3.6.31 \(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^6} \, dx\)

Optimal. Leaf size=152 \[ \frac {1}{2} b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {b^2 x \sqrt {a+b x^2} (5 a B+2 A b)}{2 a}-\frac {b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac {\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

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Rubi [A]  time = 0.06, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {453, 277, 195, 217, 206} \begin {gather*} \frac {b^2 x \sqrt {a+b x^2} (5 a B+2 A b)}{2 a}+\frac {1}{2} b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac {b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

(b^2*(2*A*b + 5*a*B)*x*Sqrt[a + b*x^2])/(2*a) - (b*(2*A*b + 5*a*B)*(a + b*x^2)^(3/2))/(3*a*x) - ((2*A*b + 5*a*
B)*(a + b*x^2)^(5/2))/(15*a*x^3) - (A*(a + b*x^2)^(7/2))/(5*a*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx &=-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}-\frac {(-2 A b-5 a B) \int \frac {\left (a+b x^2\right )^{5/2}}{x^4} \, dx}{5 a}\\ &=-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {(b (2 A b+5 a B)) \int \frac {\left (a+b x^2\right )^{3/2}}{x^2} \, dx}{3 a}\\ &=-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {\left (b^2 (2 A b+5 a B)\right ) \int \sqrt {a+b x^2} \, dx}{a}\\ &=\frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} \left (b^2 (2 A b+5 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} \left (b^2 (2 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 84, normalized size = 0.55 \begin {gather*} \frac {a \sqrt {a+b x^2} (-5 a B-2 A b) \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x^2}{a}\right )}{15 x^3 \sqrt {\frac {b x^2}{a}+1}}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

-1/5*(A*(a + b*x^2)^(7/2))/(a*x^5) + (a*(-2*A*b - 5*a*B)*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/2, -3/2, -1/2, -
((b*x^2)/a)])/(15*x^3*Sqrt[1 + (b*x^2)/a])

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IntegrateAlgebraic [A]  time = 0.29, size = 112, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-6 a^2 A-10 a^2 B x^2-22 a A b x^2-70 a b B x^4-46 A b^2 x^4+15 b^2 B x^6\right )}{30 x^5}+\frac {1}{2} \left (-5 a b^{3/2} B-2 A b^{5/2}\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

(Sqrt[a + b*x^2]*(-6*a^2*A - 22*a*A*b*x^2 - 10*a^2*B*x^2 - 46*A*b^2*x^4 - 70*a*b*B*x^4 + 15*b^2*B*x^6))/(30*x^
5) + ((-2*A*b^(5/2) - 5*a*b^(3/2)*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/2

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fricas [A]  time = 1.12, size = 220, normalized size = 1.45 \begin {gather*} \left [\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {b} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (15 \, B b^{2} x^{6} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{60 \, x^{5}}, -\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (15 \, B b^{2} x^{6} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="fricas")

[Out]

[1/60*(15*(5*B*a*b + 2*A*b^2)*sqrt(b)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(15*B*b^2*x^6 -
2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5, -1/30*(15*(5*B*a*b +
2*A*b^2)*sqrt(-b)*x^5*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (15*B*b^2*x^6 - 2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*
a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5]

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giac [B]  time = 0.57, size = 321, normalized size = 2.11 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a} B b^{2} x - \frac {1}{4} \, {\left (5 \, B a b^{\frac {3}{2}} + 2 \, A b^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{2} b^{\frac {3}{2}} + 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a b^{\frac {5}{2}} - 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{3} b^{\frac {3}{2}} - 90 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A a^{2} b^{\frac {5}{2}} + 200 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{4} b^{\frac {3}{2}} + 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{3} b^{\frac {5}{2}} - 130 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{5} b^{\frac {3}{2}} - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{4} b^{\frac {5}{2}} + 35 \, B a^{6} b^{\frac {3}{2}} + 23 \, A a^{5} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*B*b^2*x - 1/4*(5*B*a*b^(3/2) + 2*A*b^(5/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/15*(4
5*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^2*b^(3/2) + 45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*a*b^(5/2) - 150*(sqrt(b
)*x - sqrt(b*x^2 + a))^6*B*a^3*b^(3/2) - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a^2*b^(5/2) + 200*(sqrt(b)*x - s
qrt(b*x^2 + a))^4*B*a^4*b^(3/2) + 140*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^3*b^(5/2) - 130*(sqrt(b)*x - sqrt(b*
x^2 + a))^2*B*a^5*b^(3/2) - 70*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^4*b^(5/2) + 35*B*a^6*b^(3/2) + 23*A*a^5*b^(
5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5

