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

Optimal. Leaf size=330 \[ \frac {\sqrt {a+b x^2} \left (-3 a^2 d^2-7 a b c d+2 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 \sqrt {c} d^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {a+b x^2} \left (\frac {3 a^2 d}{c}+7 a b-\frac {2 b^2 c}{d}\right )}{3 \sqrt {c+d x^2}}-\frac {b \sqrt {c} \sqrt {a+b x^2} (b c-9 a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}+\frac {b x \sqrt {a+b x^2} \sqrt {c+d x^2} (3 a d+b c)}{3 c d} \]

[Out]

1/3*(7*a*b-2*b^2*c/d+3*a^2*d/c)*x*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)+1/3*(-3*a^2*d^2-7*a*b*c*d+2*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))*(b*x^2+a)^(1
/2)/d^(3/2)/c^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-1/3*b*(-9*a*d+b*c)*(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^(3
/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-a*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/c/x+1/3*b*(3*a*d+b*c)*x*
(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/c/d

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Rubi [A]  time = 0.29, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {474, 528, 531, 418, 492, 411} \[ \frac {\sqrt {a+b x^2} \left (-3 a^2 d^2-7 a b c d+2 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 \sqrt {c} d^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {a+b x^2} \left (\frac {3 a^2 d}{c}+7 a b-\frac {2 b^2 c}{d}\right )}{3 \sqrt {c+d x^2}}-\frac {b \sqrt {c} \sqrt {a+b x^2} (b c-9 a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}+\frac {b x \sqrt {a+b x^2} \sqrt {c+d x^2} (3 a d+b c)}{3 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*a*b - (2*b^2*c)/d + (3*a^2*d)/c)*x*Sqrt[a + b*x^2])/(3*Sqrt[c + d*x^2]) + (b*(b*c + 3*a*d)*x*Sqrt[a + b*x^
2]*Sqrt[c + d*x^2])/(3*c*d) - (a*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(c*x) + ((2*b^2*c^2 - 7*a*b*c*d - 3*a^2*d^
2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*Sqrt[c]*d^(3/2)*Sqrt[(c*(a + b*
x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (b*Sqrt[c]*(b*c - 9*a*d)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x
)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

Rule 411

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

Rule 418

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

Rule 474

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[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 528

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 531

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}}{x^2 \sqrt {c+d x^2}} \, dx &=-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}+\frac {\int \frac {\sqrt {a+b x^2} \left (4 a b c+b (b c+3 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{c}\\ &=\frac {b (b c+3 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c d}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}+\frac {\int \frac {-a b c (b c-9 a d)-b \left (2 b^2 c^2-7 a b c d-3 a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 c d}\\ &=\frac {b (b c+3 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c d}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}-\frac {(a b (b c-9 a d)) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 d}-\frac {\left (b \left (2 b^2 c^2-7 a b c d-3 a^2 d^2\right )\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 c d}\\ &=\frac {\left (7 a b-\frac {2 b^2 c}{d}+\frac {3 a^2 d}{c}\right ) x \sqrt {a+b x^2}}{3 \sqrt {c+d x^2}}+\frac {b (b c+3 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c d}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}-\frac {b \sqrt {c} (b c-9 a d) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\left (2 b^2 c^2-7 a b c d-3 a^2 d^2\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac {\left (7 a b-\frac {2 b^2 c}{d}+\frac {3 a^2 d}{c}\right ) x \sqrt {a+b x^2}}{3 \sqrt {c+d x^2}}+\frac {b (b c+3 a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c d}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{c x}+\frac {\left (2 b^2 c^2-7 a b c d-3 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 )}{3 \sqrt {c} d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {b \sqrt {c} (b c-9 a d) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \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]  time = 0.48, size = 254, normalized size = 0.77 \[ \frac {-2 i b c x \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (3 a^2 d^2-4 a b c d+b^2 c^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i b c x \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (3 a^2 d^2+7 a b c d-2 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+d \left (-\sqrt {\frac {b}{a}}\right ) \left (a+b x^2\right ) \left (c+d x^2\right ) \left (3 a^2 d-b^2 c x^2\right )}{3 c d^2 x \sqrt {\frac {b}{a}} \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[b/a]*d*(a + b*x^2)*(3*a^2*d - b^2*c*x^2)*(c + d*x^2)) - I*b*c*(-2*b^2*c^2 + 7*a*b*c*d + 3*a^2*d^2)*x*S
qrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - (2*I)*b*c*(b^2*c^2 - 4
*a*b*c*d + 3*a^2*d^2)*x*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]
)/(3*Sqrt[b/a]*c*d^2*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{d x^{4} + c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(d*x^4 + c*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{\sqrt {d x^{2} + c} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 568, normalized size = 1.72 \[ \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (\sqrt {-\frac {b}{a}}\, b^{3} c \,d^{2} x^{6}-3 \sqrt {-\frac {b}{a}}\, a^{2} b \,d^{3} x^{4}+\sqrt {-\frac {b}{a}}\, a \,b^{2} c \,d^{2} x^{4}+\sqrt {-\frac {b}{a}}\, b^{3} c^{2} d \,x^{4}-3 \sqrt {-\frac {b}{a}}\, a^{3} d^{3} x^{2}-3 \sqrt {-\frac {b}{a}}\, a^{2} b c \,d^{2} x^{2}+3 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} b c \,d^{2} x \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+6 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} b c \,d^{2} x \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {-\frac {b}{a}}\, a \,b^{2} c^{2} d \,x^{2}+7 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,b^{2} c^{2} d x \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,b^{2} c^{2} d x \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{3} c^{3} x \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{3} c^{3} x \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-3 \sqrt {-\frac {b}{a}}\, a^{3} c \,d^{2}\right )}{3 \left (x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c \right ) \sqrt {-\frac {b}{a}}\, c \,d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*((-1/a*b)^(1/2)*x^6*b^3*c*d^2-3*(-1/a*b)^(1/2)*x^4*a^2*b*d^3+(-1/a*b)^(1/2
)*x^4*a*b^2*c*d^2+(-1/a*b)^(1/2)*x^4*b^3*c^2*d+6*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1
/2)*x,(a/b/c*d)^(1/2))*x*a^2*b*c*d^2-8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b
/c*d)^(1/2))*x*a*b^2*c^2*d+2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2
))*x*b^3*c^3+3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x*a^2*b*c*d
^2+7*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x*a*b^2*c^2*d-2*((b*x
^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x*b^3*c^3-3*(-1/a*b)^(1/2)*x^2*
a^3*d^3-3*(-1/a*b)^(1/2)*x^2*a^2*b*c*d^2+(-1/a*b)^(1/2)*x^2*a*b^2*c^2*d-3*(-1/a*b)^(1/2)*a^3*c*d^2)/(b*d*x^4+a
*d*x^2+b*c*x^2+a*c)/d^2/(-1/a*b)^(1/2)/c/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{\sqrt {d x^{2} + c} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^2\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{x^{2} \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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