3.123 \(\int \frac {(d x)^m}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=71 \[ -\frac {2 c x^2 (b+c x) (d x)^m \left (-\frac {c x}{b}\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {c x}{b}+1\right )}{3 b^2 \left (b x+c x^2\right )^{5/2}} \]

[Out]

-2/3*c*x^2*(-c*x/b)^(1/2-m)*(d*x)^m*(c*x+b)*hypergeom([-3/2, 5/2-m],[-1/2],c*x/b+1)/b^2/(c*x^2+b*x)^(5/2)

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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {674, 67, 65} \[ -\frac {2 c x^2 (b+c x) (d x)^m \left (-\frac {c x}{b}\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {c x}{b}+1\right )}{3 b^2 \left (b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*c*x^2*(-((c*x)/b))^(1/2 - m)*(d*x)^m*(b + c*x)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, 1 + (c*x)/b])/(3*b^2
*(b*x + c*x^2)^(5/2))

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])

Rule 67

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

Rule 674

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[((e*x)^m*(b*x + c*x^2)^p)/(x^(m + p)
*(b + c*x)^p), Int[x^(m + p)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, m}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {(d x)^m}{\left (b x+c x^2\right )^{5/2}} \, dx &=\frac {\left (x^{\frac {5}{2}-m} (d x)^m (b+c x)^{5/2}\right ) \int \frac {x^{-\frac {5}{2}+m}}{(b+c x)^{5/2}} \, dx}{\left (b x+c x^2\right )^{5/2}}\\ &=\frac {\left (c^2 x^2 \left (-\frac {c x}{b}\right )^{\frac {1}{2}-m} (d x)^m (b+c x)^{5/2}\right ) \int \frac {\left (-\frac {c x}{b}\right )^{-\frac {5}{2}+m}}{(b+c x)^{5/2}} \, dx}{b^2 \left (b x+c x^2\right )^{5/2}}\\ &=-\frac {2 c x^2 \left (-\frac {c x}{b}\right )^{\frac {1}{2}-m} (d x)^m (b+c x) \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};1+\frac {c x}{b}\right )}{3 b^2 \left (b x+c x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 60, normalized size = 0.85 \[ \frac {2 (d x)^m \left (-\frac {c x}{b}\right )^{\frac {3}{2}-m} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {c x}{b}+1\right )}{3 b (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m/(b*x + c*x^2)^(5/2),x]

[Out]

(2*(-((c*x)/b))^(3/2 - m)*(d*x)^m*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, 1 + (c*x)/b])/(3*b*(x*(b + c*x))^(3/2
))

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x} \left (d x\right )^{m}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, b^{2} c x^{4} + b^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x)*(d*x)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

integrate((d*x)^m/(c*x^2 + b*x)^(5/2), x)

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(c*x^2+b*x)^(5/2),x)

[Out]

int((d*x)^m/(c*x^2+b*x)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((d*x)^m/(c*x^2 + b*x)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x\right )}^m}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(b*x + c*x^2)^(5/2),x)

[Out]

int((d*x)^m/(b*x + c*x^2)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d*x)**m/(x*(b + c*x))**(5/2), x)

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