3.2.100 \(\int \frac {(a+b x^2)^{5/2}}{\sqrt {c+d x^2}} \, dx\) [200]

Optimal. Leaf size=344 \[ \frac {\left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 d^2 \sqrt {c+d x^2}}-\frac {4 b (b c-2 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 d^2}+\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d}-\frac {\sqrt {c} \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\sqrt {c} \left (4 b^2 c^2-11 a b c d+15 a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

1/15*(23*a^2*d^2-23*a*b*c*d+8*b^2*c^2)*x*(b*x^2+a)^(1/2)/d^2/(d*x^2+c)^(1/2)-1/15*(23*a^2*d^2-23*a*b*c*d+8*b^2
*c^2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))
*c^(1/2)*(b*x^2+a)^(1/2)/d^(5/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+1/15*(15*a^2*d^2-11*a*b*c*d+4
*b^2*c^2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1
/2))*c^(1/2)*(b*x^2+a)^(1/2)/d^(5/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+1/5*b*x*(b*x^2+a)^(3/2)*(
d*x^2+c)^(1/2)/d-4/15*b*(-2*a*d+b*c)*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^2

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Rubi [A]
time = 0.18, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {427, 542, 545, 429, 506, 422} \begin {gather*} \frac {\sqrt {c} \sqrt {a+b x^2} \left (15 a^2 d^2-11 a b c d+4 b^2 c^2\right ) F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\sqrt {c} \sqrt {a+b x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right ) E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {a+b x^2} \left (23 a^2 d^2-23 a b c d+8 b^2 c^2\right )}{15 d^2 \sqrt {c+d x^2}}-\frac {4 b x \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-2 a d)}{15 d^2}+\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b^2*c^2 - 23*a*b*c*d + 23*a^2*d^2)*x*Sqrt[a + b*x^2])/(15*d^2*Sqrt[c + d*x^2]) - (4*b*(b*c - 2*a*d)*x*Sqrt
[a + b*x^2]*Sqrt[c + d*x^2])/(15*d^2) + (b*x*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(5*d) - (Sqrt[c]*(8*b^2*c^2 -
23*a*b*c*d + 23*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*d^(5/2)*
Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (Sqrt[c]*(4*b^2*c^2 - 11*a*b*c*d + 15*a^2*d^2)*Sqrt[a
+ b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*d^(5/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*
x^2))]*Sqrt[c + d*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 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

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 542

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(n*(p + q + 1) + 1))), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 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 )^{5/2}}{\sqrt {c+d x^2}} \, dx &=\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d}+\frac {\int \frac {\sqrt {a+b x^2} \left (-a (b c-5 a d)-4 b (b c-2 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{5 d}\\ &=-\frac {4 b (b c-2 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 d^2}+\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d}+\frac {\int \frac {a \left (4 b^2 c^2-11 a b c d+15 a^2 d^2\right )+b \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 d^2}\\ &=-\frac {4 b (b c-2 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 d^2}+\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d}+\frac {\left (a \left (4 b^2 c^2-11 a b c d+15 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 d^2}+\frac {\left (b \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right )\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 d^2}\\ &=\frac {\left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 d^2 \sqrt {c+d x^2}}-\frac {4 b (b c-2 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 d^2}+\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d}+\frac {\sqrt {c} \left (4 b^2 c^2-11 a b c d+15 a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {\left (c \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {\left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 d^2 \sqrt {c+d x^2}}-\frac {4 b (b c-2 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 d^2}+\frac {b x \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{5 d}-\frac {\sqrt {c} \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\sqrt {c} \left (4 b^2 c^2-11 a b c d+15 a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.91, size = 260, normalized size = 0.76 \begin {gather*} \frac {b \sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (-4 b c+11 a d+3 b d x^2\right )-i b c \left (8 b^2 c^2-23 a b c d+23 a^2 d^2\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i \left (-8 b^3 c^3+27 a b^2 c^2 d-34 a^2 b c d^2+15 a^3 d^3\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{15 \sqrt {\frac {b}{a}} d^3 \sqrt {a+b x^2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(-4*b*c + 11*a*d + 3*b*d*x^2) - I*b*c*(8*b^2*c^2 - 23*a*b*c*d + 23*a^
2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*(-8*b^3*c^3
+ 27*a*b^2*c^2*d - 34*a^2*b*c*d^2 + 15*a^3*d^3)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sq
rt[b/a]*x], (a*d)/(b*c)])/(15*Sqrt[b/a]*d^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]
time = 0.10, size = 615, normalized size = 1.79

