3.363 \(\int \frac{x^{7/2} (A+B x)}{(a+b x)^3} \, dx\)

Optimal. Leaf size=169 \[ \frac{7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{11/2}}+\frac{x^{7/2} (5 A b-9 a B)}{4 a b^2 (a+b x)}-\frac{7 x^{5/2} (5 A b-9 a B)}{20 a b^3}+\frac{7 x^{3/2} (5 A b-9 a B)}{12 b^4}-\frac{7 a \sqrt{x} (5 A b-9 a B)}{4 b^5}+\frac{x^{9/2} (A b-a B)}{2 a b (a+b x)^2} \]

[Out]

(-7*a*(5*A*b - 9*a*B)*Sqrt[x])/(4*b^5) + (7*(5*A*b - 9*a*B)*x^(3/2))/(12*b^4) - (7*(5*A*b - 9*a*B)*x^(5/2))/(2
0*a*b^3) + ((A*b - a*B)*x^(9/2))/(2*a*b*(a + b*x)^2) + ((5*A*b - 9*a*B)*x^(7/2))/(4*a*b^2*(a + b*x)) + (7*a^(3
/2)*(5*A*b - 9*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(11/2))

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Rubi [A]  time = 0.0773488, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 205} \[ \frac{7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{11/2}}+\frac{x^{7/2} (5 A b-9 a B)}{4 a b^2 (a+b x)}-\frac{7 x^{5/2} (5 A b-9 a B)}{20 a b^3}+\frac{7 x^{3/2} (5 A b-9 a B)}{12 b^4}-\frac{7 a \sqrt{x} (5 A b-9 a B)}{4 b^5}+\frac{x^{9/2} (A b-a B)}{2 a b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(a + b*x)^3,x]

[Out]

(-7*a*(5*A*b - 9*a*B)*Sqrt[x])/(4*b^5) + (7*(5*A*b - 9*a*B)*x^(3/2))/(12*b^4) - (7*(5*A*b - 9*a*B)*x^(5/2))/(2
0*a*b^3) + ((A*b - a*B)*x^(9/2))/(2*a*b*(a + b*x)^2) + ((5*A*b - 9*a*B)*x^(7/2))/(4*a*b^2*(a + b*x)) + (7*a^(3
/2)*(5*A*b - 9*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(11/2))

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 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+b x)^3} \, dx &=\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}-\frac{\left (\frac{5 A b}{2}-\frac{9 a B}{2}\right ) \int \frac{x^{7/2}}{(a+b x)^2} \, dx}{2 a b}\\ &=\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac{(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}-\frac{(7 (5 A b-9 a B)) \int \frac{x^{5/2}}{a+b x} \, dx}{8 a b^2}\\ &=-\frac{7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac{(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac{(7 (5 A b-9 a B)) \int \frac{x^{3/2}}{a+b x} \, dx}{8 b^3}\\ &=\frac{7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac{7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac{(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}-\frac{(7 a (5 A b-9 a B)) \int \frac{\sqrt{x}}{a+b x} \, dx}{8 b^4}\\ &=-\frac{7 a (5 A b-9 a B) \sqrt{x}}{4 b^5}+\frac{7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac{7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac{(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac{\left (7 a^2 (5 A b-9 a B)\right ) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{8 b^5}\\ &=-\frac{7 a (5 A b-9 a B) \sqrt{x}}{4 b^5}+\frac{7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac{7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac{(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac{\left (7 a^2 (5 A b-9 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{4 b^5}\\ &=-\frac{7 a (5 A b-9 a B) \sqrt{x}}{4 b^5}+\frac{7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac{7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac{(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac{(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac{7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0263418, size = 61, normalized size = 0.36 \[ \frac{x^{9/2} \left (\frac{9 a^2 (A b-a B)}{(a+b x)^2}+(9 a B-5 A b) \, _2F_1\left (2,\frac{9}{2};\frac{11}{2};-\frac{b x}{a}\right )\right )}{18 a^3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x))/(a + b*x)^3,x]

[Out]

(x^(9/2)*((9*a^2*(A*b - a*B))/(a + b*x)^2 + (-5*A*b + 9*a*B)*Hypergeometric2F1[2, 9/2, 11/2, -((b*x)/a)]))/(18
*a^3*b)

