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

Optimal. Leaf size=154 \[ \frac{a^2 \sqrt{x} (7 A b-9 a B)}{b^5}-\frac{a^{5/2} (7 A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{11/2}}-\frac{x^{7/2} (7 A b-9 a B)}{7 a b^2}+\frac{x^{5/2} (7 A b-9 a B)}{5 b^3}-\frac{a x^{3/2} (7 A b-9 a B)}{3 b^4}+\frac{x^{9/2} (A b-a B)}{a b (a+b x)} \]

[Out]

(a^2*(7*A*b - 9*a*B)*Sqrt[x])/b^5 - (a*(7*A*b - 9*a*B)*x^(3/2))/(3*b^4) + ((7*A*b - 9*a*B)*x^(5/2))/(5*b^3) -
((7*A*b - 9*a*B)*x^(7/2))/(7*a*b^2) + ((A*b - a*B)*x^(9/2))/(a*b*(a + b*x)) - (a^(5/2)*(7*A*b - 9*a*B)*ArcTan[
(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(11/2)

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Rubi [A]  time = 0.0833223, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {27, 78, 50, 63, 205} \[ \frac{a^2 \sqrt{x} (7 A b-9 a B)}{b^5}-\frac{a^{5/2} (7 A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{11/2}}-\frac{x^{7/2} (7 A b-9 a B)}{7 a b^2}+\frac{x^{5/2} (7 A b-9 a B)}{5 b^3}-\frac{a x^{3/2} (7 A b-9 a B)}{3 b^4}+\frac{x^{9/2} (A b-a B)}{a b (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*A*b - 9*a*B)*Sqrt[x])/b^5 - (a*(7*A*b - 9*a*B)*x^(3/2))/(3*b^4) + ((7*A*b - 9*a*B)*x^(5/2))/(5*b^3) -
((7*A*b - 9*a*B)*x^(7/2))/(7*a*b^2) + ((A*b - a*B)*x^(9/2))/(a*b*(a + b*x)) - (a^(5/2)*(7*A*b - 9*a*B)*ArcTan[
(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(11/2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 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{x^{7/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{x^{7/2} (A+B x)}{(a+b x)^2} \, dx\\ &=\frac{(A b-a B) x^{9/2}}{a b (a+b x)}-\frac{\left (\frac{7 A b}{2}-\frac{9 a B}{2}\right ) \int \frac{x^{7/2}}{a+b x} \, dx}{a b}\\ &=-\frac{(7 A b-9 a B) x^{7/2}}{7 a b^2}+\frac{(A b-a B) x^{9/2}}{a b (a+b x)}+\frac{(7 A b-9 a B) \int \frac{x^{5/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac{(7 A b-9 a B) x^{5/2}}{5 b^3}-\frac{(7 A b-9 a B) x^{7/2}}{7 a b^2}+\frac{(A b-a B) x^{9/2}}{a b (a+b x)}-\frac{(a (7 A b-9 a B)) \int \frac{x^{3/2}}{a+b x} \, dx}{2 b^3}\\ &=-\frac{a (7 A b-9 a B) x^{3/2}}{3 b^4}+\frac{(7 A b-9 a B) x^{5/2}}{5 b^3}-\frac{(7 A b-9 a B) x^{7/2}}{7 a b^2}+\frac{(A b-a B) x^{9/2}}{a b (a+b x)}+\frac{\left (a^2 (7 A b-9 a B)\right ) \int \frac{\sqrt{x}}{a+b x} \, dx}{2 b^4}\\ &=\frac{a^2 (7 A b-9 a B) \sqrt{x}}{b^5}-\frac{a (7 A b-9 a B) x^{3/2}}{3 b^4}+\frac{(7 A b-9 a B) x^{5/2}}{5 b^3}-\frac{(7 A b-9 a B) x^{7/2}}{7 a b^2}+\frac{(A b-a B) x^{9/2}}{a b (a+b x)}-\frac{\left (a^3 (7 A b-9 a B)\right ) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{2 b^5}\\ &=\frac{a^2 (7 A b-9 a B) \sqrt{x}}{b^5}-\frac{a (7 A b-9 a B) x^{3/2}}{3 b^4}+\frac{(7 A b-9 a B) x^{5/2}}{5 b^3}-\frac{(7 A b-9 a B) x^{7/2}}{7 a b^2}+\frac{(A b-a B) x^{9/2}}{a b (a+b x)}-\frac{\left (a^3 (7 A b-9 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{b^5}\\ &=\frac{a^2 (7 A b-9 a B) \sqrt{x}}{b^5}-\frac{a (7 A b-9 a B) x^{3/2}}{3 b^4}+\frac{(7 A b-9 a B) x^{5/2}}{5 b^3}-\frac{(7 A b-9 a B) x^{7/2}}{7 a b^2}+\frac{(A b-a B) x^{9/2}}{a b (a+b x)}-\frac{a^{5/2} (7 A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0921146, size = 128, normalized size = 0.83 \[ \frac{\sqrt{x} \left (14 a^2 b^2 x (35 A+9 B x)+105 a^3 b (7 A-6 B x)-945 a^4 B-2 a b^3 x^2 (49 A+27 B x)+6 b^4 x^3 (7 A+5 B x)\right )}{105 b^5 (a+b x)}+\frac{a^{5/2} (9 a B-7 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{11/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-945*a^4*B + 105*a^3*b*(7*A - 6*B*x) + 6*b^4*x^3*(7*A + 5*B*x) + 14*a^2*b^2*x*(35*A + 9*B*x) - 2*a*b
^3*x^2*(49*A + 27*B*x)))/(105*b^5*(a + b*x)) + (a^(5/2)*(-7*A*b + 9*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^
(11/2)

