\(\int \frac {(a x+b x^n)^{3/2}}{(c x)^{5/2}} \, dx\) [376]

Optimal result

Integrand size = 21, antiderivative size = 122 \[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=-\frac {2 a \sqrt {a x+b x^n}}{c^2 (1-n) \sqrt {c x}}-\frac {2 \left (a x+b x^n\right )^{3/2}}{3 c (1-n) (c x)^{3/2}}+\frac {2 a^{3/2} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^n}}\right )}{c^2 (1-n) \sqrt {c x}} \]

[Out]

-2/3*(a*x+b*x^n)^(3/2)/c/(1-n)/(c*x)^(3/2)+2*a^(3/2)*arctanh(a^(1/2)*x^(1/2)/(a*x+b*x^n)^(1/2))*x^(1/2)/c^2/(1
-n)/(c*x)^(1/2)-2*a*(a*x+b*x^n)^(1/2)/c^2/(1-n)/(c*x)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2053, 2056, 2054, 212} \[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\frac {2 a^{3/2} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^n}}\right )}{c^2 (1-n) \sqrt {c x}}-\frac {2 a \sqrt {a x+b x^n}}{c^2 (1-n) \sqrt {c x}}-\frac {2 \left (a x+b x^n\right )^{3/2}}{3 c (1-n) (c x)^{3/2}} \]

[In]

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

[Out]

(-2*a*Sqrt[a*x + b*x^n])/(c^2*(1 - n)*Sqrt[c*x]) - (2*(a*x + b*x^n)^(3/2))/(3*c*(1 - n)*(c*x)^(3/2)) + (2*a^(3
/2)*Sqrt[x]*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[a*x + b*x^n]])/(c^2*(1 - n)*Sqrt[c*x])

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 2053

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*p*(n - j))), x] + Dist[a/c^j, Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c,
j, m, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rule 2056

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[c^IntPart[m]*((c*x)^FracPar
t[m]/x^FracPart[m]), Int[x^m*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2]
&& NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]

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

Mathematica [A] (verified)

Time = 1.62 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\frac {x \left (8 a^2 x^2+2 b^2 x^{2 n}+10 a b x^{1+n}-6 a^{3/2} \sqrt {b} x^{\frac {3+n}{2}} \sqrt {1+\frac {a x^{1-n}}{b}} \text {arcsinh}\left (\frac {\sqrt {a} x^{\frac {1}{2}-\frac {n}{2}}}{\sqrt {b}}\right )\right )}{3 (-1+n) (c x)^{5/2} \sqrt {a x+b x^n}} \]

[In]

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

[Out]

(x*(8*a^2*x^2 + 2*b^2*x^(2*n) + 10*a*b*x^(1 + n) - 6*a^(3/2)*Sqrt[b]*x^((3 + n)/2)*Sqrt[1 + (a*x^(1 - n))/b]*A
rcSinh[(Sqrt[a]*x^(1/2 - n/2))/Sqrt[b]]))/(3*(-1 + n)*(c*x)^(5/2)*Sqrt[a*x + b*x^n])

Maple [F]

\[\int \frac {\left (a x +b \,x^{n}\right )^{\frac {3}{2}}}{\left (c x \right )^{\frac {5}{2}}}d x\]

[In]

int((a*x+b*x^n)^(3/2)/(c*x)^(5/2),x)

[Out]

int((a*x+b*x^n)^(3/2)/(c*x)^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\int \frac {\left (a x + b x^{n}\right )^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a*x+b*x**n)**(3/2)/(c*x)**(5/2),x)

[Out]

Integral((a*x + b*x**n)**(3/2)/(c*x)**(5/2), x)

Maxima [F]

\[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (a x + b x^{n}\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a*x+b*x^n)^(3/2)/(c*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + b*x^n)^(3/2)/(c*x)^(5/2), x)

Giac [F]

\[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (a x + b x^{n}\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate((a*x + b*x^n)^(3/2)/(c*x)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^n\right )^{3/2}}{(c x)^{5/2}} \, dx=\int \frac {{\left (b\,x^n+a\,x\right )}^{3/2}}{{\left (c\,x\right )}^{5/2}} \,d x \]

[In]

int((b*x^n + a*x)^(3/2)/(c*x)^(5/2),x)

[Out]

int((b*x^n + a*x)^(3/2)/(c*x)^(5/2), x)