3.1.32 \(\int \frac {(a+b x^2) (e+f x^2)^{3/2}}{(c+d x^2)^{5/2}} \, dx\) [32]

Optimal. Leaf size=373 \[ -\frac {f (b c (d e-8 c f)+2 a d (d e+c f)) x \sqrt {c+d x^2}}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {\sqrt {e} \sqrt {f} (b c (d e-8 c f)+2 a d (d e+c f)) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^3 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {(4 b c-a d) e^{3/2} \sqrt {f} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}} \]

[Out]

-1/3*(-a*d+b*c)*x*(f*x^2+e)^(3/2)/c/d/(d*x^2+c)^(3/2)-1/3*f*(b*c*(-8*c*f+d*e)+2*a*d*(c*f+d*e))*x*(d*x^2+c)^(1/
2)/c^2/d^3/(f*x^2+e)^(1/2)+1/3*(-a*d+4*b*c)*e^(3/2)*(1/(1+f*x^2/e))^(1/2)*(1+f*x^2/e)^(1/2)*EllipticF(x*f^(1/2
)/e^(1/2)/(1+f*x^2/e)^(1/2),(1-d*e/c/f)^(1/2))*f^(1/2)*(d*x^2+c)^(1/2)/c^2/d^2/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)
/(f*x^2+e)^(1/2)+1/3*(b*c*(-8*c*f+d*e)+2*a*d*(c*f+d*e))*(1/(1+f*x^2/e))^(1/2)*(1+f*x^2/e)^(1/2)*EllipticE(x*f^
(1/2)/e^(1/2)/(1+f*x^2/e)^(1/2),(1-d*e/c/f)^(1/2))*e^(1/2)*f^(1/2)*(d*x^2+c)^(1/2)/c^2/d^3/(e*(d*x^2+c)/c/(f*x
^2+e))^(1/2)/(f*x^2+e)^(1/2)+1/3*(b*c*(-4*c*f+d*e)+a*d*(c*f+2*d*e))*x*(f*x^2+e)^(1/2)/c^2/d^2/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 373, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {540, 545, 429, 506, 422} \begin {gather*} \frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (2 a d (c f+d e)+b c (d e-8 c f)) E\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^3 \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} (4 b c-a d) F\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^2 \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {f x \sqrt {c+d x^2} (2 a d (c f+d e)+b c (d e-8 c f))}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {x \sqrt {e+f x^2} (a d (c f+2 d e)+b c (d e-4 c f))}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {x \left (e+f x^2\right )^{3/2} (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(5/2),x]

[Out]

-1/3*(f*(b*c*(d*e - 8*c*f) + 2*a*d*(d*e + c*f))*x*Sqrt[c + d*x^2])/(c^2*d^3*Sqrt[e + f*x^2]) + ((b*c*(d*e - 4*
c*f) + a*d*(2*d*e + c*f))*x*Sqrt[e + f*x^2])/(3*c^2*d^2*Sqrt[c + d*x^2]) - ((b*c - a*d)*x*(e + f*x^2)^(3/2))/(
3*c*d*(c + d*x^2)^(3/2)) + (Sqrt[e]*Sqrt[f]*(b*c*(d*e - 8*c*f) + 2*a*d*(d*e + c*f))*Sqrt[c + d*x^2]*EllipticE[
ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*d^3*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2
]) + ((4*b*c - a*d)*e^(3/2)*Sqrt[f]*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(
3*c^2*d^2*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sq
rt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[c + d*x^2])), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 540

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*n*(p + 1))), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x
^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))
*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 545

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{5/2}} \, dx &=-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {\sqrt {e+f x^2} \left (-(b c+2 a d) e-(4 b c-a d) f x^2\right )}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d}\\ &=\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {c (4 b c-a d) e f-f (b c (d e-8 c f)+2 a d (d e+c f)) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c^2 d^2}\\ &=\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {((4 b c-a d) e f) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c d^2}-\frac {(f (b c (d e-8 c f)+2 a d (d e+c f))) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c^2 d^2}\\ &=-\frac {f (b c (d e-8 c f)+2 a d (d e+c f)) x \sqrt {c+d x^2}}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {(4 b c-a d) e^{3/2} \sqrt {f} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {(e f (b c (d e-8 c f)+2 a d (d e+c f))) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{3 c^2 d^3}\\ &=-\frac {f (b c (d e-8 c f)+2 a d (d e+c f)) x \sqrt {c+d x^2}}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {\sqrt {e} \sqrt {f} (b c (d e-8 c f)+2 a d (d e+c f)) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^3 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {(4 b c-a d) e^{3/2} \sqrt {f} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 5.85, size = 296, normalized size = 0.79 \begin {gather*} \frac {\left (\frac {d}{c}\right )^{3/2} \left (\sqrt {\frac {d}{c}} x \left (e+f x^2\right ) \left (b c \left (-4 c^2 f+d^2 e x^2-5 c d f x^2\right )+a d \left (c^2 f+2 d^2 e x^2+c d \left (3 e+2 f x^2\right )\right )\right )-i e (-2 a d (d e+c f)+b c (-d e+8 c f)) \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} \sqrt {1+\frac {f x^2}{e}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i e (-a d (2 d e+c f)+b c (-d e+4 c f)) \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} \sqrt {1+\frac {f x^2}{e}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )\right )}{3 d^4 \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(5/2),x]

