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

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

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Rubi [A]  time = 0.08, antiderivative size = 169, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.36 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.21, size = 146, normalized size = 0.86 \begin {gather*} \frac {\sqrt {x} \left (945 a^4 B-525 a^3 A b+1575 a^3 b B x-875 a^2 A b^2 x+504 a^2 b^2 B x^2-280 a A b^3 x^2-72 a b^3 B x^3+40 A b^4 x^3+24 b^4 B x^4\right )}{60 b^5 (a+b x)^2}-\frac {7 \left (9 a^{5/2} B-5 a^{3/2} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-525*a^3*A*b + 945*a^4*B - 875*a^2*A*b^2*x + 1575*a^3*b*B*x - 280*a*A*b^3*x^2 + 504*a^2*b^2*B*x^2 +
40*A*b^4*x^3 - 72*a*b^3*B*x^3 + 24*b^4*B*x^4))/(60*b^5*(a + b*x)^2) - (7*(-5*a^(3/2)*A*b + 9*a^(5/2)*B)*ArcTan
[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(11/2))

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fricas [A]  time = 0.98, size = 408, normalized size = 2.41 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.26, size = 146, normalized size = 0.86 \begin {gather*} -\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 {gather*}

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

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maple [A]  time = 0.02, size = 178, normalized size = 1.05 \begin {gather*} -\frac {13 A \,a^{2} x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} b^{3}}+\frac {17 B \,a^{3} x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} b^{4}}-\frac {11 A \,a^{3} \sqrt {x}}{4 \left (b x +a \right )^{2} b^{4}}+\frac {15 B \,a^{4} \sqrt {x}}{4 \left (b x +a \right )^{2} b^{5}}+\frac {2 B \,x^{\frac {5}{2}}}{5 b^{3}}+\frac {35 A \,a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{4}}-\frac {63 B \,a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{5}}+\frac {2 A \,x^{\frac {3}{2}}}{3 b^{3}}-\frac {2 B a \,x^{\frac {3}{2}}}{b^{4}}-\frac {6 A a \sqrt {x}}{b^{4}}+\frac {12 B \,a^{2} \sqrt {x}}{b^{5}} \end {gather*}

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(1/(a*b)^(1/2)*b*x^(1/2))*A-63/4*a^3/b^5/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*
b*x^(1/2))*B

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maxima [A]  time = 2.00, size = 151, normalized size = 0.89 \begin {gather*} \frac {{\left (17 \, B a^{3} b - 13 \, A a^{2} b^{2}\right )} x^{\frac {3}{2}} + {\left (15 \, B a^{4} - 11 \, A a^{3} b\right )} \sqrt {x}}{4 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} - \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 {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (3 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} + 45 \, {\left (2 \, B a^{2} - A a b\right )} \sqrt {x}\right )}}{15 \, 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)^3,x, algorithm="maxima")

[Out]

1/4*((17*B*a^3*b - 13*A*a^2*b^2)*x^(3/2) + (15*B*a^4 - 11*A*a^3*b)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5) -
7/4*(9*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 2/15*(3*B*b^2*x^(5/2) - 5*(3*B*a*b - A
*b^2)*x^(3/2) + 45*(2*B*a^2 - A*a*b)*sqrt(x))/b^5

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mupad [B]  time = 0.41, size = 183, normalized size = 1.08 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,b^3}-\frac {2\,B\,a}{b^4}\right )-\frac {x^{3/2}\,\left (\frac {13\,A\,a^2\,b^2}{4}-\frac {17\,B\,a^3\,b}{4}\right )-\sqrt {x}\,\left (\frac {15\,B\,a^4}{4}-\frac {11\,A\,a^3\,b}{4}\right )}{a^2\,b^5+2\,a\,b^6\,x+b^7\,x^2}-\sqrt {x}\,\left (\frac {3\,a\,\left (\frac {2\,A}{b^3}-\frac {6\,B\,a}{b^4}\right )}{b}+\frac {6\,B\,a^2}{b^5}\right )+\frac {2\,B\,x^{5/2}}{5\,b^3}-\frac {7\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (5\,A\,b-9\,B\,a\right )}{9\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-9\,B\,a\right )}{4\,b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(3/2)*((2*A)/(3*b^3) - (2*B*a)/b^4) - (x^(3/2)*((13*A*a^2*b^2)/4 - (17*B*a^3*b)/4) - x^(1/2)*((15*B*a^4)/4 -
(11*A*a^3*b)/4))/(a^2*b^5 + b^7*x^2 + 2*a*b^6*x) - x^(1/2)*((3*a*((2*A)/b^3 - (6*B*a)/b^4))/b + (6*B*a^2)/b^5
) + (2*B*x^(5/2))/(5*b^3) - (7*a^(3/2)*atan((a^(3/2)*b^(1/2)*x^(1/2)*(5*A*b - 9*B*a))/(9*B*a^3 - 5*A*a^2*b))*(
5*A*b - 9*B*a))/(4*b^(11/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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