3.5.18 \(\int \frac {A+B x}{x^5 (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac {35 b^3 (9 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2}}+\frac {35 b^3 (9 A b-8 a B)}{64 a^5 \sqrt {a+b x}}+\frac {35 b^2 (9 A b-8 a B)}{192 a^4 x \sqrt {a+b x}}-\frac {7 b (9 A b-8 a B)}{96 a^3 x^2 \sqrt {a+b x}}+\frac {9 A b-8 a B}{24 a^2 x^3 \sqrt {a+b x}}-\frac {A}{4 a x^4 \sqrt {a+b x}} \]

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Rubi [A]  time = 0.08, antiderivative size = 174, 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 \sqrt {a+b x} (9 A b-8 a B)}{64 a^5 x}-\frac {35 b^3 (9 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2}}-\frac {35 b \sqrt {a+b x} (9 A b-8 a B)}{96 a^4 x^2}+\frac {7 \sqrt {a+b x} (9 A b-8 a B)}{24 a^3 x^3}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}-\frac {A}{4 a x^4 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^5*(a + b*x)^(3/2)),x]

[Out]

-A/(4*a*x^4*Sqrt[a + b*x]) - (9*A*b - 8*a*B)/(4*a^2*x^3*Sqrt[a + b*x]) + (7*(9*A*b - 8*a*B)*Sqrt[a + b*x])/(24
*a^3*x^3) - (35*b*(9*A*b - 8*a*B)*Sqrt[a + b*x])/(96*a^4*x^2) + (35*b^2*(9*A*b - 8*a*B)*Sqrt[a + b*x])/(64*a^5
*x) - (35*b^3*(9*A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*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^5 (a+b x)^{3/2}} \, dx &=-\frac {A}{4 a x^4 \sqrt {a+b x}}+\frac {\left (-\frac {9 A b}{2}+4 a B\right ) \int \frac {1}{x^4 (a+b x)^{3/2}} \, dx}{4 a}\\ &=-\frac {A}{4 a x^4 \sqrt {a+b x}}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}-\frac {(7 (9 A b-8 a B)) \int \frac {1}{x^4 \sqrt {a+b x}} \, dx}{8 a^2}\\ &=-\frac {A}{4 a x^4 \sqrt {a+b x}}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}+\frac {7 (9 A b-8 a B) \sqrt {a+b x}}{24 a^3 x^3}+\frac {(35 b (9 A b-8 a B)) \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{48 a^3}\\ &=-\frac {A}{4 a x^4 \sqrt {a+b x}}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}+\frac {7 (9 A b-8 a B) \sqrt {a+b x}}{24 a^3 x^3}-\frac {35 b (9 A b-8 a B) \sqrt {a+b x}}{96 a^4 x^2}-\frac {\left (35 b^2 (9 A b-8 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{64 a^4}\\ &=-\frac {A}{4 a x^4 \sqrt {a+b x}}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}+\frac {7 (9 A b-8 a B) \sqrt {a+b x}}{24 a^3 x^3}-\frac {35 b (9 A b-8 a B) \sqrt {a+b x}}{96 a^4 x^2}+\frac {35 b^2 (9 A b-8 a B) \sqrt {a+b x}}{64 a^5 x}+\frac {\left (35 b^3 (9 A b-8 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a^5}\\ &=-\frac {A}{4 a x^4 \sqrt {a+b x}}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}+\frac {7 (9 A b-8 a B) \sqrt {a+b x}}{24 a^3 x^3}-\frac {35 b (9 A b-8 a B) \sqrt {a+b x}}{96 a^4 x^2}+\frac {35 b^2 (9 A b-8 a B) \sqrt {a+b x}}{64 a^5 x}+\frac {\left (35 b^2 (9 A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a^5}\\ &=-\frac {A}{4 a x^4 \sqrt {a+b x}}-\frac {9 A b-8 a B}{4 a^2 x^3 \sqrt {a+b x}}+\frac {7 (9 A b-8 a B) \sqrt {a+b x}}{24 a^3 x^3}-\frac {35 b (9 A b-8 a B) \sqrt {a+b x}}{96 a^4 x^2}+\frac {35 b^2 (9 A b-8 a B) \sqrt {a+b x}}{64 a^5 x}-\frac {35 b^3 (9 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 58, normalized size = 0.33 \begin {gather*} \frac {b^3 x^4 (9 A b-8 a B) \, _2F_1\left (-\frac {1}{2},4;\frac {1}{2};\frac {b x}{a}+1\right )-a^4 A}{4 a^5 x^4 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^5*(a + b*x)^(3/2)),x]

[Out]

