\(\int \frac {b+c x}{(a+2 b x+c x^2)^{3/7}} \, dx\) [1448]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7}{8} \left (a+2 b x+c x^2\right )^{4/7} \]

[Out]

7/8*(c*x^2+2*b*x+a)^(4/7)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {643} \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7}{8} \left (a+2 b x+c x^2\right )^{4/7} \]

[In]

Int[(b + c*x)/(a + 2*b*x + c*x^2)^(3/7),x]

[Out]

(7*(a + 2*b*x + c*x^2)^(4/7))/8

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {7}{8} \left (a+2 b x+c x^2\right )^{4/7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7}{8} (a+x (2 b+c x))^{4/7} \]

[In]

Integrate[(b + c*x)/(a + 2*b*x + c*x^2)^(3/7),x]

[Out]

(7*(a + x*(2*b + c*x))^(4/7))/8

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {7 \left (c \,x^{2}+2 b x +a \right )^{\frac {4}{7}}}{8}\)	\(16\)
default	\(\frac {7 \left (c \,x^{2}+2 b x +a \right )^{\frac {4}{7}}}{8}\)	\(16\)
trager	\(\frac {7 \left (c \,x^{2}+2 b x +a \right )^{\frac {4}{7}}}{8}\)	\(16\)
risch	\(\frac {7 \left (c \,x^{2}+2 b x +a \right )^{\frac {4}{7}}}{8}\)	\(16\)
pseudoelliptic	\(\frac {7 \left (c \,x^{2}+2 b x +a \right )^{\frac {4}{7}}}{8}\)	\(16\)

[In]

int((c*x+b)/(c*x^2+2*b*x+a)^(3/7),x,method=_RETURNVERBOSE)

[Out]

7/8*(c*x^2+2*b*x+a)^(4/7)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7}{8} \, {\left (c x^{2} + 2 \, b x + a\right )}^{\frac {4}{7}} \]

[In]

integrate((c*x+b)/(c*x^2+2*b*x+a)^(3/7),x, algorithm="fricas")

[Out]

7/8*(c*x^2 + 2*b*x + a)^(4/7)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7 \left (a + 2 b x + c x^{2}\right )^{\frac {4}{7}}}{8} \]

[In]

integrate((c*x+b)/(c*x**2+2*b*x+a)**(3/7),x)

[Out]

7*(a + 2*b*x + c*x**2)**(4/7)/8

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7}{8} \, {\left (c x^{2} + 2 \, b x + a\right )}^{\frac {4}{7}} \]

[In]

integrate((c*x+b)/(c*x^2+2*b*x+a)^(3/7),x, algorithm="maxima")

[Out]

7/8*(c*x^2 + 2*b*x + a)^(4/7)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7}{8} \, {\left (c x^{2} + 2 \, b x + a\right )}^{\frac {4}{7}} \]

[In]

integrate((c*x+b)/(c*x^2+2*b*x+a)^(3/7),x, algorithm="giac")

[Out]

7/8*(c*x^2 + 2*b*x + a)^(4/7)

Mupad [B] (verification not implemented)

Time = 9.69 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {b+c x}{\left (a+2 b x+c x^2\right )^{3/7}} \, dx=\frac {7\,{\left (c\,x^2+2\,b\,x+a\right )}^{4/7}}{8} \]

[In]

int((b + c*x)/(a + 2*b*x + c*x^2)^(3/7),x)

[Out]

(7*(a + 2*b*x + c*x^2)^(4/7))/8