3.952 \(\int \frac {(a+b x)^n}{x^2 \sqrt {c x^2}} \, dx\)

Optimal. Leaf size=48 \[ -\frac {b^2 x (a+b x)^{n+1} \, _2F_1\left (3,n+1;n+2;\frac {b x}{a}+1\right )}{a^3 (n+1) \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 65} \[ -\frac {b^2 x (a+b x)^{n+1} \, _2F_1\left (3,n+1;n+2;\frac {b x}{a}+1\right )}{a^3 (n+1) \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(x^2*Sqrt[c*x^2]),x]

[Out]

-((b^2*x*(a + b*x)^(1 + n)*Hypergeometric2F1[3, 1 + n, 2 + n, 1 + (b*x)/a])/(a^3*(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 65

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 51, normalized size = 1.06 \[ -\frac {b^2 c x^3 (a+b x)^{n+1} \, _2F_1\left (3,n+1;n+2;\frac {b x}{a}+1\right )}{a^3 (n+1) \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{c x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n/x^2/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2)*(b*x + a)^n/(c*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n}}{\sqrt {c x^{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n/x^2/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{n}}{\sqrt {c \,x^{2}}\, x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^n/x^2/(c*x^2)^(1/2),x)

[Out]

int((b*x+a)^n/x^2/(c*x^2)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n}}{\sqrt {c x^{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n/x^2/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+b\,x\right )}^n}{x^2\,\sqrt {c\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^n/(x^2*(c*x^2)^(1/2)),x)

[Out]

int((a + b*x)^n/(x^2*(c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{n}}{x^{2} \sqrt {c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**n/x**2/(c*x**2)**(1/2),x)

[Out]

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

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