[Out]

((d/c)^(3/2)*(Sqrt[d/c]*x*(e + f*x^2)*(b*c*(-4*c^2*f + d^2*e*x^2 - 5*c*d*f*x^2) + a*d*(c^2*f + 2*d^2*e*x^2 + c
*d*(3*e + 2*f*x^2))) - I*e*(-2*a*d*(d*e + c*f) + b*c*(-(d*e) + 8*c*f))*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[1
+ (f*x^2)/e]*EllipticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + I*e*(-(a*d*(2*d*e + c*f)) + b*c*(-(d*e) + 4*c*f)
)*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)]))/(3*d^4*
(c + d*x^2)^(3/2)*Sqrt[e + f*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1224\) vs. \(2(401)=802\).
time = 0.14, size = 1225, normalized size = 3.28

method	result	size
elliptic	\(\frac {\sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\, \left (-\frac {\left (a c d f -a \,d^{2} e -b \,c^{2} f +b c d e \right ) x \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}{3 d^{4} c \left (x^{2}+\frac {c}{d}\right )^{2}}+\frac {\left (d f \,x^{2}+d e \right ) \left (2 a c d f +2 a \,d^{2} e -5 b \,c^{2} f +b c d e \right ) x}{3 d^{3} c^{2} \sqrt {\left (x^{2}+\frac {c}{d}\right ) \left (d f \,x^{2}+d e \right )}}+\frac {\left (\frac {f \left (a d f -2 b c f +2 b d e \right )}{d^{3}}-\frac {\left (a c d f -a \,d^{2} e -b \,c^{2} f +b c d e \right ) f}{3 d^{3} c}-\frac {\left (2 a c d f +2 a \,d^{2} e -5 b \,c^{2} f +b c d e \right ) \left (c f -d e \right )}{3 d^{3} c^{2}}-\frac {e \left (2 a c d f +2 a \,d^{2} e -5 b \,c^{2} f +b c d e \right )}{3 d^{2} c^{2}}\right ) \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}-\frac {\left (\frac {b \,f^{2}}{d^{2}}-\frac {\left (2 a c d f +2 a \,d^{2} e -5 b \,c^{2} f +b c d e \right ) f}{3 d^{2} c^{2}}\right ) e \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}\, f}\right )}{\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}\)	\(559\)
default	\(\text {Expression too large to display}\)	\(1225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((-d/c)^(1/2)*a*c^2*d*e*f*x+2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(
1/2))*a*d^3*e^2*x^2-2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*d^3*
e^2*x^2-4*(-d/c)^(1/2)*b*c^3*e*f*x+(-d/c)^(1/2)*b*c*d^2*e*f*x^5+5*(-d/c)^(1/2)*a*c*d^2*e*f*x^3-5*(-d/c)^(1/2)*
b*c^2*d*e*f*x^3-5*(-d/c)^(1/2)*b*c^2*d*f^2*x^5+(-d/c)^(1/2)*a*c^2*d*f^2*x^3+(-d/c)^(1/2)*b*c*d^2*e^2*x^3+3*(-d
/c)^(1/2)*a*c*d^2*e^2*x+((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c*
d^2*e^2*x^2-((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c*d^2*e^2*x^2+
((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c^2*d*e*f-2*((d*x^2+c)/c)^
(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c^2*d*e*f+2*((d*x^2+c)/c)^(1/2)*((f*x^2+
e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d^2*e^2-4*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*El
lipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^3*e*f+((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(
1/2),(c*f/d/e)^(1/2))*b*c^2*d*e^2-2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)
^(1/2))*a*c*d^2*e^2+8*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^3*
e*f-((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*d*e^2+2*(-d/c)^(1/
2)*a*d^3*e^2*x^3-4*(-d/c)^(1/2)*b*c^3*f^2*x^3+2*(-d/c)^(1/2)*a*c*d^2*f^2*x^5+2*(-d/c)^(1/2)*a*d^3*e*f*x^5+((d*
x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d^2*e*f*x^2+8*((d*x^2+c)/c)^
(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*d*e*f*x^2-2*((d*x^2+c)/c)^(1/2)*((f*
x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d^2*e*f*x^2-4*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^
(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*d*e*f*x^2)/(f*x^2+e)^(1/2)/c^2/(-d/c)^(1/2)/(d*x^2+c)^(3
/2)/d^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*(f*x^2 + e)^(3/2)/(d*x^2 + c)^(5/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right ) \left (e + f x^{2}\right )^{\frac {3}{2}}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(f*x**2+e)**(3/2)/(d*x**2+c)**(5/2),x)

[Out]

Integral((a + b*x**2)*(e + f*x**2)**(3/2)/(c + d*x**2)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)*(f*x^2 + e)^(3/2)/(d*x^2 + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (b\,x^2+a\right )\,{\left (f\,x^2+e\right )}^{3/2}}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(5/2),x)

[Out]

int(((a + b*x^2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(5/2), x)

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