^4\right ) \left (-108 b^3 e+12 b^2 c \left (19 d+7 e x^2\right )+b c \left (624 a e+7 c x^2 \left (323 d+231 e x^2\right )\right )+c^2 \left (77 c x^4 \left (19 d+15 e x^2\right )+a \left (3971 d+2415 e x^2\right )\right )\right )+28 a \left (-57 b^2 c d+836 a c^2 d+27 b^3 e-156 a b c e\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+12 \left (-95 b^3 c d+684 a b c^2 d+45 b^4 e-309 a b^2 c e+420 a^2 c^2 e\right ) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{153615 c^2 \sqrt {a+b x^2+c x^4}} \]

input

Integrate[Sqrt[f*x]*(d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2),x]
 

output

(2*x*Sqrt[f*x]*(7*(a + b*x^2 + c*x^4)*(-108*b^3*e + 12*b^2*c*(19*d + 7*e*x 
^2) + b*c*(624*a*e + 7*c*x^2*(323*d + 231*e*x^2)) + c^2*(77*c*x^4*(19*d + 
15*e*x^2) + a*(3971*d + 2415*e*x^2))) + 28*a*(-57*b^2*c*d + 836*a*c^2*d + 
27*b^3*e - 156*a*b*c*e)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b 
^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c 
])]*AppellF1[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c* 
x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 12*(-95*b^3*c*d + 684*a*b*c^2*d + 45*b^4* 
e - 309*a*b^2*c*e + 420*a^2*c^2*e)*x^2*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x 
^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + S 
qrt[b^2 - 4*a*c])]*AppellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b + Sqrt[b^2 
- 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(153615*c^2*Sqrt[a + b*x^ 
2 + c*x^4])
 

3.3.9.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1674, 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.

\(\displaystyle \int \sqrt {f x} \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1674

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 a d (f x)^{3/2} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {3}{2},-\frac {3}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}+\frac {2 a e (f x)^{7/2} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {7}{4},-\frac {3}{2},-\frac {3}{2},\frac {11}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{7 f^3 \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}\)

input

Int[Sqrt[f*x]*(d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2),x]
 

output

(2*a*d*(f*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[3/4, -3/2, -3/2, 7/4, 
(-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/( 
3*f*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sq 
rt[b^2 - 4*a*c])]) + (2*a*e*(f*x)^(7/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[7 
/4, -3/2, -3/2, 11/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + 
Sqrt[b^2 - 4*a*c])])/(7*f^3*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sq 
rt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])
 

3.3.9.3.1 Defintions of rubi rules used

rule 1674

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N 
eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
 

rule 2009

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

3.3.9.4 Maple [F]

\[\int \sqrt {f x}\, \left (e \,x^{2}+d \right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}d x\]

input

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

output

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

3.3.9.5 Fricas [F]

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

input

integrate((f*x)^(1/2)*(e*x^2+d)*(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas 
")
 

output

integral((c*e*x^6 + (c*d + b*e)*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(c*x^4 + 
b*x^2 + a)*sqrt(f*x), x)
 

3.3.9.6 Sympy [F]

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

input

integrate((f*x)**(1/2)*(e*x**2+d)*(c*x**4+b*x**2+a)**(3/2),x)
 

output

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

3.3.9.7 Maxima [F]

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

input

integrate((f*x)^(1/2)*(e*x^2+d)*(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima 
")
 

output

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

3.3.9.8 Giac [F]

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

input

integrate((f*x)^(1/2)*(e*x^2+d)*(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")
 

output

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

3.3.9.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {f x} \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int \sqrt {f\,x}\,\left (e\,x^2+d\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2} \,d x \]

input

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

output

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