3.2.71 \(\int \frac {\sqrt {c+d x^2}}{(a+b x^2)^{7/2}} \, dx\) [171]

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

[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 = 0.15, 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 = {423, 541, 539, 429, 422} \begin {gather*} -\frac {2 c^{3/2} \sqrt {d} \sqrt {a+b x^2} (2 b c-3 a d) F\left (\text {ArcTan}\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 (\text {ArcTan}\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}} \end {gather*}

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 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 423

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 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 539

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 541

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] Result contains complex when optimal does not.
time = 3.12, size = 285, normalized size = 0.92 \begin {gather*} \frac {\sqrt {\frac {b}{a}} x \left (c+d x^2\right ) \left (3 a^2 (b c-a d)^2+a (-b c+a d) (-4 b c+3 a d) \left (a+b x^2\right )+\left (8 b^2 c^2-13 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^2\right )+i c \left (a+b x^2\right )^2 \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \left (\left (8 b^2 c^2-13 a b c d+3 a^2 d^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+\left (-8 b^2 c^2+17 a b c d-9 a^2 d^2\right ) F\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}} (b c-a d)^2 \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1410\) vs. \(2(343)=686\).
time = 0.09, size = 1411, normalized size = 4.57

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {x \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{5 b^{3} a \left (x^{2}+\frac {a}{b}\right )^{3}}+\frac {\left (3 a d -4 b c \right ) x \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{15 a^{2} \left (a d -b c \right ) b^{2} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {\left (b d \,x^{2}+b c \right ) x \left (3 a^{2} d^{2}-13 a b c d +8 b^{2} c^{2}\right )}{15 b \,a^{3} \left (a d -b c \right )^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) \left (b d \,x^{2}+b c \right )}}+\frac {\left (\frac {d \left (3 a d -4 b c \right )}{15 b \,a^{2} \left (a d -b c \right )}-\frac {3 a^{2} d^{2}-13 a b c d +8 b^{2} c^{2}}{15 \left (a d -b c \right ) b \,a^{3}}-\frac {c \left (3 a^{2} d^{2}-13 a b c d +8 b^{2} c^{2}\right )}{15 a^{3} \left (a d -b c \right )^{2}}\right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}+\frac {\left (3 a^{2} d^{2}-13 a b c d +8 b^{2} c^{2}\right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{15 a^{3} \left (a d -b c \right )^{2} \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(570\)
default	\(\text {Expression too large to display}\)	\(1411\)

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,method=_RETURNVERBOSE)

[Out]

1/15*(9*((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^2*c*d^2*x^4-17
*((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^3*c^2*d*x^4-3*((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^2*c*d^2*x^4+9*(-b/a)^(1/2)*a^4*
d^3*x^3+8*(-b/a)^(1/2)*b^4*c^3*x^5+16*((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^3*c^3*x^2-16*((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^3*c^3*x^2-17*((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*b*c^2*d+
13*((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^3*b*c^2*d+8*((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))*b^4*c^3*x^4-8*((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^4*c^3*x^4+9*((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^4*c*d^2+8*((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^2*c^3-3*((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^4*c*d^2-8*((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))*a^2
*b^2*c^3-26*(-b/a)^(1/2)*a^3*b*c^2*d*x+13*((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^3*c^2*d*x^4+18*((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*b*c*d^2*x^2-34*((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^2*c^2*d*x^2-6*((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^3*b*c*d^2*
x^2+26*((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^2*c^2*d*x^2+3*(
-b/a)^(1/2)*a^2*b^2*d^3*x^7+8*(-b/a)^(1/2)*b^4*c^2*d*x^7+9*(-b/a)^(1/2)*a^3*b*d^3*x^5+20*(-b/a)^(1/2)*a*b^3*c^
3*x^3+9*(-b/a)^(1/2)*a^4*c*d^2*x+15*(-b/a)^(1/2)*a^2*b^2*c^3*x-13*(-b/a)^(1/2)*a*b^3*c*d^2*x^7-30*(-b/a)^(1/2)
*a^2*b^2*c*d^2*x^5+7*(-b/a)^(1/2)*a*b^3*c^2*d*x^5-17*(-b/a)^(1/2)*a^3*b*c*d^2*x^3-18*(-b/a)^(1/2)*a^2*b^2*c^2*
d*x^3)/(d*x^2+c)^(1/2)/(-b/a)^(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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((d*x^2+c)^(1/2)/(b*x^2+a)^(7/2),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 {\sqrt {c + d x^{2}}}{\left (a + b x^{2}\right )^{\frac {7}{2}}}\, dx \end {gather*}

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

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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