\(\int \frac {(a+b x)^n}{x (c x^2)^{3/2}} \, dx\) [960]

Optimal result

Integrand size = 20, antiderivative size = 50 \[ \int \frac {(a+b x)^n}{x \left (c x^2\right )^{3/2}} \, dx=\frac {b^3 x (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (4,1+n,2+n,1+\frac {b x}{a}\right )}{a^4 c (1+n) \sqrt {c x^2}} \]

[Out]

b^3*x*(b*x+a)^(1+n)*hypergeom([4, 1+n],[2+n],1+b*x/a)/a^4/c/(1+n)/(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 67} \[ \int \frac {(a+b x)^n}{x \left (c x^2\right )^{3/2}} \, dx=\frac {b^3 x (a+b x)^{n+1} \operatorname {Hypergeometric2F1}\left (4,n+1,n+2,\frac {b x}{a}+1\right )}{a^4 c (n+1) \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^n/(x*(c*x^2)^(3/2)),x]

[Out]

(b^3*x*(a + b*x)^(1 + n)*Hypergeometric2F1[4, 1 + n, 2 + n, 1 + (b*x)/a])/(a^4*c*(1 + n)*Sqrt[c*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {(a+b x)^n}{x^4} \, dx}{c \sqrt {c x^2}} \\ & = \frac {b^3 x (a+b x)^{1+n} \, _2F_1\left (4,1+n;2+n;1+\frac {b x}{a}\right )}{a^4 c (1+n) \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^n}{x \left (c x^2\right )^{3/2}} \, dx=\frac {b^3 c x^5 (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (4,1+n,2+n,1+\frac {b x}{a}\right )}{a^4 (1+n) \left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

(b^3*c*x^5*(a + b*x)^(1 + n)*Hypergeometric2F1[4, 1 + n, 2 + n, 1 + (b*x)/a])/(a^4*(1 + n)*(c*x^2)^(5/2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(c*x^2)*(b*x + a)^n/(c^2*x^5), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*x)^n/(x*(c*x^2)^(3/2)),x)

[Out]

int((a + b*x)^n/(x*(c*x^2)^(3/2)), x)