3.171 \(\int \frac {\sqrt {c+d x^2}}{(a+b x^2)^{7/2}} \, dx\)

Optimal. Leaf size=309 \[ -\frac {2 c^{3/2} \sqrt {d} \sqrt {a+b x^2} (2 b c-3 a d) \operatorname {EllipticF}\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{15 a^3 \sqrt {c+d x^2} (b c-a d)^2 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {c+d x^2} (4 b c-3 a d)}{15 a^2 \left (a+b x^2\right )^{3/2} (b c-a d)}+\frac {\sqrt {c+d x^2} \left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{15 a^{5/2} \sqrt {b} \sqrt {a+b x^2} (b c-a d)^2 \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac {x \sqrt {c+d x^2}}{5 a \left (a+b x^2\right )^{5/2}} \]

[Out]

-2/15*c^(3/2)*(-3*a*d+2*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))*d^(1/2)*(b*x^2+a)^(1/2)/a^3/(-a*d+b*c)^2/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/
2)+1/5*x*(d*x^2+c)^(1/2)/a/(b*x^2+a)^(5/2)+1/15*(-3*a*d+4*b*c)*x*(d*x^2+c)^(1/2)/a^2/(-a*d+b*c)/(b*x^2+a)^(3/2
)+1/15*(3*a^2*d^2-13*a*b*c*d+8*b^2*c^2)*(1/(1+b*x^2/a))^(1/2)*(1+b*x^2/a)^(1/2)*EllipticE(x*b^(1/2)/a^(1/2)/(1
+b*x^2/a)^(1/2),(1-a*d/b/c)^(1/2))*(d*x^2+c)^(1/2)/a^(5/2)/(-a*d+b*c)^2/b^(1/2)/(b*x^2+a)^(1/2)/(a*(d*x^2+c)/c
/(b*x^2+a))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {412, 527, 525, 418, 411} \[ \frac {\sqrt {c+d x^2} \left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{15 a^{5/2} \sqrt {b} \sqrt {a+b x^2} (b c-a d)^2 \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac {2 c^{3/2} \sqrt {d} \sqrt {a+b x^2} (2 b c-3 a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 a^3 \sqrt {c+d x^2} (b c-a d)^2 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {c+d x^2} (4 b c-3 a d)}{15 a^2 \left (a+b x^2\right )^{3/2} (b c-a d)}+\frac {x \sqrt {c+d x^2}}{5 a \left (a+b x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]/(a + b*x^2)^(7/2),x]

[Out]

(x*Sqrt[c + d*x^2])/(5*a*(a + b*x^2)^(5/2)) + ((4*b*c - 3*a*d)*x*Sqrt[c + d*x^2])/(15*a^2*(b*c - a*d)*(a + b*x
^2)^(3/2)) + ((8*b^2*c^2 - 13*a*b*c*d + 3*a^2*d^2)*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]], 1 -
(a*d)/(b*c)])/(15*a^(5/2)*Sqrt[b]*(b*c - a*d)^2*Sqrt[a + b*x^2]*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]) - (2*c^
(3/2)*Sqrt[d]*(2*b*c - 3*a*d)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*a^3
*(b*c - a*d)^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 412

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

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 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 527

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 + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^{7/2}} \, dx &=\frac {x \sqrt {c+d x^2}}{5 a \left (a+b x^2\right )^{5/2}}-\frac {\int \frac {-4 c-3 d x^2}{\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}} \, dx}{5 a}\\ &=\frac {x \sqrt {c+d x^2}}{5 a \left (a+b x^2\right )^{5/2}}+\frac {(4 b c-3 a d) x \sqrt {c+d x^2}}{15 a^2 (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {\int \frac {c (8 b c-9 a d)+d (4 b c-3 a d) x^2}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx}{15 a^2 (b c-a d)}\\ &=\frac {x \sqrt {c+d x^2}}{5 a \left (a+b x^2\right )^{5/2}}+\frac {(4 b c-3 a d) x \sqrt {c+d x^2}}{15 a^2 (b c-a d) \left (a+b x^2\right )^{3/2}}-\frac {(2 c d (2 b c-3 a d)) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 a^2 (b c-a d)^2}+\frac {\left (8 b^2 c^2-13 a b c d+3 a^2 d^2\right ) \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^{3/2}} \, dx}{15 a^2 (b c-a d)^2}\\ &=\frac {x \sqrt {c+d x^2}}{5 a \left (a+b x^2\right )^{5/2}}+\frac {(4 b c-3 a d) x \sqrt {c+d x^2}}{15 a^2 (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {\left (8 b^2 c^2-13 a b c d+3 a^2 d^2\right ) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{15 a^{5/2} \sqrt {b} (b c-a d)^2 \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac {2 c^{3/2} \sqrt {d} (2 b c-3 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 )}{15 a^3 (b c-a d)^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 = 0.55, size = 285, normalized size = 0.92 \[ \frac {x \sqrt {\frac {b}{a}} \left (c+d x^2\right ) \left (\left (a+b x^2\right )^2 \left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right )+3 a^2 (b c-a d)^2+a \left (a+b x^2\right ) (a d-b c) (3 a d-4 b c)\right )+i c \sqrt {\frac {b x^2}{a}+1} \left (a+b x^2\right )^2 \sqrt {\frac {d x^2}{c}+1} \left (\left (-9 a^2 d^2+17 a b c d-8 b^2 c^2\right ) \operatorname {EllipticF}\left (i \sinh ^{-1}\left (x \sqrt {\frac {b}{a}}\right ),\frac {a d}{b c}\right )+\left (3 a^2 d^2-13 a b c d+8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )\right )}{15 a^3 \sqrt {\frac {b}{a}} \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^2]/(a + b*x^2)^(7/2),x]