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Maple [A]  time = 0.014, size = 178, normalized size = 1.1 \begin{align*}{\frac{2\,B}{5\,{b}^{3}}{x}^{{\frac{5}{2}}}}+{\frac{2\,A}{3\,{b}^{3}}{x}^{{\frac{3}{2}}}}-2\,{\frac{B{x}^{3/2}a}{{b}^{4}}}-6\,{\frac{aA\sqrt{x}}{{b}^{4}}}+12\,{\frac{B{a}^{2}\sqrt{x}}{{b}^{5}}}-{\frac{13\,A{a}^{2}}{4\,{b}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{17\,B{a}^{3}}{4\,{b}^{4} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{11\,A{a}^{3}}{4\,{b}^{4} \left ( bx+a \right ) ^{2}}\sqrt{x}}+{\frac{15\,B{a}^{4}}{4\,{b}^{5} \left ( bx+a \right ) ^{2}}\sqrt{x}}+{\frac{35\,A{a}^{2}}{4\,{b}^{4}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{63\,B{a}^{3}}{4\,{b}^{5}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/(b*x+a)^3,x)

[Out]

2/5/b^3*B*x^(5/2)+2/3/b^3*A*x^(3/2)-2/b^4*B*x^(3/2)*a-6/b^4*a*A*x^(1/2)+12/b^5*a^2*B*x^(1/2)-13/4*a^2/b^3/(b*x
+a)^2*A*x^(3/2)+17/4*a^3/b^4/(b*x+a)^2*B*x^(3/2)-11/4*a^3/b^4/(b*x+a)^2*A*x^(1/2)+15/4*a^4/b^5/(b*x+a)^2*B*x^(
1/2)+35/4*a^2/b^4/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))*A-63/4*a^3/b^5/(a*b)^(1/2)*arctan(b*x^(1/2)/(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*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.68815, size = 905, normalized size = 5.36 \begin{align*} \left [-\frac{105 \,{\left (9 \, B a^{4} - 5 \, A a^{3} b +{\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \,{\left (9 \, B a^{3} b - 5 \, 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 (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \,{\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \,{\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \,{\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{120 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac{105 \,{\left (9 \, B a^{4} - 5 \, A a^{3} b +{\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \,{\left (9 \, B a^{3} b - 5 \, 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 (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \,{\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \,{\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \,{\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{60 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} 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*x+a)^3,x, algorithm="fricas")

[Out]

[-1/120*(105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(-a/b)*
log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(24*B*b^4*x^4 + 945*B*a^4 - 525*A*a^3*b - 8*(9*B*a*b^3 -
5*A*b^4)*x^3 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 175*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(b^7*x^2 + 2*a*b^
6*x + a^2*b^5), -1/60*(105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*
x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (24*B*b^4*x^4 + 945*B*a^4 - 525*A*a^3*b - 8*(9*B*a*b^3 - 5*A*b^4)
*x^3 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 175*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2
*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*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.19546, size = 197, normalized size = 1.17 \begin{align*} -\frac{7 \,{\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{5}} + \frac{17 \, B a^{3} b x^{\frac{3}{2}} - 13 \, A a^{2} b^{2} x^{\frac{3}{2}} + 15 \, B a^{4} \sqrt{x} - 11 \, A a^{3} b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{5}} + \frac{2 \,{\left (3 \, B b^{12} x^{\frac{5}{2}} - 15 \, B a b^{11} x^{\frac{3}{2}} + 5 \, A b^{12} x^{\frac{3}{2}} + 90 \, B a^{2} b^{10} \sqrt{x} - 45 \, A a b^{11} \sqrt{x}\right )}}{15 \, b^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^3,x, algorithm="giac")

[Out]

-7/4*(9*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/4*(17*B*a^3*b*x^(3/2) - 13*A*a^2*b^
2*x^(3/2) + 15*B*a^4*sqrt(x) - 11*A*a^3*b*sqrt(x))/((b*x + a)^2*b^5) + 2/15*(3*B*b^12*x^(5/2) - 15*B*a*b^11*x^
(3/2) + 5*A*b^12*x^(3/2) + 90*B*a^2*b^10*sqrt(x) - 45*A*a*b^11*sqrt(x))/b^15