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

Optimal. Leaf size=102 \[ -\frac {(b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} b}-\frac {c \sqrt {c+d x^2}}{a x}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {474, 523, 217, 206, 377, 205} \[ -\frac {(b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} b}-\frac {c \sqrt {c+d x^2}}{a x}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*Sqrt[c + d*x^2])/(a*x)) - ((b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(3
/2)*b) + (d^(3/2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/b

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]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 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 523

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x^2\right )^{3/2}}{x^2 \left (a+b x^2\right )} \, dx &=-\frac {c \sqrt {c+d x^2}}{a x}+\frac {\int \frac {-c (b c-2 a d)+a d^2 x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{a}\\ &=-\frac {c \sqrt {c+d x^2}}{a x}+\frac {d^2 \int \frac {1}{\sqrt {c+d x^2}} \, dx}{b}-\frac {(b c-a d)^2 \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{a b}\\ &=-\frac {c \sqrt {c+d x^2}}{a x}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b}-\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 b}\\ &=-\frac {c \sqrt {c+d x^2}}{a 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^{3/2} b}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 105, normalized size = 1.03 \[ -\frac {(b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} b}-\frac {c \sqrt {c+d x^2}}{a x}+\frac {d^{3/2} \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*Sqrt[c + d*x^2])/(a*x)) - ((b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(3
/2)*b) + (d^(3/2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/b

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fricas [A]  time = 1.29, size = 718, normalized size = 7.04 \[ \left [\frac {2 \, a d^{\frac {3}{2}} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (b c - a d\right )} x \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 \, \sqrt {d x^{2} + c} b c}{4 \, a b x}, -\frac {4 \, a \sqrt {-d} d x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b c - a d\right )} x \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 \, \sqrt {d x^{2} + c} b c}{4 \, a b x}, \frac {a d^{\frac {3}{2}} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (b c - a d\right )} x \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 \, \sqrt {d x^{2} + c} b c}{2 \, a b x}, -\frac {2 \, a \sqrt {-d} d x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b c - a d\right )} x \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 \, \sqrt {d x^{2} + c} b c}{2 \, a b x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*a*d^(3/2)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - (b*c - a*d)*x*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*sqrt(d*x^2 + c)*b*c)/(a*b*x), -
1/4*(4*a*sqrt(-d)*d*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (b*c - a*d)*x*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)*sqr
t(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*sqrt(d*x^2 + c)*b*c)/(a*b*x), 1/2*(a*d^(3/
2)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - (b*c - a*d)*x*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*sqrt(d*x^2
+ c)*b*c)/(a*b*x), -1/2*(2*a*sqrt(-d)*d*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (b*c - a*d)*x*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*sqrt(d*x^2 + c)*b*c)/(a*b*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.02, size = 1956, normalized size = 19.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x^2+c\right )}^{3/2}}{x^2\,\left (b\,x^2+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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