3.377 \(\int \frac{a+b x^2}{x^5 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=166 \[ \frac{5 a d^2+4 b c^2}{8 c^4 x^2 \sqrt{d x-c} \sqrt{c+d x}}-\frac{3 d^2 \left (5 a d^2+4 b c^2\right )}{8 c^6 \sqrt{d x-c} \sqrt{c+d x}}-\frac{3 d^2 \left (5 a d^2+4 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{8 c^7}+\frac{a}{4 c^2 x^4 \sqrt{d x-c} \sqrt{c+d x}} \]

[Out]

(-3*d^2*(4*b*c^2 + 5*a*d^2))/(8*c^6*Sqrt[-c + d*x]*Sqrt[c + d*x]) + a/(4*c^2*x^4*Sqrt[-c + d*x]*Sqrt[c + d*x])
+ (4*b*c^2 + 5*a*d^2)/(8*c^4*x^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) - (3*d^2*(4*b*c^2 + 5*a*d^2)*ArcTan[(Sqrt[-c +
d*x]*Sqrt[c + d*x])/c])/(8*c^7)

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Rubi [A]  time = 0.120213, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {454, 103, 12, 104, 21, 92, 205} \[ \frac{5 a d^2+4 b c^2}{8 c^4 x^2 \sqrt{d x-c} \sqrt{c+d x}}-\frac{3 d^2 \left (5 a d^2+4 b c^2\right )}{8 c^6 \sqrt{d x-c} \sqrt{c+d x}}-\frac{3 d^2 \left (5 a d^2+4 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{8 c^7}+\frac{a}{4 c^2 x^4 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(x^5*(-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

(-3*d^2*(4*b*c^2 + 5*a*d^2))/(8*c^6*Sqrt[-c + d*x]*Sqrt[c + d*x]) + a/(4*c^2*x^4*Sqrt[-c + d*x]*Sqrt[c + d*x])
+ (4*b*c^2 + 5*a*d^2)/(8*c^4*x^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) - (3*d^2*(4*b*c^2 + 5*a*d^2)*ArcTan[(Sqrt[-c +
d*x]*Sqrt[c + d*x])/c])/(8*c^7)

Rule 454

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(a1*a2*e*
(m + 1)), x] + Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*e^n*(m + 1)), Int[(e*x)^(m + n)*(a1
+ b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, p}, x] && EqQ[non2, n/2] && Eq
Q[a2*b1 + a1*b2, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1
])) &&  !ILtQ[p, -1]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 12

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

Rule 104

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

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{a+b x^2}{x^5 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx &=\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{1}{4} \left (4 b+\frac{5 a d^2}{c^2}\right ) \int \frac{1}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\\ &=\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{4 b c^2+5 a d^2}{8 c^4 x^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{\left (4 b c^2+5 a d^2\right ) \int \frac{3 d^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx}{8 c^4}\\ &=\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{4 b c^2+5 a d^2}{8 c^4 x^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{\left (3 d^2 \left (4 b c^2+5 a d^2\right )\right ) \int \frac{1}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx}{8 c^4}\\ &=-\frac{3 d^2 \left (4 b c^2+5 a d^2\right )}{8 c^6 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{4 b c^2+5 a d^2}{8 c^4 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (3 d \left (4 b c^2+5 a d^2\right )\right ) \int \frac{c d+d^2 x}{x \sqrt{-c+d x} (c+d x)^{3/2}} \, dx}{8 c^6}\\ &=-\frac{3 d^2 \left (4 b c^2+5 a d^2\right )}{8 c^6 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{4 b c^2+5 a d^2}{8 c^4 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (3 d^2 \left (4 b c^2+5 a d^2\right )\right ) \int \frac{1}{x \sqrt{-c+d x} \sqrt{c+d x}} \, dx}{8 c^6}\\ &=-\frac{3 d^2 \left (4 b c^2+5 a d^2\right )}{8 c^6 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{4 b c^2+5 a d^2}{8 c^4 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (3 d^3 \left (4 b c^2+5 a d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c^2 d+d x^2} \, dx,x,\sqrt{-c+d x} \sqrt{c+d x}\right )}{8 c^6}\\ &=-\frac{3 d^2 \left (4 b c^2+5 a d^2\right )}{8 c^6 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{4 c^2 x^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{4 b c^2+5 a d^2}{8 c^4 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{3 d^2 \left (4 b c^2+5 a d^2\right ) \tan ^{-1}\left (\frac{\sqrt{-c+d x} \sqrt{c+d x}}{c}\right )}{8 c^7}\\ \end{align*}

Mathematica [C]  time = 0.0299248, size = 78, normalized size = 0.47 \[ \frac{a c^4-d^2 x^4 \left (5 a d^2+4 b c^2\right ) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};1-\frac{d^2 x^2}{c^2}\right )}{4 c^6 x^4 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)/(x^5*(-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

(a*c^4 - d^2*(4*b*c^2 + 5*a*d^2)*x^4*Hypergeometric2F1[-1/2, 2, 1/2, 1 - (d^2*x^2)/c^2])/(4*c^6*x^4*Sqrt[-c +
d*x]*Sqrt[c + d*x])

