\(\int x^{3/2} (a+b x) \, dx\) [430]

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int x^{3/2} (a+b x) \, dx=\frac {2}{5} a x^{5/2}+\frac {2}{7} b x^{7/2} \]

[Out]

2/5*a*x^(5/2)+2/7*b*x^(7/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int x^{3/2} (a+b x) \, dx=\frac {2}{5} a x^{5/2}+\frac {2}{7} b x^{7/2} \]

[In]

Int[x^(3/2)*(a + b*x),x]

[Out]

(2*a*x^(5/2))/5 + (2*b*x^(7/2))/7

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (a x^{3/2}+b x^{5/2}\right ) \, dx \\ & = \frac {2}{5} a x^{5/2}+\frac {2}{7} b x^{7/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int x^{3/2} (a+b x) \, dx=\frac {2}{35} x^{5/2} (7 a+5 b x) \]

[In]

Integrate[x^(3/2)*(a + b*x),x]

[Out]

(2*x^(5/2)*(7*a + 5*b*x))/35

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {2 x^{\frac {5}{2}} \left (5 b x +7 a \right )}{35}\)	\(14\)
derivativedivides	\(\frac {2 a \,x^{\frac {5}{2}}}{5}+\frac {2 b \,x^{\frac {7}{2}}}{7}\)	\(14\)
default	\(\frac {2 a \,x^{\frac {5}{2}}}{5}+\frac {2 b \,x^{\frac {7}{2}}}{7}\)	\(14\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (5 b x +7 a \right )}{35}\)	\(14\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (5 b x +7 a \right )}{35}\)	\(14\)

[In]

int(x^(3/2)*(b*x+a),x,method=_RETURNVERBOSE)

[Out]

2/35*x^(5/2)*(5*b*x+7*a)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int x^{3/2} (a+b x) \, dx=\frac {2}{35} \, {\left (5 \, b x^{3} + 7 \, a x^{2}\right )} \sqrt {x} \]

[In]

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

[Out]

2/35*(5*b*x^3 + 7*a*x^2)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int x^{3/2} (a+b x) \, dx=\frac {2 a x^{\frac {5}{2}}}{5} + \frac {2 b x^{\frac {7}{2}}}{7} \]

[In]

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

[Out]

2*a*x**(5/2)/5 + 2*b*x**(7/2)/7

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{3/2} (a+b x) \, dx=\frac {2}{7} \, b x^{\frac {7}{2}} + \frac {2}{5} \, a x^{\frac {5}{2}} \]

[In]

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

[Out]

2/7*b*x^(7/2) + 2/5*a*x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{3/2} (a+b x) \, dx=\frac {2}{7} \, b x^{\frac {7}{2}} + \frac {2}{5} \, a x^{\frac {5}{2}} \]

[In]

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

[Out]

2/7*b*x^(7/2) + 2/5*a*x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{3/2} (a+b x) \, dx=\frac {2\,x^{5/2}\,\left (7\,a+5\,b\,x\right )}{35} \]

[In]

int(x^(3/2)*(a + b*x),x)

[Out]

(2*x^(5/2)*(7*a + 5*b*x))/35