(-(a^4*A) + b^3*(9*A*b - 8*a*B)*x^4*Hypergeometric2F1[-1/2, 4, 1/2, 1 + (b*x)/a])/(4*a^5*x^4*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.28, size = 173, normalized size = 0.99 \begin {gather*} \frac {35 \left (8 a b^3 B-9 A b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2}}-\frac {384 a^5 B-384 a^4 A b-2232 a^4 B (a+b x)+2511 a^3 A b (a+b x)+4088 a^3 B (a+b x)^2-4599 a^2 A b (a+b x)^2-3080 a^2 B (a+b x)^3+3465 a A b (a+b x)^3-945 A b (a+b x)^4+840 a B (a+b x)^4}{192 a^5 b x^4 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^5*(a + b*x)^(3/2)),x]

[Out]

-1/192*(-384*a^4*A*b + 384*a^5*B + 2511*a^3*A*b*(a + b*x) - 2232*a^4*B*(a + b*x) - 4599*a^2*A*b*(a + b*x)^2 +
4088*a^3*B*(a + b*x)^2 + 3465*a*A*b*(a + b*x)^3 - 3080*a^2*B*(a + b*x)^3 - 945*A*b*(a + b*x)^4 + 840*a*B*(a +
b*x)^4)/(a^5*b*x^4*Sqrt[a + b*x]) + (35*(-9*A*b^4 + 8*a*b^3*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(11/2))

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fricas [A]  time = 0.97, size = 377, normalized size = 2.17 \begin {gather*} \left [-\frac {105 \, {\left ({\left (8 \, B a b^{4} - 9 \, A b^{5}\right )} x^{5} + {\left (8 \, B a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{4}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{5} + 105 \, {\left (8 \, B a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{4} + 35 \, {\left (8 \, B a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{3} - 14 \, {\left (8 \, B a^{4} b - 9 \, A a^{3} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{5} - 9 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{384 \, {\left (a^{6} b x^{5} + a^{7} x^{4}\right )}}, -\frac {105 \, {\left ({\left (8 \, B a b^{4} - 9 \, A b^{5}\right )} x^{5} + {\left (8 \, B a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, A a^{5} + 105 \, {\left (8 \, B a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{4} + 35 \, {\left (8 \, B a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{3} - 14 \, {\left (8 \, B a^{4} b - 9 \, A a^{3} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{5} - 9 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{192 \, {\left (a^{6} b x^{5} + a^{7} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/384*(105*((8*B*a*b^4 - 9*A*b^5)*x^5 + (8*B*a^2*b^3 - 9*A*a*b^4)*x^4)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sq
rt(a) + 2*a)/x) + 2*(48*A*a^5 + 105*(8*B*a^2*b^3 - 9*A*a*b^4)*x^4 + 35*(8*B*a^3*b^2 - 9*A*a^2*b^3)*x^3 - 14*(8
*B*a^4*b - 9*A*a^3*b^2)*x^2 + 8*(8*B*a^5 - 9*A*a^4*b)*x)*sqrt(b*x + a))/(a^6*b*x^5 + a^7*x^4), -1/192*(105*((8
*B*a*b^4 - 9*A*b^5)*x^5 + (8*B*a^2*b^3 - 9*A*a*b^4)*x^4)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*A*a^5
+ 105*(8*B*a^2*b^3 - 9*A*a*b^4)*x^4 + 35*(8*B*a^3*b^2 - 9*A*a^2*b^3)*x^3 - 14*(8*B*a^4*b - 9*A*a^3*b^2)*x^2 +
8*(8*B*a^5 - 9*A*a^4*b)*x)*sqrt(b*x + a))/(a^6*b*x^5 + a^7*x^4)]

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giac [A]  time = 1.30, size = 197, normalized size = 1.13 \begin {gather*} -\frac {35 \, {\left (8 \, B a b^{3} - 9 \, A b^{4}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{64 \, \sqrt {-a} a^{5}} - \frac {2 \, {\left (B a b^{3} - A b^{4}\right )}}{\sqrt {b x + a} a^{5}} - \frac {456 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{3} - 1544 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{3} + 1784 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{3} - 696 \, \sqrt {b x + a} B a^{4} b^{3} - 561 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{4} + 1929 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{4} - 2295 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{4} + 975 \, \sqrt {b x + a} A a^{3} b^{4}}{192 \, a^{5} b^{4} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

-35/64*(8*B*a*b^3 - 9*A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) - 2*(B*a*b^3 - A*b^4)/(sqrt(b*x + a
)*a^5) - 1/192*(456*(b*x + a)^(7/2)*B*a*b^3 - 1544*(b*x + a)^(5/2)*B*a^2*b^3 + 1784*(b*x + a)^(3/2)*B*a^3*b^3
- 696*sqrt(b*x + a)*B*a^4*b^3 - 561*(b*x + a)^(7/2)*A*b^4 + 1929*(b*x + a)^(5/2)*A*a*b^4 - 2295*(b*x + a)^(3/2
)*A*a^2*b^4 + 975*sqrt(b*x + a)*A*a^3*b^4)/(a^5*b^4*x^4)