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Maple [B]  time = 0.026, size = 387, normalized size = 2.3 \begin{align*}{\frac{1}{8\,{c}^{6}{x}^{4}} \left ( 15\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{6}a{d}^{6}+12\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{6}b{c}^{2}{d}^{4}-15\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{4}a{c}^{2}{d}^{4}-12\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{4}b{c}^{4}{d}^{2}-15\,\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}{x}^{4}a{d}^{4}-12\,\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}{x}^{4}b{c}^{2}{d}^{2}+5\,{x}^{2}a{c}^{2}{d}^{2}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}\sqrt{-{c}^{2}}+4\,{x}^{2}b{c}^{4}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}\sqrt{-{c}^{2}}+2\,a{c}^{4}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}\sqrt{-{c}^{2}} \right ){\frac{1}{\sqrt{-{c}^{2}}}}{\frac{1}{\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{dx-c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)/x^5/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/8/c^6*(15*ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^6*a*d^6+12*ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)
^(1/2))/x)*x^6*b*c^2*d^4-15*ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^4*a*c^2*d^4-12*ln(-2*(c^2-(-c^2)
^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^4*b*c^4*d^2-15*(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2)*x^4*a*d^4-12*(-c^2)^(1/2)*(d^
2*x^2-c^2)^(1/2)*x^4*b*c^2*d^2+5*x^2*a*c^2*d^2*(d^2*x^2-c^2)^(1/2)*(-c^2)^(1/2)+4*x^2*b*c^4*(d^2*x^2-c^2)^(1/2
)*(-c^2)^(1/2)+2*a*c^4*(d^2*x^2-c^2)^(1/2)*(-c^2)^(1/2))/x^4/(-c^2)^(1/2)/(d^2*x^2-c^2)^(1/2)/(d*x+c)^(1/2)/(d
*x-c)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^5/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62221, size = 335, normalized size = 2.02 \begin{align*} \frac{{\left (2 \, a c^{5} - 3 \,{\left (4 \, b c^{3} d^{2} + 5 \, a c d^{4}\right )} x^{4} +{\left (4 \, b c^{5} + 5 \, a c^{3} d^{2}\right )} x^{2}\right )} \sqrt{d x + c} \sqrt{d x - c} - 6 \,{\left ({\left (4 \, b c^{2} d^{4} + 5 \, a d^{6}\right )} x^{6} -{\left (4 \, b c^{4} d^{2} + 5 \, a c^{2} d^{4}\right )} x^{4}\right )} \arctan \left (-\frac{d x - \sqrt{d x + c} \sqrt{d x - c}}{c}\right )}{8 \,{\left (c^{7} d^{2} x^{6} - c^{9} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^5/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/8*((2*a*c^5 - 3*(4*b*c^3*d^2 + 5*a*c*d^4)*x^4 + (4*b*c^5 + 5*a*c^3*d^2)*x^2)*sqrt(d*x + c)*sqrt(d*x - c) - 6
*((4*b*c^2*d^4 + 5*a*d^6)*x^6 - (4*b*c^4*d^2 + 5*a*c^2*d^4)*x^4)*arctan(-(d*x - sqrt(d*x + c)*sqrt(d*x - c))/c
))/(c^7*d^2*x^6 - c^9*x^4)

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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((b*x**2+a)/x**5/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.61696, size = 543, normalized size = 3.27 \begin{align*} \frac{3 \,{\left (4 \, b c^{2} d^{2} + 5 \, a d^{4}\right )} \arctan \left (\frac{{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}}{2 \, c}\right )}{4 \, c^{7}} - \frac{{\left (b c^{2} d^{2} + a d^{4}\right )} \sqrt{d x + c}}{2 \, \sqrt{d x - c} c^{7}} + \frac{2 \,{\left (b c^{2} d^{2} + a d^{4}\right )}}{{\left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2} + 2 \, c\right )} c^{6}} + \frac{4 \, b c^{2} d^{2}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{14} + 7 \, a d^{4}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{14} + 16 \, b c^{4} d^{2}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{10} + 60 \, a c^{2} d^{4}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{10} - 64 \, b c^{6} d^{2}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{6} - 240 \, a c^{4} d^{4}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{6} - 256 \, b c^{8} d^{2}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2} - 448 \, a c^{6} d^{4}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}}{2 \,{\left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{4} c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^5/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

3/4*(4*b*c^2*d^2 + 5*a*d^4)*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2/c)/c^7 - 1/2*(b*c^2*d^2 + a*d^4)*sqrt
(d*x + c)/(sqrt(d*x - c)*c^7) + 2*(b*c^2*d^2 + a*d^4)/(((sqrt(d*x + c) - sqrt(d*x - c))^2 + 2*c)*c^6) + 1/2*(4
*b*c^2*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^14 + 7*a*d^4*(sqrt(d*x + c) - sqrt(d*x - c))^14 + 16*b*c^4*d^2*(sqr
t(d*x + c) - sqrt(d*x - c))^10 + 60*a*c^2*d^4*(sqrt(d*x + c) - sqrt(d*x - c))^10 - 64*b*c^6*d^2*(sqrt(d*x + c)
- sqrt(d*x - c))^6 - 240*a*c^4*d^4*(sqrt(d*x + c) - sqrt(d*x - c))^6 - 256*b*c^8*d^2*(sqrt(d*x + c) - sqrt(d*
x - c))^2 - 448*a*c^6*d^4*(sqrt(d*x + c) - sqrt(d*x - c))^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^4*c^
6)