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Maple [A]  time = 0.018, size = 163, normalized size = 1.1 \begin{align*}{\frac{2\,B}{7\,{b}^{2}}{x}^{{\frac{7}{2}}}}+{\frac{2\,A}{5\,{b}^{2}}{x}^{{\frac{5}{2}}}}-{\frac{4\,aB}{5\,{b}^{3}}{x}^{{\frac{5}{2}}}}-{\frac{4\,aA}{3\,{b}^{3}}{x}^{{\frac{3}{2}}}}+2\,{\frac{{a}^{2}B{x}^{3/2}}{{b}^{4}}}+6\,{\frac{{a}^{2}A\sqrt{x}}{{b}^{4}}}-8\,{\frac{{a}^{3}B\sqrt{x}}{{b}^{5}}}+{\frac{A{a}^{3}}{{b}^{4} \left ( bx+a \right ) }\sqrt{x}}-{\frac{B{a}^{4}}{{b}^{5} \left ( bx+a \right ) }\sqrt{x}}-7\,{\frac{A{a}^{3}}{{b}^{4}\sqrt{ab}}\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) }+9\,{\frac{B{a}^{4}}{{b}^{5}\sqrt{ab}}\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61693, size = 767, normalized size = 4.98 \begin{align*} \left [-\frac{105 \,{\left (9 \, B a^{4} - 7 \, A a^{3} b +{\left (9 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) - 2 \,{\left (30 \, B b^{4} x^{4} - 945 \, B a^{4} + 735 \, A a^{3} b - 6 \,{\left (9 \, B a b^{3} - 7 \, A b^{4}\right )} x^{3} + 14 \,{\left (9 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{2} - 70 \,{\left (9 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{210 \,{\left (b^{6} x + a b^{5}\right )}}, \frac{105 \,{\left (9 \, B a^{4} - 7 \, A a^{3} b +{\left (9 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (30 \, B b^{4} x^{4} - 945 \, B a^{4} + 735 \, A a^{3} b - 6 \,{\left (9 \, B a b^{3} - 7 \, A b^{4}\right )} x^{3} + 14 \,{\left (9 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{2} - 70 \,{\left (9 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{105 \,{\left (b^{6} x + a b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/210*(105*(9*B*a^4 - 7*A*a^3*b + (9*B*a^3*b - 7*A*a^2*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b)
- a)/(b*x + a)) - 2*(30*B*b^4*x^4 - 945*B*a^4 + 735*A*a^3*b - 6*(9*B*a*b^3 - 7*A*b^4)*x^3 + 14*(9*B*a^2*b^2 -
7*A*a*b^3)*x^2 - 70*(9*B*a^3*b - 7*A*a^2*b^2)*x)*sqrt(x))/(b^6*x + a*b^5), 1/105*(105*(9*B*a^4 - 7*A*a^3*b + (
9*B*a^3*b - 7*A*a^2*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (30*B*b^4*x^4 - 945*B*a^4 + 735*A*a^3*b
- 6*(9*B*a*b^3 - 7*A*b^4)*x^3 + 14*(9*B*a^2*b^2 - 7*A*a*b^3)*x^2 - 70*(9*B*a^3*b - 7*A*a^2*b^2)*x)*sqrt(x))/(b
^6*x + a*b^5)]

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

[Out]

Timed out

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Giac [A]  time = 1.16006, size = 197, normalized size = 1.28 \begin{align*} \frac{{\left (9 \, B a^{4} - 7 \, A a^{3} b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{5}} - \frac{B a^{4} \sqrt{x} - A a^{3} b \sqrt{x}}{{\left (b x + a\right )} b^{5}} + \frac{2 \,{\left (15 \, B b^{12} x^{\frac{7}{2}} - 42 \, B a b^{11} x^{\frac{5}{2}} + 21 \, A b^{12} x^{\frac{5}{2}} + 105 \, B a^{2} b^{10} x^{\frac{3}{2}} - 70 \, A a b^{11} x^{\frac{3}{2}} - 420 \, B a^{3} b^{9} \sqrt{x} + 315 \, A a^{2} b^{10} \sqrt{x}\right )}}{105 \, b^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*B*a^4 - 7*A*a^3*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - (B*a^4*sqrt(x) - A*a^3*b*sqrt(x))/((b*x +
a)*b^5) + 2/105*(15*B*b^12*x^(7/2) - 42*B*a*b^11*x^(5/2) + 21*A*b^12*x^(5/2) + 105*B*a^2*b^10*x^(3/2) - 70*A*a
*b^11*x^(3/2) - 420*B*a^3*b^9*sqrt(x) + 315*A*a^2*b^10*sqrt(x))/b^14