3.5.27 \(\int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=171 \[ \frac {35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}-\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 171, 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, 208} \begin {gather*} \frac {35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}}+\frac {35 \sqrt {a+b x} (3 A b-2 a B)}{12 a^4 x^2}-\frac {7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt {a+b x}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {35 b \sqrt {a+b x} (3 A b-2 a B)}{8 a^5 x}-\frac {A}{3 a x^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]

[Out]

-A/(3*a*x^3*(a + b*x)^(3/2)) - (3*A*b - 2*a*B)/(3*a^2*x^2*(a + b*x)^(3/2)) - (7*(3*A*b - 2*a*B))/(3*a^3*x^2*Sq
rt[a + b*x]) + (35*(3*A*b - 2*a*B)*Sqrt[a + b*x])/(12*a^4*x^2) - (35*b*(3*A*b - 2*a*B)*Sqrt[a + b*x])/(8*a^5*x
) + (35*b^2*(3*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*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 208

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

Rubi steps

\begin {align*} \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {\left (-\frac {9 A b}{2}+3 a B\right ) \int \frac {1}{x^3 (a+b x)^{5/2}} \, dx}{3 a}\\ &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {(7 (3 A b-2 a B)) \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx}{6 a^2}\\ &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt {a+b x}}-\frac {(35 (3 A b-2 a B)) \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{6 a^3}\\ &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt {a+b x}}+\frac {35 (3 A b-2 a B) \sqrt {a+b x}}{12 a^4 x^2}+\frac {(35 b (3 A b-2 a B)) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{8 a^4}\\ &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt {a+b x}}+\frac {35 (3 A b-2 a B) \sqrt {a+b x}}{12 a^4 x^2}-\frac {35 b (3 A b-2 a B) \sqrt {a+b x}}{8 a^5 x}-\frac {\left (35 b^2 (3 A b-2 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{16 a^5}\\ &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt {a+b x}}+\frac {35 (3 A b-2 a B) \sqrt {a+b x}}{12 a^4 x^2}-\frac {35 b (3 A b-2 a B) \sqrt {a+b x}}{8 a^5 x}-\frac {(35 b (3 A b-2 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{8 a^5}\\ &=-\frac {A}{3 a x^3 (a+b x)^{3/2}}-\frac {3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac {7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt {a+b x}}+\frac {35 (3 A b-2 a B) \sqrt {a+b x}}{12 a^4 x^2}-\frac {35 b (3 A b-2 a B) \sqrt {a+b x}}{8 a^5 x}+\frac {35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 58, normalized size = 0.34 \begin {gather*} \frac {b^2 x^3 (2 a B-3 A b) \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\frac {b x}{a}+1\right )-a^3 A}{3 a^4 x^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]

[Out]

(-(a^3*A) + b^2*(-3*A*b + 2*a*B)*x^3*Hypergeometric2F1[-3/2, 3, -1/2, 1 + (b*x)/a])/(3*a^4*x^3*(a + b*x)^(3/2)
)

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IntegrateAlgebraic [A]  time = 0.26, size = 173, normalized size = 1.01 \begin {gather*} -\frac {35 \left (2 a b^2 B-3 A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}}-\frac {16 a^5 B-16 a^4 A b+96 a^4 B (a+b x)-144 a^3 A b (a+b x)-462 a^3 B (a+b x)^2+693 a^2 A b (a+b x)^2+560 a^2 B (a+b x)^3-840 a A b (a+b x)^3+315 A b (a+b x)^4-210 a B (a+b x)^4}{24 a^5 b x^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]

[Out]

-1/24*(-16*a^4*A*b + 16*a^5*B - 144*a^3*A*b*(a + b*x) + 96*a^4*B*(a + b*x) + 693*a^2*A*b*(a + b*x)^2 - 462*a^3
*B*(a + b*x)^2 - 840*a*A*b*(a + b*x)^3 + 560*a^2*B*(a + b*x)^3 + 315*A*b*(a + b*x)^4 - 210*a*B*(a + b*x)^4)/(a
^5*b*x^3*(a + b*x)^(3/2)) - (35*(-3*A*b^3 + 2*a*b^2*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(11/2))

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fricas [A]  time = 1.38, size = 448, normalized size = 2.62 \begin {gather*} \left [-\frac {105 \, {\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{5} - 105 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \, {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{48 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, \frac {105 \, {\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (8 \, A a^{5} - 105 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \, {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{24 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(105*((2*B*a*b^4 - 3*A*b^5)*x^5 + 2*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 + (2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3)*sq
rt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*A*a^5 - 105*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 - 140*(2*B
*a^3*b^2 - 3*A*a^2*b^3)*x^3 - 21*(2*B*a^4*b - 3*A*a^3*b^2)*x^2 + 6*(2*B*a^5 - 3*A*a^4*b)*x)*sqrt(b*x + a))/(a^
6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3), 1/24*(105*((2*B*a*b^4 - 3*A*b^5)*x^5 + 2*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 + (
2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8*A*a^5 - 105*(2*B*a^2*b^3 - 3*A*
a*b^4)*x^4 - 140*(2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3 - 21*(2*B*a^4*b - 3*A*a^3*b^2)*x^2 + 6*(2*B*a^5 - 3*A*a^4*b)*
x)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3)]

