3.7.75 \(\int \frac {(c+d x^2)^{3/2}}{x^4 (a+b x^2)} \, dx\)

Optimal. Leaf size=102 \[ \frac {(b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2}}+\frac {\sqrt {c+d x^2} (3 b c-4 a d)}{3 a^2 x}-\frac {c \sqrt {c+d x^2}}{3 a x^3} \]

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Rubi [A]  time = 0.13, antiderivative size = 102, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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]

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*}

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Mathematica [C]  time = 0.02, size = 53, normalized size = 0.52 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)),x]

[Out]

-1/3*((c + d*x^2)^(3/2)*Hypergeometric2F1[-3/2, 1, -1/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(a*x^3)

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IntegrateAlgebraic [A]  time = 0.31, size = 144, normalized size = 1.41 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-a c-4 a d x^2+3 b c x^2\right )}{3 a^2 x^3}-\frac {(b c-a d)^{3/2} \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)),x]

[Out]

(Sqrt[c + d*x^2]*(-(a*c) + 3*b*c*x^2 - 4*a*d*x^2))/(3*a^2*x^3) - ((b*c - a*d)^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d])/S
qrt[b*c - a*d] + (b*Sqrt[d]*x^2)/(Sqrt[a]*Sqrt[b*c - a*d]) - (b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sqrt[b*c - a*d])])
/a^(5/2)

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fricas [A]  time = 2.06, size = 331, normalized size = 3.25 \begin {gather*} \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 {gather*}

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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giac [B]  time = 5.35, size = 256, normalized size = 2.51 \begin {gather*} -\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 {gather*}

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)

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maple [B]  time = 0.01, size = 2089, normalized size = 20.48

result too large to display

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )} x^{4}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^{3/2}}{x^4\,\left (b\,x^2+a\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^(3/2)/(x^4*(a + b*x^2)),x)

[Out]

int((c + d*x^2)^(3/2)/(x^4*(a + b*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{x^{4} \left (a + b x^{2}\right )}\, dx \end {gather*}

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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