3.10.59 \(\int \frac {(a+b x)^n}{(c x^2)^{3/2}} \, dx\) [959]

Optimal. Leaf size=51 \[ -\frac {b^2 x (a+b x)^{1+n} \, _2F_1\left (3,1+n;2+n;1+\frac {b x}{a}\right )}{a^3 c (1+n) \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 67} \begin {gather*} -\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 c (n+1) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(c*x^2)^(3/2),x]

[Out]

-((b^2*x*(a + b*x)^(1 + n)*Hypergeometric2F1[3, 1 + n, 2 + n, 1 + (b*x)/a])/(a^3*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*} \int \frac {(a+b x)^n}{\left (c x^2\right )^{3/2}} \, dx &=\frac {x \int \frac {(a+b x)^n}{x^3} \, dx}{c \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 c (1+n) \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 50, normalized size = 0.98 \begin {gather*} -\frac {b^2 x^3 (a+b x)^{1+n} \, _2F_1\left (3,1+n;2+n;1+\frac {b x}{a}\right )}{a^3 (1+n) \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{n}}{\left (c \,x^{2}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{n}}{\left (c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^n}{{\left (c\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^n/(c*x^2)^(3/2),x)

[Out]

int((a + b*x)^n/(c*x^2)^(3/2), x)

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