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giac [A]  time = 1.29, size = 200, normalized size = 1.17 \begin {gather*} \frac {35 \, {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5}} + \frac {210 \, {\left (b x + a\right )}^{4} B a b^{2} - 560 \, {\left (b x + a\right )}^{3} B a^{2} b^{2} + 462 \, {\left (b x + a\right )}^{2} B a^{3} b^{2} - 96 \, {\left (b x + a\right )} B a^{4} b^{2} - 16 \, B a^{5} b^{2} - 315 \, {\left (b x + a\right )}^{4} A b^{3} + 840 \, {\left (b x + a\right )}^{3} A a b^{3} - 693 \, {\left (b x + a\right )}^{2} A a^{2} b^{3} + 144 \, {\left (b x + a\right )} A a^{3} b^{3} + 16 \, A a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

35/8*(2*B*a*b^2 - 3*A*b^3)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) + 1/24*(210*(b*x + a)^4*B*a*b^2 - 560
*(b*x + a)^3*B*a^2*b^2 + 462*(b*x + a)^2*B*a^3*b^2 - 96*(b*x + a)*B*a^4*b^2 - 16*B*a^5*b^2 - 315*(b*x + a)^4*A
*b^3 + 840*(b*x + a)^3*A*a*b^3 - 693*(b*x + a)^2*A*a^2*b^3 + 144*(b*x + a)*A*a^3*b^3 + 16*A*a^4*b^3)/(((b*x +
a)^(3/2) - sqrt(b*x + a)*a)^3*a^5)

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maple [A]  time = 0.02, size = 147, normalized size = 0.86 \begin {gather*} 2 \left (-\frac {A b -B a}{3 \left (b x +a \right )^{\frac {3}{2}} a^{4}}-\frac {4 A b -3 B a}{\sqrt {b x +a}\, a^{5}}-\frac {-\frac {35 \left (3 A b -2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}+\frac {\left (\frac {41 A b}{16}-\frac {11 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {35}{6} A a b +3 B \,a^{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {55}{16} A \,a^{2} b -\frac {13}{8} B \,a^{3}\right ) \sqrt {b x +a}}{b^{3} x^{3}}}{a^{5}}\right ) b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^4/(b*x+a)^(5/2),x)

[Out]

2*b^2*(-1/3*(A*b-B*a)/a^4/(b*x+a)^(3/2)-(4*A*b-3*B*a)/a^5/(b*x+a)^(1/2)-1/a^5*(((41/16*A*b-11/8*B*a)*(b*x+a)^(
5/2)+(-35/6*A*a*b+3*B*a^2)*(b*x+a)^(3/2)+(55/16*A*a^2*b-13/8*B*a^3)*(b*x+a)^(1/2))/x^3/b^3-35/16*(3*A*b-2*B*a)
/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))))

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maxima [A]  time = 2.04, size = 204, normalized size = 1.19 \begin {gather*} -\frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (16 \, B a^{5} - 16 \, A a^{4} b - 105 \, {\left (2 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{4} + 280 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} {\left (b x + a\right )}^{3} - 231 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{2} + 48 \, {\left (2 \, B a^{4} - 3 \, A a^{3} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {9}{2}} a^{5} b - 3 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{6} b + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{7} b - {\left (b x + a\right )}^{\frac {3}{2}} a^{8} b} - \frac {105 \, {\left (2 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {11}{2}} b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

-1/48*b^3*(2*(16*B*a^5 - 16*A*a^4*b - 105*(2*B*a - 3*A*b)*(b*x + a)^4 + 280*(2*B*a^2 - 3*A*a*b)*(b*x + a)^3 -
231*(2*B*a^3 - 3*A*a^2*b)*(b*x + a)^2 + 48*(2*B*a^4 - 3*A*a^3*b)*(b*x + a))/((b*x + a)^(9/2)*a^5*b - 3*(b*x +
a)^(7/2)*a^6*b + 3*(b*x + a)^(5/2)*a^7*b - (b*x + a)^(3/2)*a^8*b) - 105*(2*B*a - 3*A*b)*log((sqrt(b*x + a) - s
qrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(11/2)*b))

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mupad [B]  time = 0.14, size = 198, normalized size = 1.16 \begin {gather*} \frac {35\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-2\,B\,a\right )}{8\,a^{11/2}}-\frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{3\,a}+\frac {2\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,\left (a+b\,x\right )}{a^2}-\frac {77\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^2}{8\,a^3}+\frac {35\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^3}{3\,a^4}-\frac {35\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^4}{8\,a^5}}{3\,a\,{\left (a+b\,x\right )}^{7/2}-{\left (a+b\,x\right )}^{9/2}+a^3\,{\left (a+b\,x\right )}^{3/2}-3\,a^2\,{\left (a+b\,x\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^4*(a + b*x)^(5/2)),x)

[Out]

(35*b^2*atanh((a + b*x)^(1/2)/a^(1/2))*(3*A*b - 2*B*a))/(8*a^(11/2)) - ((2*(A*b^3 - B*a*b^2))/(3*a) + (2*(3*A*
b^3 - 2*B*a*b^2)*(a + b*x))/a^2 - (77*(3*A*b^3 - 2*B*a*b^2)*(a + b*x)^2)/(8*a^3) + (35*(3*A*b^3 - 2*B*a*b^2)*(
a + b*x)^3)/(3*a^4) - (35*(3*A*b^3 - 2*B*a*b^2)*(a + b*x)^4)/(8*a^5))/(3*a*(a + b*x)^(7/2) - (a + b*x)^(9/2) +
a^3*(a + b*x)^(3/2) - 3*a^2*(a + b*x)^(5/2))

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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**4/(b*x+a)**(5/2),x)

[Out]

Timed out

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