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maple [A]  time = 0.02, size = 147, normalized size = 0.84 \begin {gather*} 2 \left (-\frac {-A b +B a}{\sqrt {b x +a}\, a^{5}}+\frac {-\frac {35 \left (9 A b -8 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 \sqrt {a}}+\frac {\left (\frac {187 A b}{128}-\frac {19 B a}{16}\right ) \left (b x +a \right )^{\frac {7}{2}}+\left (-\frac {643}{128} A a b +\frac {193}{48} B \,a^{2}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (\frac {765}{128} A \,a^{2} b -\frac {223}{48} B \,a^{3}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {325}{128} A \,a^{3} b +\frac {29}{16} B \,a^{4}\right ) \sqrt {b x +a}}{b^{4} x^{4}}}{a^{5}}\right ) b^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^5/(b*x+a)^(3/2),x)

[Out]

2*b^3*(-1/a^5*(-A*b+B*a)/(b*x+a)^(1/2)+1/a^5*(((187/128*A*b-19/16*B*a)*(b*x+a)^(7/2)+(-643/128*A*a*b+193/48*B*
a^2)*(b*x+a)^(5/2)+(765/128*A*a^2*b-223/48*B*a^3)*(b*x+a)^(3/2)+(-325/128*A*a^3*b+29/16*B*a^4)*(b*x+a)^(1/2))/
x^4/b^4-35/128*(9*A*b-8*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))))

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maxima [A]  time = 2.03, size = 216, normalized size = 1.24 \begin {gather*} -\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (384 \, B a^{5} - 384 \, A a^{4} b + 105 \, {\left (8 \, B a - 9 \, A b\right )} {\left (b x + a\right )}^{4} - 385 \, {\left (8 \, B a^{2} - 9 \, A a b\right )} {\left (b x + a\right )}^{3} + 511 \, {\left (8 \, B a^{3} - 9 \, A a^{2} b\right )} {\left (b x + a\right )}^{2} - 279 \, {\left (8 \, B a^{4} - 9 \, A a^{3} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {9}{2}} a^{5} b - 4 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{6} b + 6 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{7} b - 4 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{8} b + \sqrt {b x + a} a^{9} b} + \frac {105 \, {\left (8 \, B a - 9 \, 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^5/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

-1/384*b^4*(2*(384*B*a^5 - 384*A*a^4*b + 105*(8*B*a - 9*A*b)*(b*x + a)^4 - 385*(8*B*a^2 - 9*A*a*b)*(b*x + a)^3
+ 511*(8*B*a^3 - 9*A*a^2*b)*(b*x + a)^2 - 279*(8*B*a^4 - 9*A*a^3*b)*(b*x + a))/((b*x + a)^(9/2)*a^5*b - 4*(b*
x + a)^(7/2)*a^6*b + 6*(b*x + a)^(5/2)*a^7*b - 4*(b*x + a)^(3/2)*a^8*b + sqrt(b*x + a)*a^9*b) + 105*(8*B*a - 9
*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(11/2)*b))

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mupad [B]  time = 0.46, size = 207, normalized size = 1.19 \begin {gather*} \frac {\frac {2\,\left (A\,b^4-B\,a\,b^3\right )}{a}-\frac {93\,\left (9\,A\,b^4-8\,B\,a\,b^3\right )\,\left (a+b\,x\right )}{64\,a^2}+\frac {511\,\left (9\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^2}{192\,a^3}-\frac {385\,\left (9\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^3}{192\,a^4}+\frac {35\,\left (9\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^4}{64\,a^5}}{{\left (a+b\,x\right )}^{9/2}-4\,a\,{\left (a+b\,x\right )}^{7/2}+a^4\,\sqrt {a+b\,x}-4\,a^3\,{\left (a+b\,x\right )}^{3/2}+6\,a^2\,{\left (a+b\,x\right )}^{5/2}}-\frac {35\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (9\,A\,b-8\,B\,a\right )}{64\,a^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^5*(a + b*x)^(3/2)),x)

[Out]

((2*(A*b^4 - B*a*b^3))/a - (93*(9*A*b^4 - 8*B*a*b^3)*(a + b*x))/(64*a^2) + (511*(9*A*b^4 - 8*B*a*b^3)*(a + b*x
)^2)/(192*a^3) - (385*(9*A*b^4 - 8*B*a*b^3)*(a + b*x)^3)/(192*a^4) + (35*(9*A*b^4 - 8*B*a*b^3)*(a + b*x)^4)/(6
4*a^5))/((a + b*x)^(9/2) - 4*a*(a + b*x)^(7/2) + a^4*(a + b*x)^(1/2) - 4*a^3*(a + b*x)^(3/2) + 6*a^2*(a + b*x)
^(5/2)) - (35*b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(9*A*b - 8*B*a))/(64*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**5/(b*x+a)**(3/2),x)

[Out]

Timed out

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