\(\int (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}) \, dx\) [369]

Optimal result

Integrand size = 34, antiderivative size = 13 \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\frac {x^m}{\sqrt {a+b x}} \]

[Out]

x^m/(b*x+a)^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 7.08, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {69, 67} \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\frac {x^m \left (-\frac {b x}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},\frac {b x}{a}+1\right )}{\sqrt {a+b x}}-\frac {2 m x^m \sqrt {a+b x} \left (-\frac {b x}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-m,\frac {3}{2},\frac {b x}{a}+1\right )}{a} \]

[In]

Int[-1/2*(b*x^m)/(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 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])

Rule 69

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\frac {x^m}{\sqrt {a+b x}} \]

[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]

Maple [F]

\[\int \left (-\frac {b \,x^{m}}{2 \left (b x +a \right )^{\frac {3}{2}}}+\frac {m \,x^{-1+m}}{\sqrt {b x +a}}\right )d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\frac {x^{m}}{\sqrt {b x + a}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.79 (sec) , antiderivative size = 80, normalized size of antiderivative = 6.15 \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\frac {a^{m} a^{- m - \frac {1}{2}} m x^{m} \Gamma \left (m\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m \\ m + 1 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\Gamma \left (m + 1\right )} - \frac {b x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (m + 2\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\frac {x^{m}}{\sqrt {b x + a}} \]

[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)

Giac [F]

\[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\int { \frac {m x^{m - 1}}{\sqrt {b x + a}} - \frac {b x^{m}}{2 \, {\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (-\frac {b x^m}{2 (a+b x)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x}}\right ) \, dx=\int \frac {m\,x^{m-1}}{\sqrt {a+b\,x}}-\frac {b\,x^m}{2\,{\left (a+b\,x\right )}^{3/2}} \,d x \]

[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)