3.4.44 \(\int \frac {x^{7/2} (A+B x)}{a+b x} \, dx\) [344]

Optimal. Leaf size=136 \[ -\frac {2 a^3 (A b-a B) \sqrt {x}}{b^5}+\frac {2 a^2 (A b-a B) x^{3/2}}{3 b^4}-\frac {2 a (A b-a B) x^{5/2}}{5 b^3}+\frac {2 (A b-a B) x^{7/2}}{7 b^2}+\frac {2 B x^{9/2}}{9 b}+\frac {2 a^{7/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2}} \]

[Out]

2/3*a^2*(A*b-B*a)*x^(3/2)/b^4-2/5*a*(A*b-B*a)*x^(5/2)/b^3+2/7*(A*b-B*a)*x^(7/2)/b^2+2/9*B*x^(9/2)/b+2*a^(7/2)*
(A*b-B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(11/2)-2*a^3*(A*b-B*a)*x^(1/2)/b^5

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Rubi [A]
time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 211} \begin {gather*} \frac {2 a^{7/2} (A b-a B) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2}}-\frac {2 a^3 \sqrt {x} (A b-a B)}{b^5}+\frac {2 a^2 x^{3/2} (A b-a B)}{3 b^4}-\frac {2 a x^{5/2} (A b-a B)}{5 b^3}+\frac {2 x^{7/2} (A b-a B)}{7 b^2}+\frac {2 B x^{9/2}}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(a + b*x),x]

[Out]

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

Rule 52

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 65

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 81

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

Rule 211

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

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 120, normalized size = 0.88 \begin {gather*} \frac {2 \sqrt {x} \left (315 a^4 B-105 a^3 b (3 A+B x)+21 a^2 b^2 x (5 A+3 B x)-9 a b^3 x^2 (7 A+5 B x)+5 b^4 x^3 (9 A+7 B x)\right )}{315 b^5}-\frac {2 a^{7/2} (-A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(315*a^4*B - 105*a^3*b*(3*A + B*x) + 21*a^2*b^2*x*(5*A + 3*B*x) - 9*a*b^3*x^2*(7*A + 5*B*x) + 5*b^4
*x^3*(9*A + 7*B*x)))/(315*b^5) - (2*a^(7/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(11/2)

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Maple [A]
time = 0.06, size = 130, normalized size = 0.96

method	result	size
derivativedivides	\(-\frac {2 \left (-\frac {B \,x^{\frac {9}{2}} b^{4}}{9}-\frac {A \,b^{4} x^{\frac {7}{2}}}{7}+\frac {B a \,b^{3} x^{\frac {7}{2}}}{7}+\frac {A a \,b^{3} x^{\frac {5}{2}}}{5}-\frac {B \,a^{2} b^{2} x^{\frac {5}{2}}}{5}-\frac {A \,a^{2} b^{2} x^{\frac {3}{2}}}{3}+\frac {B \,a^{3} b \,x^{\frac {3}{2}}}{3}+A \,a^{3} b \sqrt {x}-B \,a^{4} \sqrt {x}\right )}{b^{5}}+\frac {2 a^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{5} \sqrt {a b}}\)	\(130\)
default	\(-\frac {2 \left (-\frac {B \,x^{\frac {9}{2}} b^{4}}{9}-\frac {A \,b^{4} x^{\frac {7}{2}}}{7}+\frac {B a \,b^{3} x^{\frac {7}{2}}}{7}+\frac {A a \,b^{3} x^{\frac {5}{2}}}{5}-\frac {B \,a^{2} b^{2} x^{\frac {5}{2}}}{5}-\frac {A \,a^{2} b^{2} x^{\frac {3}{2}}}{3}+\frac {B \,a^{3} b \,x^{\frac {3}{2}}}{3}+A \,a^{3} b \sqrt {x}-B \,a^{4} \sqrt {x}\right )}{b^{5}}+\frac {2 a^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{5} \sqrt {a b}}\)	\(130\)
risch	\(-\frac {2 \left (-35 B \,b^{4} x^{4}-45 A \,b^{4} x^{3}+45 B a \,b^{3} x^{3}+63 A a \,b^{3} x^{2}-63 B \,a^{2} b^{2} x^{2}-105 A \,a^{2} b^{2} x +105 B \,a^{3} b x +315 A \,a^{3} b -315 B \,a^{4}\right ) \sqrt {x}}{315 b^{5}}+\frac {2 a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) A}{b^{4} \sqrt {a b}}-\frac {2 a^{5} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) B}{b^{5} \sqrt {a b}}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 128, normalized size = 0.94 \begin {gather*} -\frac {2 \, {\left (B a^{5} - A a^{4} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} + \frac {2 \, {\left (35 \, B b^{4} x^{\frac {9}{2}} - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{\frac {7}{2}} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{\frac {5}{2}} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{\frac {3}{2}} + 315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {x}\right )}}{315 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a^5 - A*a^4*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 2/315*(35*B*b^4*x^(9/2) - 45*(B*a*b^3 - A*b
^4)*x^(7/2) + 63*(B*a^2*b^2 - A*a*b^3)*x^(5/2) - 105*(B*a^3*b - A*a^2*b^2)*x^(3/2) + 315*(B*a^4 - A*a^3*b)*sqr
t(x))/b^5