[Out]

(Sqrt[b/a]*x*(c + d*x^2)*(3*a^2*(b*c - a*d)^2 + a*(-(b*c) + a*d)*(-4*b*c + 3*a*d)*(a + b*x^2) + (8*b^2*c^2 - 1
3*a*b*c*d + 3*a^2*d^2)*(a + b*x^2)^2) + I*c*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*((8*b^2*c^2
- 13*a*b*c*d + 3*a^2*d^2)*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (-8*b^2*c^2 + 17*a*b*c*d - 9*a^2*d^
2)*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(15*a^3*Sqrt[b/a]*(b*c - a*d)^2*(a + b*x^2)^(5/2)*Sqrt[c +
d*x^2])

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)/(b*x^2+a)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + c}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)/(b*x^2+a)^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*x^2 + c)/(b*x^2 + a)^(7/2), x)

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maple [B]  time = 0.06, size = 1411, normalized size = 4.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(1/2)/(b*x^2+a)^(7/2),x)

[Out]

-1/15*(30*x^5*a^2*b^2*c*d^2*(-1/a*b)^(1/2)-7*x^5*a*b^3*c^2*d*(-1/a*b)^(1/2)+17*x^3*a^3*b*c*d^2*(-1/a*b)^(1/2)+
18*x^3*a^2*b^2*c^2*d*(-1/a*b)^(1/2)+26*x*a^3*b*c^2*d*(-1/a*b)^(1/2)+8*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/
2))*x^4*b^4*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-8*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^4*b^4*
c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+3*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^4*c*d^2*((b*x^2+a)
/a)^(1/2)*((d*x^2+c)/c)^(1/2)+8*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*b^2*c^3*((b*x^2+a)/a)^(1/2)*((
d*x^2+c)/c)^(1/2)-9*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^4*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1
/2)-8*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*b^2*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-8*x^5*b^
4*c^3*(-1/a*b)^(1/2)-9*x^3*a^4*d^3*(-1/a*b)^(1/2)+13*x^7*a*b^3*c*d^2*(-1/a*b)^(1/2)-3*x^7*a^2*b^2*d^3*(-1/a*b)
^(1/2)-13*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^4*a*b^3*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+
3*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^4*a^2*b^2*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+34*Ell
ipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^2*a^2*b^2*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+6*EllipticE
((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^2*a^3*b*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+17*EllipticF((-1/a*
b)^(1/2)*x,(a/b/c*d)^(1/2))*x^4*a*b^3*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-9*EllipticF((-1/a*b)^(1/2)
*x,(a/b/c*d)^(1/2))*x^4*a^2*b^2*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-18*EllipticF((-1/a*b)^(1/2)*x,(a
/b/c*d)^(1/2))*x^2*a^3*b*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-26*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)
^(1/2))*x^2*a^2*b^2*c^2*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+16*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2
))*x^2*a*b^3*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-16*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^2*a*
b^3*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-13*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^3*b*c^2*d*((b
*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+17*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^3*b*c^2*d*((b*x^2+a)/a)^
(1/2)*((d*x^2+c)/c)^(1/2)-8*x^7*b^4*c^2*d*(-1/a*b)^(1/2)-9*x^5*a^3*b*d^3*(-1/a*b)^(1/2)-20*x^3*a*b^3*c^3*(-1/a
*b)^(1/2)-9*x*a^4*c*d^2*(-1/a*b)^(1/2)-15*x*a^2*b^2*c^3*(-1/a*b)^(1/2))/(d*x^2+c)^(1/2)/(-1/a*b)^(1/2)/(a*d-b*
c)^2/a^3/(b*x^2+a)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + c}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)/(b*x^2+a)^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^2 + c)/(b*x^2 + a)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {d\,x^2+c}}{{\left (b\,x^2+a\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^(1/2)/(a + b*x^2)^(7/2),x)

[Out]

int((c + d*x^2)^(1/2)/(a + b*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c + d x^{2}}}{\left (a + b x^{2}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(1/2)/(b*x**2+a)**(7/2),x)

[Out]

Integral(sqrt(c + d*x**2)/(a + b*x**2)**(7/2), x)

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