3.953 \(\int \frac {(a+b x^2)^{3/2}}{x^4 \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=311 \[ -\frac {2 \sqrt {d} \sqrt {a+b x^2} (2 b c-a d) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 c^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {2 \sqrt {a+b x^2} \sqrt {c+d x^2} (2 b c-a d)}{3 c^2 x}+\frac {2 d x \sqrt {a+b x^2} (2 b c-a d)}{3 c^2 \sqrt {c+d x^2}}+\frac {b \sqrt {a+b x^2} (3 b c-a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a \sqrt {c} \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 311, 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, 583, 531, 418, 492, 411} \[ -\frac {2 \sqrt {a+b x^2} \sqrt {c+d x^2} (2 b c-a d)}{3 c^2 x}+\frac {2 d x \sqrt {a+b x^2} (2 b c-a d)}{3 c^2 \sqrt {c+d x^2}}-\frac {2 \sqrt {d} \sqrt {a+b x^2} (2 b c-a d) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 c^{3/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3}+\frac {b \sqrt {a+b x^2} (3 b c-a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a \sqrt {c} \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 583

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{3/2}}{x^4 \sqrt {c+d x^2}} \, dx &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3}+\frac {\int \frac {2 a (2 b c-a d)+b (3 b c-a d) x^2}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 c}\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3}-\frac {2 (2 b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c^2 x}-\frac {\int \frac {-a b c (3 b c-a d)-2 a b d (2 b c-a d) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 a c^2}\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3}-\frac {2 (2 b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c^2 x}+\frac {(2 b d (2 b c-a d)) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 c^2}+\frac {(b (3 b c-a d)) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 c}\\ &=\frac {2 d (2 b c-a d) x \sqrt {a+b x^2}}{3 c^2 \sqrt {c+d x^2}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3}-\frac {2 (2 b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c^2 x}+\frac {b (3 b c-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 a \sqrt {c} \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {(2 d (2 b c-a d)) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c}\\ &=\frac {2 d (2 b c-a d) x \sqrt {a+b x^2}}{3 c^2 \sqrt {c+d x^2}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c x^3}-\frac {2 (2 b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 c^2 x}-\frac {2 \sqrt {d} (2 b c-a d) \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 c^{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 (3 b c-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 a \sqrt {c} \sqrt {d} \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 = 0.38, size = 227, normalized size = 0.73 \[ \frac {\sqrt {\frac {b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (-a c+2 a d x^2-4 b c x^2\right )-i b c x^3 \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+2 i b c x^3 \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{3 c^2 x^3 \sqrt {\frac {b}{a}} \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x^{2} + c}}{d x^{6} + c x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(3/2)*sqrt(d*x^2 + c)/(d*x^6 + c*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\sqrt {d x^{2} + c} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 433, normalized size = 1.39 \[ \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (2 \sqrt {-\frac {b}{a}}\, a b \,d^{2} x^{6}-4 \sqrt {-\frac {b}{a}}\, b^{2} c d \,x^{6}+2 \sqrt {-\frac {b}{a}}\, a^{2} d^{2} x^{4}-3 \sqrt {-\frac {b}{a}}\, a b c d \,x^{4}-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c d \,x^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c d \,x^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-4 \sqrt {-\frac {b}{a}}\, b^{2} c^{2} x^{4}+4 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{2} x^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{2} x^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {-\frac {b}{a}}\, a^{2} c d \,x^{2}-5 \sqrt {-\frac {b}{a}}\, a b \,c^{2} x^{2}-\sqrt {-\frac {b}{a}}\, a^{2} c^{2}\right )}{3 \left (x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c \right ) \sqrt {-\frac {b}{a}}\, c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(2*(-1/a*b)^(1/2)*a*b*d^2*x^6-4*(-1/a*b)^(1/2)*b^2*c*d*x^6+b*d*((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^3*a*c-((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^3*b^2*c^2-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*E
llipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^3*a*b*c*d+4*((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^3*b^2*c^2+2*(-1/a*b)^(1/2)*a^2*d^2*x^4-3*(-1/a*b)^(1/2)*x^4*a*b*c*d-4*(-1/a*b
)^(1/2)*b^2*c^2*x^4+(-1/a*b)^(1/2)*a^2*c*d*x^2-5*(-1/a*b)^(1/2)*a*b*c^2*x^2-(-1/a*b)^(1/2)*a^2*c^2)/(b*d*x^4+a
*d*x^2+b*c*x^2+a*c)/c^2/x^3/(-1/a*b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\sqrt {d x^{2} + c} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{x^4\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{x^{4} \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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