3.693 \(\int \frac{(c+d x^2)^{3/2}}{x^4 (a+b x^2)} \, dx\)
Optimal. Leaf size=102 \[ \frac{\sqrt{c+d x^2} (3 b c-4 a d)}{3 a^2 x}+\frac{(b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2}}-\frac{c \sqrt{c+d x^2}}{3 a x^3} \]
[Out]
-(c*Sqrt[c + d*x^2])/(3*a*x^3) + ((3*b*c - 4*a*d)*Sqrt[c + d*x^2])/(3*a^2*x) + ((b*c - a*d)^(3/2)*ArcTan[(Sqrt
[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/a^(5/2)
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Rubi [A] time = 0.126267, antiderivative size = 102, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used =
{474, 583, 12, 377, 205} \[ \frac{\sqrt{c+d x^2} (3 b c-4 a d)}{3 a^2 x}+\frac{(b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2}}-\frac{c \sqrt{c+d x^2}}{3 a x^3} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)),x]
[Out]
-(c*Sqrt[c + d*x^2])/(3*a*x^3) + ((3*b*c - 4*a*d)*Sqrt[c + d*x^2])/(3*a^2*x) + ((b*c - a*d)^(3/2)*ArcTan[(Sqrt
[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/a^(5/2)
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 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]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx &=-\frac{c \sqrt{c+d x^2}}{3 a x^3}+\frac{\int \frac{-c (3 b c-4 a d)-d (2 b c-3 a d) x^2}{x^2 \left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{3 a}\\ &=-\frac{c \sqrt{c+d x^2}}{3 a x^3}+\frac{(3 b c-4 a d) \sqrt{c+d x^2}}{3 a^2 x}-\frac{\int -\frac{3 c (b c-a d)^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{3 a^2 c}\\ &=-\frac{c \sqrt{c+d x^2}}{3 a x^3}+\frac{(3 b c-4 a d) \sqrt{c+d x^2}}{3 a^2 x}+\frac{(b c-a d)^2 \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{a^2}\\ &=-\frac{c \sqrt{c+d x^2}}{3 a x^3}+\frac{(3 b c-4 a d) \sqrt{c+d x^2}}{3 a^2 x}+\frac{(b c-a d)^2 \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{a^2}\\ &=-\frac{c \sqrt{c+d x^2}}{3 a x^3}+\frac{(3 b c-4 a d) \sqrt{c+d x^2}}{3 a^2 x}+\frac{(b c-a d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.0166746, size = 53, normalized size = 0.52 \[ -\frac{\left (c+d x^2\right )^{3/2} \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{(a d-b c) x^2}{a \left (d x^2+c\right )}\right )}{3 a x^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)),x]
[Out]
-((c + d*x^2)^(3/2)*Hypergeometric2F1[-3/2, 1, -1/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(3*a*x^3)
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Maple [B] time = 0.013, size = 2089, normalized size = 20.5 \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)/x^4/(b*x^2+a),x)
[Out]
-1/2/a*d^(3/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)
/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/a*d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-1/2/a*d^(3/2)*ln((-d*(-a
*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-
(a*d-b*c)/b)^(1/2))+3/4*b/a^2*d^(1/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2
))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*c-1/2*b/a/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2)
)^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*d+1/2*b^2/a^2/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/
2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*c+1/2*b/a/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2)
)^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*d-1/2*b^2/a^2/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/
2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*c+1/4*b/a^2*d*((x+1/b*(-a*b)^(1/2))^2*d-2*d
*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x+3/4*b/a^2*d^(1/2)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b
)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*c-2/
3/a*d/c^2/x*(d*x^2+c)^(5/2)+2/3/a*d^2/c^2*x*(d*x^2+c)^(3/2)+1/a*d^2/c*x*(d*x^2+c)^(1/2)-3/2*b/a^2*d^(1/2)*c*ln
(x*d^(1/2)+(d*x^2+c)^(1/2))+1/4*b/a^2*d*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d
-b*c)/b)^(1/2)*x-b/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/
2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2
))/(x+1/b*(-a*b)^(1/2)))*d*c+b/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/
b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-
b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*d*c-b/a^2*d/c*x*(d*x^2+c)^(3/2)-1/2*b^2/a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(
1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2
*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*c^2+1/2*b^2/a^2/(-a*b)^(1
/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x
+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*c^2-1
/6*b^2/a^2/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+1
/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/
b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(
1/2)))*d^2-1/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*
(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-
1/b*(-a*b)^(1/2)))*d^2+1/6*b^2/a^2/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/
2))-(a*d-b*c)/b)^(3/2)-1/3/a/c/x^3*(d*x^2+c)^(5/2)+b/a^2/c/x*(d*x^2+c)^(5/2)-3/2*b/a^2*d*x*(d*x^2+c)^(1/2)
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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 )} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(3/2)/x^4/(b*x^2+a),x, algorithm="maxima")
[Out]
integrate((d*x^2 + c)^(3/2)/((b*x^2 + a)*x^4), x)
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Fricas [A] time = 1.82501, size = 701, normalized size = 6.87 \begin{align*} \left [-\frac{3 \,{\left (b c - a d\right )} x^{3} \sqrt{-\frac{b c - a d}{a}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left (a^{2} c x -{\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \,{\left ({\left (3 \, b c - 4 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{12 \, a^{2} x^{3}}, \frac{3 \,{\left (b c - a d\right )} x^{3} \sqrt{\frac{b c - a d}{a}} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{a}}}{2 \,{\left ({\left (b c d - a d^{2}\right )} x^{3} +{\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \,{\left ({\left (3 \, b c - 4 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{6 \, a^{2} x^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(3/2)/x^4/(b*x^2+a),x, algorithm="fricas")
[Out]
[-1/12*(3*(b*c - a*d)*x^3*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b
*c^2 - 4*a^2*c*d)*x^2 + 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2
*a*b*x^2 + a^2)) - 4*((3*b*c - 4*a*d)*x^2 - a*c)*sqrt(d*x^2 + c))/(a^2*x^3), 1/6*(3*(b*c - a*d)*x^3*sqrt((b*c
- a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c
^2 - a*c*d)*x)) + 2*((3*b*c - 4*a*d)*x^2 - a*c)*sqrt(d*x^2 + c))/(a^2*x^3)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{2}\right )^{\frac{3}{2}}}{x^{4} \left (a + b x^{2}\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x**2+c)**(3/2)/x**4/(b*x**2+a),x)
[Out]
Integral((c + d*x**2)**(3/2)/(x**4*(a + b*x**2)), x)
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Giac [B] time = 2.86217, size = 346, normalized size = 3.39 \begin{align*} -\frac{{\left (b^{2} c^{2} \sqrt{d} - 2 \, a b c d^{\frac{3}{2}} + a^{2} d^{\frac{5}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{\sqrt{a b c d - a^{2} d^{2}} a^{2}} - \frac{2 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b c^{2} \sqrt{d} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a c d^{\frac{3}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c^{3} \sqrt{d} + 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a c^{2} d^{\frac{3}{2}} + 3 \, b c^{4} \sqrt{d} - 4 \, a c^{3} d^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(3/2)/x^4/(b*x^2+a),x, algorithm="giac")
[Out]
-(b^2*c^2*sqrt(d) - 2*a*b*c*d^(3/2) + a^2*d^(5/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d
)/sqrt(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*a^2) - 2/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c^2*sqrt(d
) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*c*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^3*sqrt(d) + 6*(sqrt(
d)*x - sqrt(d*x^2 + c))^2*a*c^2*d^(3/2) + 3*b*c^4*sqrt(d) - 4*a*c^3*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2
- c)^3*a^2)