3.10.85 \(\int \frac {(d x)^m (a+b x)^n}{(c x^2)^{3/2}} \, dx\) [985]

Optimal. Leaf size=68 \[ -\frac {d^2 x (d x)^{-2+m} (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \, _2F_1\left (-2+m,-n;-1+m;-\frac {b x}{a}\right )}{c (2-m) \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 16, 68, 66} \begin {gather*} -\frac {d^2 x (d x)^{m-2} (a+b x)^n \left (\frac {b x}{a}+1\right )^{-n} \, _2F_1\left (m-2,-n;m-1;-\frac {b x}{a}\right )}{c (2-m) \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(
x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 57, normalized size = 0.84 \begin {gather*} \frac {x (d x)^m (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \, _2F_1\left (-2+m,-n;-1+m;-\frac {b x}{a}\right )}{(-2+m) \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (d x \right )^{m} \left (b x +a \right )^{n}}{\left (c \,x^{2}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sqrt(c*x^2)*(b*x + a)^n*(d*x)^m/(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 (d x\right )^{m} \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((d*x)**m*(b*x+a)**n/(c*x**2)**(3/2),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x\right )}^m\,{\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(((d*x)^m*(a + b*x)^n)/(c*x^2)^(3/2),x)

[Out]

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

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