3.550 \(\int \frac{(a+b x^2)^{5/2} (A+B x^2)}{x^7} \, dx\)

Optimal. Leaf size=149 \[ \frac{5 b^2 \sqrt{a+b x^2} (6 a B+A b)}{16 a}-\frac{5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{\left (a+b x^2\right )^{5/2} (6 a B+A b)}{24 a x^4}-\frac{5 b \left (a+b x^2\right )^{3/2} (6 a B+A b)}{48 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6} \]

[Out]

(5*b^2*(A*b + 6*a*B)*Sqrt[a + b*x^2])/(16*a) - (5*b*(A*b + 6*a*B)*(a + b*x^2)^(3/2))/(48*a*x^2) - ((A*b + 6*a*
B)*(a + b*x^2)^(5/2))/(24*a*x^4) - (A*(a + b*x^2)^(7/2))/(6*a*x^6) - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x
^2]/Sqrt[a]])/(16*Sqrt[a])

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Rubi [A]  time = 0.110362, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {446, 78, 47, 50, 63, 208} \[ \frac{5 b^2 \sqrt{a+b x^2} (6 a B+A b)}{16 a}-\frac{5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{\left (a+b x^2\right )^{5/2} (6 a B+A b)}{24 a x^4}-\frac{5 b \left (a+b x^2\right )^{3/2} (6 a B+A b)}{48 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^7,x]

[Out]

(5*b^2*(A*b + 6*a*B)*Sqrt[a + b*x^2])/(16*a) - (5*b*(A*b + 6*a*B)*(a + b*x^2)^(3/2))/(48*a*x^2) - ((A*b + 6*a*
B)*(a + b*x^2)^(5/2))/(24*a*x^4) - (A*(a + b*x^2)^(7/2))/(6*a*x^6) - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x
^2]/Sqrt[a]])/(16*Sqrt[a])

Rule 446

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} (A+B x)}{x^4} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac{(A b+6 a B) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^3} \, dx,x,x^2\right )}{12 a}\\ &=-\frac{(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac{(5 b (A b+6 a B)) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )}{48 a}\\ &=-\frac{5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac{(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac{\left (5 b^2 (A b+6 a B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )}{32 a}\\ &=\frac{5 b^2 (A b+6 a B) \sqrt{a+b x^2}}{16 a}-\frac{5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac{(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac{1}{32} \left (5 b^2 (A b+6 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{5 b^2 (A b+6 a B) \sqrt{a+b x^2}}{16 a}-\frac{5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac{(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac{1}{16} (5 b (A b+6 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{5 b^2 (A b+6 a B) \sqrt{a+b x^2}}{16 a}-\frac{5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac{(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac{A \left (a+b x^2\right )^{7/2}}{6 a x^6}-\frac{5 b^2 (A b+6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 \sqrt{a}}\\ \end{align*}

Mathematica [C]  time = 0.0278098, size = 61, normalized size = 0.41 \[ -\frac{\left (a+b x^2\right )^{7/2} \left (7 a^3 A+b^2 x^6 (6 a B+A b) \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b x^2}{a}+1\right )\right )}{42 a^4 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^7,x]

[Out]

-((a + b*x^2)^(7/2)*(7*a^3*A + b^2*(A*b + 6*a*B)*x^6*Hypergeometric2F1[3, 7/2, 9/2, 1 + (b*x^2)/a]))/(42*a^4*x
^6)

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Maple [B]  time = 0.012, size = 266, normalized size = 1.8 \begin{align*} -{\frac{A}{6\,a{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{Ab}{24\,{a}^{2}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{A{b}^{2}}{16\,{a}^{3}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{A{b}^{3}}{16\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,A{b}^{3}}{48\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,A{b}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{5\,A{b}^{3}}{16\,a}\sqrt{b{x}^{2}+a}}-{\frac{B}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,Bb}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{3\,B{b}^{2}}{8\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,B{b}^{2}}{8\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{15\,B{b}^{2}}{8}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{15\,B{b}^{2}}{8}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x)

[Out]

