3.177 \(\int \frac{(c+d x^2)^{3/2}}{(a+b x^2)^{7/2}} \, dx\)

Optimal. Leaf size=315 \[ -\frac{c^{3/2} \sqrt{d} \sqrt{a+b x^2} (4 b c-a d) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{15 a^3 b \sqrt{c+d x^2} (b c-a d) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{\sqrt{c+d x^2} \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{15 a^{5/2} b^{3/2} \sqrt{a+b x^2} (b c-a d) \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac{2 x \sqrt{c+d x^2} (a d+2 b c)}{15 a^2 b \left (a+b x^2\right )^{3/2}}+\frac{x \sqrt{c+d x^2} (b c-a d)}{5 a b \left (a+b x^2\right )^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.282366, antiderivative size = 315, normalized size of antiderivative = 1., 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 = {413, 527, 525, 418, 411} \[ \frac{\sqrt{c+d x^2} \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )|1-\frac{a d}{b c}\right )}{15 a^{5/2} b^{3/2} \sqrt{a+b x^2} (b c-a d) \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac{c^{3/2} \sqrt{d} \sqrt{a+b x^2} (4 b c-a d) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 a^3 b \sqrt{c+d x^2} (b c-a d) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{2 x \sqrt{c+d x^2} (a d+2 b c)}{15 a^2 b \left (a+b x^2\right )^{3/2}}+\frac{x \sqrt{c+d x^2} (b c-a d)}{5 a b \left (a+b x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(3/2)/(a + b*x^2)^(7/2),x]

[Out]

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

Rule 413

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^{7/2}} \, dx &=\frac{(b c-a d) x \sqrt{c+d x^2}}{5 a b \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{c (4 b c+a d)+d (3 b c+2 a d) x^2}{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}} \, dx}{5 a b}\\ &=\frac{(b c-a d) x \sqrt{c+d x^2}}{5 a b \left (a+b x^2\right )^{5/2}}+\frac{2 (2 b c+a d) x \sqrt{c+d x^2}}{15 a^2 b \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{-c (b c-a d) (8 b c+a d)-2 d (b c-a d) (2 b c+a d) x^2}{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}} \, dx}{15 a^2 b (b c-a d)}\\ &=\frac{(b c-a d) x \sqrt{c+d x^2}}{5 a b \left (a+b x^2\right )^{5/2}}+\frac{2 (2 b c+a d) x \sqrt{c+d x^2}}{15 a^2 b \left (a+b x^2\right )^{3/2}}-\frac{(c d (4 b c-a d)) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{15 a^2 b (b c-a d)}+\frac{\left (8 b^2 c^2-3 a b c d-2 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 (b c-a d)}\\ &=\frac{(b c-a d) x \sqrt{c+d x^2}}{5 a b \left (a+b x^2\right )^{5/2}}+\frac{2 (2 b c+a d) x \sqrt{c+d x^2}}{15 a^2 b \left (a+b x^2\right )^{3/2}}+\frac{\left (8 b^2 c^2-3 a b c d-2 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} b^{3/2} (b c-a d) \sqrt{a+b x^2} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac{c^{3/2} \sqrt{d} (4 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 )}{15 a^3 b (b c-a d) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.580328, size = 285, normalized size = 0.9 \[ \frac{x \sqrt{\frac{b}{a}} \left (c+d x^2\right ) \left (\left (a+b x^2\right )^2 \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right )+3 a^2 (b c-a d)^2+2 a \left (a+b x^2\right ) (a d+2 b c) (b c-a d)\right )-i c \left (a+b x^2\right )^2 \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (\left (-a^2 d^2-7 a b c d+8 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+\left (2 a^2 d^2+3 a b c d-8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )\right )}{15 a^4 \left (\frac{b}{a}\right )^{3/2} \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(3/2)/(a + b*x^2)^(7/2),x]

[Out]

(Sqrt[b/a]*x*(c + d*x^2)*(3*a^2*(b*c - a*d)^2 + 2*a*(b*c - a*d)*(2*b*c + a*d)*(a + b*x^2) + (8*b^2*c^2 - 3*a*b
*c*d - 2*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 + 3*
a*b*c*d + 2*a^2*d^2)*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (8*b^2*c^2 - 7*a*b*c*d - a^2*d^2)*Ellipt
icF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(15*a^4*(b/a)^(3/2)*(b*c - a*d)*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])

________________________________________________________________________________________

Maple [B]  time = 0.027, size = 1414, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(3/2)/(b*x^2+a)^(7/2),x)

[Out]

-1/15*(-2*x^7*a^2*b^2*d^3*(-b/a)^(1/2)-3*x^7*a*b^3*c*d^2*(-b/a)^(1/2)+7*x^3*a^2*b^2*c^2*d*(-b/a)^(1/2)-11*x*a^
3*b*c^2*d*(-b/a)^(1/2)+8*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*b^4*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/
c)^(1/2)-8*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*b^4*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-Ellip
ticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^4*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+8*EllipticF(x*(-b/a)^(1
/2),(a*d/b/c)^(1/2))*a^2*b^2*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+2*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^
(1/2))*a^4*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-8*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(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*(-b/a)^(1/2)-x^3*a^4*d^3*(-b/a)^(1/2)-EllipticF(x*(-b
/a)^(1/2),(a*d/b/c)^(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)-10*x^5*a^2*b^2*c*d^2*(-b/
a)^(1/2)+17*x^5*a*b^3*c^2*d*(-b/a)^(1/2)-17*x^3*a^3*b*c*d^2*(-b/a)^(1/2)+16*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)
^(1/2))*x^2*a*b^3*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-16*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2
*a*b^3*c^3*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)-7*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*b*c^2*d*((b
*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)+3*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^3*b*c^2*d*((b*x^2+a)/a)^(1/
2)*((d*x^2+c)/c)^(1/2)+2*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(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)-7*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a*b^3*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*(-b/a)^(1/2)+20*x^3*a*b^3*c^3*(-b/a)^(1/2)-x*a^4*c*d^2*(-b/a)^(1/2)+15*x*a^2*b^2*c^3*(-b/
a)^(1/2)-6*x^5*a^3*b*d^3*(-b/a)^(1/2)+3*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^4*a*b^3*c^2*d*((b*x^2+a)/a
)^(1/2)*((d*x^2+c)/c)^(1/2)-2*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2*a^3*b*c*d^2*((b*x^2+a)/a)^(1/2)*((
d*x^2+c)/c)^(1/2)-14*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(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)+4*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^2*a^3*b*c*d^2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)
+6*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(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))/(d*x^2
+c)^(1/2)/a^3/(a*d-b*c)/(-b/a)^(1/2)/(b*x^2+a)^(5/2)/b

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(3/2)/(b*x^2+a)^(7/2),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(3/2)/(b*x^2 + a)^(7/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(3/2)/(b*x^2+a)^(7/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(3/2)/(b*x**2+a)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(3/2)/(b*x^2+a)^(7/2),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)^(3/2)/(b*x^2 + a)^(7/2), x)