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

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

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Rubi [A]  time = 0.07, antiderivative size = 169, 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 = {78, 51, 63, 205} \begin {gather*} -\frac {7 b^{3/2} (9 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{11/2}}+\frac {7 (9 A b-5 a B)}{12 a^4 x^{3/2}}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}-\frac {7 (9 A b-5 a B)}{20 a^3 b x^{5/2}}-\frac {7 b (9 A b-5 a B)}{4 a^5 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*(9*A*b - 5*a*B))/(20*a^3*b*x^(5/2)) + (7*(9*A*b - 5*a*B))/(12*a^4*x^(3/2)) - (7*b*(9*A*b - 5*a*B))/(4*a^5*
Sqrt[x]) + (A*b - a*B)/(2*a*b*x^(5/2)*(a + b*x)^2) + (9*A*b - 5*a*B)/(4*a^2*b*x^(5/2)*(a + b*x)) - (7*b^(3/2)*
(9*A*b - 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(11/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {A+B x}{x^{7/2} (a+b x)^3} \, dx &=\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}-\frac {\left (-\frac {9 A b}{2}+\frac {5 a B}{2}\right ) \int \frac {1}{x^{7/2} (a+b x)^2} \, dx}{2 a b}\\ &=\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}+\frac {(7 (9 A b-5 a B)) \int \frac {1}{x^{7/2} (a+b x)} \, dx}{8 a^2 b}\\ &=-\frac {7 (9 A b-5 a B)}{20 a^3 b x^{5/2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}-\frac {(7 (9 A b-5 a B)) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{8 a^3}\\ &=-\frac {7 (9 A b-5 a B)}{20 a^3 b x^{5/2}}+\frac {7 (9 A b-5 a B)}{12 a^4 x^{3/2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}+\frac {(7 b (9 A b-5 a B)) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{8 a^4}\\ &=-\frac {7 (9 A b-5 a B)}{20 a^3 b x^{5/2}}+\frac {7 (9 A b-5 a B)}{12 a^4 x^{3/2}}-\frac {7 b (9 A b-5 a B)}{4 a^5 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}-\frac {\left (7 b^2 (9 A b-5 a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 a^5}\\ &=-\frac {7 (9 A b-5 a B)}{20 a^3 b x^{5/2}}+\frac {7 (9 A b-5 a B)}{12 a^4 x^{3/2}}-\frac {7 b (9 A b-5 a B)}{4 a^5 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}-\frac {\left (7 b^2 (9 A b-5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 a^5}\\ &=-\frac {7 (9 A b-5 a B)}{20 a^3 b x^{5/2}}+\frac {7 (9 A b-5 a B)}{12 a^4 x^{3/2}}-\frac {7 b (9 A b-5 a B)}{4 a^5 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x)^2}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} (a+b x)}-\frac {7 b^{3/2} (9 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.36 \begin {gather*} \frac {\frac {5 a^2 (A b-a B)}{(a+b x)^2}+(5 a B-9 A b) \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};-\frac {b x}{a}\right )}{10 a^3 b x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^2*(A*b - a*B))/(a + b*x)^2 + (-9*A*b + 5*a*B)*Hypergeometric2F1[-5/2, 2, -3/2, -((b*x)/a)])/(10*a^3*b*x^
(5/2))

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IntegrateAlgebraic [A]  time = 0.20, size = 149, normalized size = 0.88 \begin {gather*} \frac {7 \left (5 a b^{3/2} B-9 A b^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{11/2}}+\frac {-24 a^4 A-40 a^4 B x+72 a^3 A b x+280 a^3 b B x^2-504 a^2 A b^2 x^2+875 a^2 b^2 B x^3-1575 a A b^3 x^3+525 a b^3 B x^4-945 A b^4 x^4}{60 a^5 x^{5/2} (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 437, normalized size = 2.59 \begin {gather*} \left [-\frac {105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{120 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, -\frac {105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{60 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.21, size = 135, normalized size = 0.80 \begin {gather*} \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} + \frac {11 \, B a b^{3} x^{\frac {3}{2}} - 15 \, A b^{4} x^{\frac {3}{2}} + 13 \, B a^{2} b^{2} \sqrt {x} - 17 \, A a b^{3} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{5}} + \frac {2 \, {\left (45 \, B a b x^{2} - 90 \, A b^{2} x^{2} - 5 \, B a^{2} x + 15 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{5} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4/a^5*b^4/(b*x+a)^2*x^(3/2)*A+11/4/a^4*b^3/(b*x+a)^2*x^(3/2)*B-17/4/a^4*b^3/(b*x+a)^2*A*x^(1/2)+13/4/a^3*b
^2/(b*x+a)^2*B*x^(1/2)-63/4/a^5*b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A+35/4/a^4*b^2/(a*b)^(1/2)*arc
tan(1/(a*b)^(1/2)*b*x^(1/2))*B-2/5*A/a^3/x^(5/2)+2/a^4/x^(3/2)*A*b-2/3/a^3/x^(3/2)*B-12*b^2/a^5/x^(1/2)*A+6*b/
a^4/x^(1/2)*B

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maxima [A]  time = 2.00, size = 154, normalized size = 0.91 \begin {gather*} -\frac {24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x}{60 \, {\left (a^{5} b^{2} x^{\frac {9}{2}} + 2 \, a^{6} b x^{\frac {7}{2}} + a^{7} x^{\frac {5}{2}}\right )}} + \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.48, size = 135, normalized size = 0.80 \begin {gather*} -\frac {\frac {2\,A}{5\,a}-\frac {2\,x\,\left (9\,A\,b-5\,B\,a\right )}{15\,a^2}+\frac {35\,b^2\,x^3\,\left (9\,A\,b-5\,B\,a\right )}{12\,a^4}+\frac {7\,b^3\,x^4\,\left (9\,A\,b-5\,B\,a\right )}{4\,a^5}+\frac {14\,b\,x^2\,\left (9\,A\,b-5\,B\,a\right )}{15\,a^3}}{a^2\,x^{5/2}+b^2\,x^{9/2}+2\,a\,b\,x^{7/2}}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (9\,A\,b-5\,B\,a\right )}{4\,a^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*A)/(5*a) - (2*x*(9*A*b - 5*B*a))/(15*a^2) + (35*b^2*x^3*(9*A*b - 5*B*a))/(12*a^4) + (7*b^3*x^4*(9*A*b -
5*B*a))/(4*a^5) + (14*b*x^2*(9*A*b - 5*B*a))/(15*a^3))/(a^2*x^(5/2) + b^2*x^(9/2) + 2*a*b*x^(7/2)) - (7*b^(3/2
)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(9*A*b - 5*B*a))/(4*a^(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((B*x+A)/x**(7/2)/(b*x+a)**3,x)

[Out]

Timed out

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