3.945 \(\int \frac {(c x^2)^{5/2} (a+b x)^n}{x^7} \, dx\)

Optimal. Leaf size=50 \[ \frac {b c^2 \sqrt {c x^2} (a+b x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 (n+1) x} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 50, 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} \[ \frac {b c^2 \sqrt {c x^2} (a+b x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 (n+1) x} \]

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(5/2)*(a + b*x)^n)/x^7,x]

[Out]

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

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 {\left (c x^2\right )^{5/2} (a+b x)^n}{x^7} \, dx &=\frac {\left (c^2 \sqrt {c x^2}\right ) \int \frac {(a+b x)^n}{x^2} \, dx}{x}\\ &=\frac {b c^2 \sqrt {c x^2} (a+b x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac {b x}{a}\right )}{a^2 (1+n) x}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 47, normalized size = 0.94 \[ \frac {b \left (c x^2\right )^{5/2} (a+b x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {b x}{a}+1\right )}{a^2 (n+1) x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((c*x^2)^(5/2)*(a + b*x)^n)/x^7,x]

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2}} {\left (b x + a\right )}^{n} c^{2}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)*(b*x+a)^n/x^7,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x^{2}\right )^{\frac {5}{2}} {\left (b x + a\right )}^{n}}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)*(b*x+a)^n/x^7,x, algorithm="giac")

[Out]

integrate((c*x^2)^(5/2)*(b*x + a)^n/x^7, x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{n}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(5/2)*(b*x+a)^n/x^7,x)

[Out]

int((c*x^2)^(5/2)*(b*x+a)^n/x^7,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x^{2}\right )^{\frac {5}{2}} {\left (b x + a\right )}^{n}}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)*(b*x+a)^n/x^7,x, algorithm="maxima")

[Out]

integrate((c*x^2)^(5/2)*(b*x + a)^n/x^7, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c\,x^2\right )}^{5/2}\,{\left (a+b\,x\right )}^n}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c*x^2)^(5/2)*(a + b*x)^n)/x^7,x)

[Out]

int(((c*x^2)^(5/2)*(a + b*x)^n)/x^7, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{n}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2)**(5/2)*(b*x+a)**n/x**7,x)

[Out]

Integral((c*x**2)**(5/2)*(a + b*x)**n/x**7, x)

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