-1/6*A*(b*x^2+a)^(7/2)/a/x^6-1/24*A*b/a^2/x^4*(b*x^2+a)^(7/2)-1/16*A*b^2/a^3/x^2*(b*x^2+a)^(7/2)+1/16*A*b^3/a^
3*(b*x^2+a)^(5/2)+5/48*A*b^3/a^2*(b*x^2+a)^(3/2)-5/16*A*b^3/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+5/16
*A*b^3/a*(b*x^2+a)^(1/2)-1/4*B/a/x^4*(b*x^2+a)^(7/2)-3/8*B*b/a^2/x^2*(b*x^2+a)^(7/2)+3/8*B*b^2/a^2*(b*x^2+a)^(
5/2)+5/8*B*b^2/a*(b*x^2+a)^(3/2)-15/8*B*b^2*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+15/8*B*b^2*(b*x^2+a)
^(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)^(5/2)*(B*x^2+A)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72656, size = 559, normalized size = 3.75 \begin{align*} \left [\frac{15 \,{\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt{a} x^{6} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (48 \, B a b^{2} x^{6} - 3 \,{\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{4} - 8 \, A a^{3} - 2 \,{\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{96 \, a x^{6}}, \frac{15 \,{\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt{-a} x^{6} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (48 \, B a b^{2} x^{6} - 3 \,{\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{4} - 8 \, A a^{3} - 2 \,{\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{48 \, a x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x, algorithm="fricas")

[Out]

[1/96*(15*(6*B*a*b^2 + A*b^3)*sqrt(a)*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(48*B*a*b^2*
x^6 - 3*(18*B*a^2*b + 11*A*a*b^2)*x^4 - 8*A*a^3 - 2*(6*B*a^3 + 13*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a*x^6), 1/48
*(15*(6*B*a*b^2 + A*b^3)*sqrt(-a)*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (48*B*a*b^2*x^6 - 3*(18*B*a^2*b + 11*
A*a*b^2)*x^4 - 8*A*a^3 - 2*(6*B*a^3 + 13*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a*x^6)]

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Sympy [B]  time = 90.2878, size = 306, normalized size = 2.05 \begin{align*} - \frac{A a^{3}}{6 \sqrt{b} x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{17 A a^{2} \sqrt{b}}{24 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{35 A a b^{\frac{3}{2}}}{48 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{A b^{\frac{5}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} - \frac{3 A b^{\frac{5}{2}}}{16 x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 A b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{16 \sqrt{a}} - \frac{15 B \sqrt{a} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8} - \frac{B a^{3}}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 B a^{2} \sqrt{b}}{8 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{B a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{x} + \frac{7 B a b^{\frac{3}{2}}}{8 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{B b^{\frac{5}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**7,x)

[Out]

-A*a**3/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 17*A*a**2*sqrt(b)/(24*x**5*sqrt(a/(b*x**2) + 1)) - 35*A*a*b**(
3/2)/(48*x**3*sqrt(a/(b*x**2) + 1)) - A*b**(5/2)*sqrt(a/(b*x**2) + 1)/(2*x) - 3*A*b**(5/2)/(16*x*sqrt(a/(b*x**
2) + 1)) - 5*A*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*sqrt(a)) - 15*B*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*x))/8 -
B*a**3/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 3*B*a**2*sqrt(b)/(8*x**3*sqrt(a/(b*x**2) + 1)) - B*a*b**(3/2)*
sqrt(a/(b*x**2) + 1)/x + 7*B*a*b**(3/2)/(8*x*sqrt(a/(b*x**2) + 1)) + B*b**(5/2)*x/sqrt(a/(b*x**2) + 1)

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Giac [A]  time = 1.14836, size = 225, normalized size = 1.51 \begin{align*} \frac{48 \, \sqrt{b x^{2} + a} B b^{3} + \frac{15 \,{\left (6 \, B a b^{3} + A b^{4}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{54 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B a b^{3} - 96 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{2} b^{3} + 42 \, \sqrt{b x^{2} + a} B a^{3} b^{3} + 33 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} A b^{4} - 40 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a b^{4} + 15 \, \sqrt{b x^{2} + a} A a^{2} b^{4}}{b^{3} x^{6}}}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x, algorithm="giac")

[Out]

1/48*(48*sqrt(b*x^2 + a)*B*b^3 + 15*(6*B*a*b^3 + A*b^4)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (54*(b*x^2
+ a)^(5/2)*B*a*b^3 - 96*(b*x^2 + a)^(3/2)*B*a^2*b^3 + 42*sqrt(b*x^2 + a)*B*a^3*b^3 + 33*(b*x^2 + a)^(5/2)*A*b
^4 - 40*(b*x^2 + a)^(3/2)*A*a*b^4 + 15*sqrt(b*x^2 + a)*A*a^2*b^4)/(b^3*x^6))/b