3.4.69 \(\int (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}) \, dx\)

Optimal. Leaf size=13 \[ \frac {x^m}{\sqrt {a+b x}} \]

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Rubi [C]  time = 0.04, antiderivative size = 92, normalized size of antiderivative = 7.08, number of steps used = 5, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {67, 65} \begin {gather*} \frac {x^m \left (-\frac {b x}{a}\right )^{-m} \, _2F_1\left (-\frac {1}{2},-m;\frac {1}{2};\frac {b x}{a}+1\right )}{\sqrt {a+b x}}-\frac {2 m x^m \sqrt {a+b x} \left (-\frac {b x}{a}\right )^{-m} \, _2F_1\left (\frac {1}{2},1-m;\frac {3}{2};\frac {b x}{a}+1\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-(b*x^m)/(2*(a + b*x)^(3/2)) + (m*x^(-1 + m))/Sqrt[a + b*x],x]

[Out]

(x^m*Hypergeometric2F1[-1/2, -m, 1/2, 1 + (b*x)/a])/((-((b*x)/a))^m*Sqrt[a + b*x]) - (2*m*x^m*Sqrt[a + b*x]*Hy
pergeometric2F1[1/2, 1 - m, 3/2, 1 + (b*x)/a])/(a*(-((b*x)/a))^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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 13, normalized size = 1.00 \begin {gather*} \frac {x^m}{\sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1/2*(b*x^m)/(a + b*x)^(3/2) + (m*x^(-1 + m))/Sqrt[a + b*x],x]

[Out]

x^m/Sqrt[a + b*x]

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IntegrateAlgebraic [F]  time = 0.59, size = 0, normalized size = 0.00 \begin {gather*} \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[-1/2*(b*x^m)/(a + b*x)^(3/2) + (m*x^(-1 + m))/Sqrt[a + b*x],x]

[Out]

Defer[IntegrateAlgebraic][-1/2*(b*x^m)/(a + b*x)^(3/2) + (m*x^(-1 + m))/Sqrt[a + b*x], x]

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fricas [A]  time = 0.94, size = 11, normalized size = 0.85 \begin {gather*} \frac {x^{m}}{\sqrt {b x + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*b*x^m/(b*x+a)^(3/2)+m*x^(-1+m)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

x^m/sqrt(b*x + a)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {m x^{m - 1}}{\sqrt {b x + a}} - \frac {b x^{m}}{2 \, {\left (b x + a\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*b*x^m/(b*x+a)^(3/2)+m*x^(-1+m)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(m*x^(m - 1)/sqrt(b*x + a) - 1/2*b*x^m/(b*x + a)^(3/2), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {b \,x^{m}}{2 \left (b x +a \right )^{\frac {3}{2}}}+\frac {m \,x^{m -1}}{\sqrt {b x +a}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/2*b*x^m/(b*x+a)^(3/2)+m*x^(-1+m)/(b*x+a)^(1/2),x)

[Out]

int(-1/2*b*x^m/(b*x+a)^(3/2)+m*x^(-1+m)/(b*x+a)^(1/2),x)

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maxima [A]  time = 1.86, size = 11, normalized size = 0.85 \begin {gather*} \frac {x^{m}}{\sqrt {b x + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*b*x^m/(b*x+a)^(3/2)+m*x^(-1+m)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

x^m/sqrt(b*x + a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.08 \begin {gather*} \int \frac {m\,x^{m-1}}{\sqrt {a+b\,x}}-\frac {b\,x^m}{2\,{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((m*x^(m - 1))/(a + b*x)^(1/2) - (b*x^m)/(2*(a + b*x)^(3/2)),x)

[Out]

int((m*x^(m - 1))/(a + b*x)^(1/2) - (b*x^m)/(2*(a + b*x)^(3/2)), x)

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sympy [C]  time = 5.34, size = 73, normalized size = 5.62 \begin {gather*} \frac {m x^{m} \Gamma \relax (m) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m \\ m + 1 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\sqrt {a} \Gamma \left (m + 1\right )} - \frac {b x x^{m} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (m + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*b*x**m/(b*x+a)**(3/2)+m*x**(-1+m)/(b*x+a)**(1/2),x)

[Out]

m*x**m*gamma(m)*hyper((1/2, m), (m + 1,), b*x*exp_polar(I*pi)/a)/(sqrt(a)*gamma(m + 1)) - b*x*x**m*gamma(m + 1
)*hyper((3/2, m + 1), (m + 2,), b*x*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(m + 2))

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