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {b^{2} x^{3} \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{5 d}+\frac {\left (3 a \,b^{2}-\frac {b^{2} \left (4 a d +4 b c \right )}{5 d}\right ) x \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{3 b d}+\frac {\left (a^{3}-\frac {\left (3 a \,b^{2}-\frac {b^{2} \left (4 a d +4 b c \right )}{5 d}\right ) a c}{3 b d}\right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}-\frac {\left (3 a^{2} b -\frac {3 a \,b^{2} c}{5 d}-\frac {\left (3 a \,b^{2}-\frac {b^{2} \left (4 a d +4 b c \right )}{5 d}\right ) \left (2 a d +2 b c \right )}{3 b d}\right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}\, d}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(435\)
risch	\(\frac {b x \left (3 b d \,x^{2}+11 a d -4 b c \right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{15 d^{2}}+\frac {\left (-\frac {\left (23 a^{2} b \,d^{2}-23 a \,b^{2} c d +8 b^{3} c^{2}\right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}\, d}+\frac {15 a^{3} d^{2} \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}-\frac {11 a^{2} b c d \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}+\frac {4 a \,b^{2} c^{2} \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{15 d^{2} \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(513\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (3 \sqrt {-\frac {b}{a}}\, b^{3} d^{3} x^{7}+14 \sqrt {-\frac {b}{a}}\, a \,b^{2} d^{3} x^{5}-\sqrt {-\frac {b}{a}}\, b^{3} c \,d^{2} x^{5}+11 \sqrt {-\frac {b}{a}}\, a^{2} b \,d^{3} x^{3}+10 \sqrt {-\frac {b}{a}}\, a \,b^{2} c \,d^{2} x^{3}-4 \sqrt {-\frac {b}{a}}\, b^{3} c^{2} d \,x^{3}+15 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a^{3} d^{3}-34 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a^{2} b c \,d^{2}+27 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a \,b^{2} c^{2} d -8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b^{3} c^{3}+23 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a^{2} b c \,d^{2}-23 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a \,b^{2} c^{2} d +8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b^{3} c^{3}+11 \sqrt {-\frac {b}{a}}\, a^{2} b c \,d^{2} x -4 \sqrt {-\frac {b}{a}}\, a \,b^{2} c^{2} d x \right )}{15 d^{3} \left (b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c \right ) \sqrt {-\frac {b}{a}}}\)	\(615\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(3*(-b/a)^(1/2)*b^3*d^3*x^7+14*(-b/a)^(1/2)*a*b^2*d^3*x^5-(-b/a)^(1/2)*b^
3*c*d^2*x^5+11*(-b/a)^(1/2)*a^2*b*d^3*x^3+10*(-b/a)^(1/2)*a*b^2*c*d^2*x^3-4*(-b/a)^(1/2)*b^3*c^2*d*x^3+15*((b*
x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*d^3-34*((b*x^2+a)/a)^(1/2)*(
(d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^2*b*c*d^2+27*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^
(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a*b^2*c^2*d-8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Elliptic
F(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^3+23*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),
(a*d/b/c)^(1/2))*a^2*b*c*d^2-23*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/
2))*a*b^2*c^2*d+8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^3*c^3+11
*(-b/a)^(1/2)*a^2*b*c*d^2*x-4*(-b/a)^(1/2)*a*b^2*c^2*d*x)/d^3/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/(-b/a)^(1/2)

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

[Out]

integrate((b*x^2 + a)^(5/2)/sqrt(d*x^2 + c), 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)^(5/2)/(d*x^2+c)^(1/2),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 )^{\frac {5}{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^2 + a)^(5/2)/sqrt(d*x^2 + c), 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 )}^{5/2}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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