\(\int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx\) [496]

Optimal result

Integrand size = 20, antiderivative size = 89 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+2 b^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

-2/3*B*(b*x+a)^(3/2)/x^(3/2)-2/5*A*(b*x+a)^(5/2)/a/x^(5/2)+2*b^(3/2)*B*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))-
2*b*B*(b*x+a)^(1/2)/x^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 49, 65, 223, 212} \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+2 b^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}} \]

[In]

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

[Out]

(-2*b*B*Sqrt[a + b*x])/Sqrt[x] - (2*B*(a + b*x)^(3/2))/(3*x^(3/2)) - (2*A*(a + b*x)^(5/2))/(5*a*x^(5/2)) + 2*b
^(3/2)*B*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]]

Rule 49

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 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 79

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] || L
tQ[p, n]))))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+B \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx \\ & = -\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+(b B) \int \frac {\sqrt {a+b x}}{x^{3/2}} \, dx \\ & = -\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+\left (b^2 B\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx \\ & = -\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+\left (2 b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+\left (2 b^2 B\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right ) \\ & = -\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+2 b^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (3 a^2 A+6 a A b x+5 a^2 B x+3 A b^2 x^2+20 a b B x^2\right )}{15 a x^{5/2}}-2 b^{3/2} B \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right ) \]

[In]

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

[Out]

(-2*Sqrt[a + b*x]*(3*a^2*A + 6*a*A*b*x + 5*a^2*B*x + 3*A*b^2*x^2 + 20*a*b*B*x^2))/(15*a*x^(5/2)) - 2*b^(3/2)*B
*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15

method	result	size
risch	\(-\frac {2 \sqrt {b x +a}\, \left (3 A \,b^{2} x^{2}+20 B a b \,x^{2}+6 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}} a}+\frac {b^{\frac {3}{2}} B \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\)	\(102\)
default	\(-\frac {\sqrt {b x +a}\, \left (-15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{2} x^{3}+6 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}+40 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a \,x^{2}+12 A a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+10 B \,a^{2} x \sqrt {x \left (b x +a \right )}\, \sqrt {b}+6 A \,a^{2} \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{15 x^{\frac {5}{2}} a \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(156\)

[In]

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

[Out]

-2/15*(b*x+a)^(1/2)*(3*A*b^2*x^2+20*B*a*b*x^2+6*A*a*b*x+5*B*a^2*x+3*A*a^2)/x^(5/2)/a+b^(3/2)*B*ln((1/2*a+b*x)/
b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.03 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=\left [\frac {15 \, B a b^{\frac {3}{2}} x^{3} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (3 \, A a^{2} + {\left (20 \, B a b + 3 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} + 6 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, a x^{3}}, -\frac {2 \, {\left (15 \, B a \sqrt {-b} b x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, A a^{2} + {\left (20 \, B a b + 3 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} + 6 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}\right )}}{15 \, a x^{3}}\right ] \]

[In]

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

[Out]

[1/15*(15*B*a*b^(3/2)*x^3*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(3*A*a^2 + (20*B*a*b + 3*A*b^2)
*x^2 + (5*B*a^2 + 6*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^3), -2/15*(15*B*a*sqrt(-b)*b*x^3*arctan(sqrt(b*x + a
)*sqrt(-b)/(b*sqrt(x))) + (3*A*a^2 + (20*B*a*b + 3*A*b^2)*x^2 + (5*B*a^2 + 6*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/
(a*x^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (85) = 170\).

Time = 4.74 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.13 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=- \frac {2 A a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {4 A b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{5 x} - \frac {2 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{5 a} - \frac {2 B \sqrt {a} b}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {2 B a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} + 2 B b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 B b^{2} \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

-2*A*a*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 4*A*b**(3/2)*sqrt(a/(b*x) + 1)/(5*x) - 2*A*b**(5/2)*sqrt(a/(b*x) +
1)/(5*a) - 2*B*sqrt(a)*b/(sqrt(x)*sqrt(1 + b*x/a)) - 2*B*a*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 2*B*b**(3/2)*sqr
t(a/(b*x) + 1)/3 + 2*B*b**(3/2)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) - 2*B*b**2*sqrt(x)/(sqrt(a)*sqrt(1 + b*x/a))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (65) = 130\).

Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.78 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=B b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {7 \, \sqrt {b x^{2} + a x} B b}{3 \, x} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{2}}{5 \, a x} - \frac {\sqrt {b x^{2} + a x} B a}{3 \, x^{2}} + \frac {\sqrt {b x^{2} + a x} A b}{5 \, x^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{3 \, x^{3}} + \frac {3 \, \sqrt {b x^{2} + a x} A a}{5 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 78.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b^{\frac {3}{2}} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) + \frac {{\left (15 \, B a^{2} b^{4} - {\left (35 \, B a b^{4} - \frac {{\left (20 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} {\left (b x + a\right )}}{a^{2}}\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}}\right )} b}{15 \, {\left | b \right |}} \]

[In]

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

[Out]

-2/15*(15*B*b^(3/2)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b))) + (15*B*a^2*b^4 - (35*B*a*b^4 -
(20*B*a^2*b^4 + 3*A*a*b^5)*(b*x + a)/a^2)*(b*x + a))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(5/2))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{7/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{x^{7/2}} \,d x \]

[In]

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

[Out]

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