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maple [A]  time = 0.01, size = 251, normalized size = 1.65 \begin {gather*} A \,b^{\frac {5}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\frac {5 B a \,b^{\frac {3}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {\sqrt {b \,x^{2}+a}\, A \,b^{3} x}{a}+\frac {5 \sqrt {b \,x^{2}+a}\, B \,b^{2} x}{2}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{3} x}{3 a^{2}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{2} x}{3 a}+\frac {8 \left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{3} x}{15 a^{3}}+\frac {4 \left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,b^{2} x}{3 a^{2}}-\frac {8 \left (b \,x^{2}+a \right )^{\frac {7}{2}} A \,b^{2}}{15 a^{3} x}-\frac {4 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B b}{3 a^{2} x}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} A b}{15 a^{2} x^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{3 a \,x^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x)

[Out]

-1/5*A*(b*x^2+a)^(7/2)/a/x^5-2/15*A*b/a^2/x^3*(b*x^2+a)^(7/2)-8/15*A*b^2/a^3/x*(b*x^2+a)^(7/2)+8/15*A*b^3/a^3*
x*(b*x^2+a)^(5/2)+2/3*A*b^3/a^2*x*(b*x^2+a)^(3/2)+A*b^3/a*x*(b*x^2+a)^(1/2)+A*b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(
1/2))-1/3*B/a/x^3*(b*x^2+a)^(7/2)-4/3*B*b/a^2/x*(b*x^2+a)^(7/2)+4/3*B*b^2/a^2*x*(b*x^2+a)^(5/2)+5/3*B*b^2/a*x*
(b*x^2+a)^(3/2)+5/2*B*b^2*x*(b*x^2+a)^(1/2)+5/2*B*b^(3/2)*a*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.12, size = 198, normalized size = 1.30 \begin {gather*} \frac {5}{2} \, \sqrt {b x^{2} + a} B b^{2} x + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2} x}{3 \, a} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} A b^{3} x}{a} + \frac {5}{2} \, B a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) + A b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{3 \, a x} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{15 \, a^{2} x} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{3 \, a x^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="maxima")

[Out]

5/2*sqrt(b*x^2 + a)*B*b^2*x + 5/3*(b*x^2 + a)^(3/2)*B*b^2*x/a + 2/3*(b*x^2 + a)^(3/2)*A*b^3*x/a^2 + sqrt(b*x^2
+ a)*A*b^3*x/a + 5/2*B*a*b^(3/2)*arcsinh(b*x/sqrt(a*b)) + A*b^(5/2)*arcsinh(b*x/sqrt(a*b)) - 4/3*(b*x^2 + a)^
(5/2)*B*b/(a*x) - 8/15*(b*x^2 + a)^(5/2)*A*b^2/(a^2*x) - 1/3*(b*x^2 + a)^(7/2)*B/(a*x^3) - 2/15*(b*x^2 + a)^(7
/2)*A*b/(a^2*x^3) - 1/5*(b*x^2 + a)^(7/2)*A/(a*x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^6,x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^6, x)

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sympy [B]  time = 11.54, size = 292, normalized size = 1.92 \begin {gather*} - \frac {A \sqrt {a} b^{2}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {8 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} + A b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b^{3} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {2 B a^{\frac {3}{2}} b}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B \sqrt {a} b^{2} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} - \frac {2 B \sqrt {a} b^{2} x}{\sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3} + \frac {5 B a b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**6,x)

[Out]

-A*sqrt(a)*b**2/(x*sqrt(1 + b*x**2/a)) - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 11*A*a*b**(3/2)*sqrt(a
/(b*x**2) + 1)/(15*x**2) - 8*A*b**(5/2)*sqrt(a/(b*x**2) + 1)/15 + A*b**(5/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**3
*x/(sqrt(a)*sqrt(1 + b*x**2/a)) - 2*B*a**(3/2)*b/(x*sqrt(1 + b*x**2/a)) + B*sqrt(a)*b**2*x*sqrt(1 + b*x**2/a)/
2 - 2*B*sqrt(a)*b**2*x/sqrt(1 + b*x**2/a) - B*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - B*a*b**(3/2)*sqrt(a
/(b*x**2) + 1)/3 + 5*B*a*b**(3/2)*asinh(sqrt(b)*x/sqrt(a))/2

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