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Fricas [A]
time = 1.35, size = 276, normalized size = 2.03 \begin {gather*} \left [-\frac {315 \, {\left (B a^{4} - A a^{3} b\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 (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{315 \, b^{5}}, -\frac {2 \, {\left (315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}\right )}}{315 \, b^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/315*(315*(B*a^4 - A*a^3*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(35*B*b^4*x^4
+ 315*B*a^4 - 315*A*a^3*b - 45*(B*a*b^3 - A*b^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2
)*x)*sqrt(x))/b^5, -2/315*(315*(B*a^4 - A*a^3*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (35*B*b^4*x^4 + 315
*B*a^4 - 315*A*a^3*b - 45*(B*a*b^3 - A*b^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2)*x)*
sqrt(x))/b^5]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (131) = 262\).
time = 16.30, size = 328, normalized size = 2.41 \begin {gather*} \begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {9}{2}}}{9} + \frac {2 B x^{\frac {11}{2}}}{11}}{a} & \text {for}\: b = 0 \\\frac {A a^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{5} \sqrt {- \frac {a}{b}}} - \frac {A a^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{5} \sqrt {- \frac {a}{b}}} - \frac {2 A a^{3} \sqrt {x}}{b^{4}} + \frac {2 A a^{2} x^{\frac {3}{2}}}{3 b^{3}} - \frac {2 A a x^{\frac {5}{2}}}{5 b^{2}} + \frac {2 A x^{\frac {7}{2}}}{7 b} - \frac {B a^{5} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{6} \sqrt {- \frac {a}{b}}} + \frac {B a^{5} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{6} \sqrt {- \frac {a}{b}}} + \frac {2 B a^{4} \sqrt {x}}{b^{5}} - \frac {2 B a^{3} x^{\frac {3}{2}}}{3 b^{4}} + \frac {2 B a^{2} x^{\frac {5}{2}}}{5 b^{3}} - \frac {2 B a x^{\frac {7}{2}}}{7 b^{2}} + \frac {2 B x^{\frac {9}{2}}}{9 b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(2*A*x**(7/2)/7 + 2*B*x**(9/2)/9), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/b,
Eq(a, 0)), ((2*A*x**(9/2)/9 + 2*B*x**(11/2)/11)/a, Eq(b, 0)), (A*a**4*log(sqrt(x) - sqrt(-a/b))/(b**5*sqrt(-a/
b)) - A*a**4*log(sqrt(x) + sqrt(-a/b))/(b**5*sqrt(-a/b)) - 2*A*a**3*sqrt(x)/b**4 + 2*A*a**2*x**(3/2)/(3*b**3)
- 2*A*a*x**(5/2)/(5*b**2) + 2*A*x**(7/2)/(7*b) - B*a**5*log(sqrt(x) - sqrt(-a/b))/(b**6*sqrt(-a/b)) + B*a**5*l
og(sqrt(x) + sqrt(-a/b))/(b**6*sqrt(-a/b)) + 2*B*a**4*sqrt(x)/b**5 - 2*B*a**3*x**(3/2)/(3*b**4) + 2*B*a**2*x**
(5/2)/(5*b**3) - 2*B*a*x**(7/2)/(7*b**2) + 2*B*x**(9/2)/(9*b), True))

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Giac [A]
time = 1.11, size = 139, normalized size = 1.02 \begin {gather*} -\frac {2 \, {\left (B a^{5} - A a^{4} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} + \frac {2 \, {\left (35 \, B b^{8} x^{\frac {9}{2}} - 45 \, B a b^{7} x^{\frac {7}{2}} + 45 \, A b^{8} x^{\frac {7}{2}} + 63 \, B a^{2} b^{6} x^{\frac {5}{2}} - 63 \, A a b^{7} x^{\frac {5}{2}} - 105 \, B a^{3} b^{5} x^{\frac {3}{2}} + 105 \, A a^{2} b^{6} x^{\frac {3}{2}} + 315 \, B a^{4} b^{4} \sqrt {x} - 315 \, A a^{3} b^{5} \sqrt {x}\right )}}{315 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a^5 - A*a^4*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 2/315*(35*B*b^8*x^(9/2) - 45*B*a*b^7*x^(7/2
) + 45*A*b^8*x^(7/2) + 63*B*a^2*b^6*x^(5/2) - 63*A*a*b^7*x^(5/2) - 105*B*a^3*b^5*x^(3/2) + 105*A*a^2*b^6*x^(3/
2) + 315*B*a^4*b^4*sqrt(x) - 315*A*a^3*b^5*sqrt(x))/b^9

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Mupad [B]
time = 0.06, size = 151, normalized size = 1.11 \begin {gather*} x^{7/2}\,\left (\frac {2\,A}{7\,b}-\frac {2\,B\,a}{7\,b^2}\right )+\frac {2\,B\,x^{9/2}}{9\,b}+\frac {a^2\,x^{3/2}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{3\,b^2}-\frac {a^3\,\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{b^3}-\frac {2\,a^{7/2}\,\mathrm {atan}\left (\frac {a^{7/2}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^5-A\,a^4\,b}\right )\,\left (A\,b-B\,a\right )}{b^{11/2}}-\frac {a\,x^{5/2}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{5\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x))/(a + b*x),x)

[Out]

x^(7/2)*((2*A)/(7*b) - (2*B*a)/(7*b^2)) + (2*B*x^(9/2))/(9*b) + (a^2*x^(3/2)*((2*A)/b - (2*B*a)/b^2))/(3*b^2)
- (a^3*x^(1/2)*((2*A)/b - (2*B*a)/b^2))/b^3 - (2*a^(7/2)*atan((a^(7/2)*b^(1/2)*x^(1/2)*(A*b - B*a))/(B*a^5 - A
*a^4*b))*(A*b - B*a))/b^(11/2) - (a*x^(5/2)*((2*A)/b - (2*B*a)/b^2))/(5*b)

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