\(\int (b+3 d x^2) (b x+d x^3)^n \, dx\) [184]

Optimal result

Integrand size = 20, antiderivative size = 19 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {\left (b x+d x^3\right )^{1+n}}{1+n} \]

[Out]

(d*x^3+b*x)^(1+n)/(1+n)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1602} \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {\left (b x+d x^3\right )^{n+1}}{n+1} \]

[In]

Int[(b + 3*d*x^2)*(b*x + d*x^3)^n,x]

[Out]

(b*x + d*x^3)^(1 + n)/(1 + n)

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b x+d x^3\right )^{1+n}}{1+n} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 5.58 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {x \left (x \left (b+d x^2\right )\right )^n \left (1+\frac {d x^2}{b}\right )^{-n} \left (b (3+n) \operatorname {Hypergeometric2F1}\left (-n,\frac {1+n}{2},\frac {3+n}{2},-\frac {d x^2}{b}\right )+3 d (1+n) x^2 \operatorname {Hypergeometric2F1}\left (-n,\frac {3+n}{2},\frac {5+n}{2},-\frac {d x^2}{b}\right )\right )}{(1+n) (3+n)} \]

[In]

Integrate[(b + 3*d*x^2)*(b*x + d*x^3)^n,x]

[Out]

(x*(x*(b + d*x^2))^n*(b*(3 + n)*Hypergeometric2F1[-n, (1 + n)/2, (3 + n)/2, -((d*x^2)/b)] + 3*d*(1 + n)*x^2*Hy
pergeometric2F1[-n, (3 + n)/2, (5 + n)/2, -((d*x^2)/b)]))/((1 + n)*(3 + n)*(1 + (d*x^2)/b)^n)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\left (x^{3} d +b x \right )^{1+n}}{1+n}\)	\(20\)
default	\(\frac {\left (x^{3} d +b x \right )^{1+n}}{1+n}\)	\(20\)
gosper	\(\frac {x \left (d \,x^{2}+b \right ) \left (x^{3} d +b x \right )^{n}}{1+n}\)	\(26\)
risch	\(\frac {x \left (d \,x^{2}+b \right ) {\left (x \left (d \,x^{2}+b \right )\right )}^{n}}{1+n}\)	\(26\)
parallelrisch	\(\frac {x^{3} {\left (x \left (d \,x^{2}+b \right )\right )}^{n} b d +x {\left (x \left (d \,x^{2}+b \right )\right )}^{n} b^{2}}{b \left (1+n \right )}\)	\(44\)
norman	\(\frac {b x \,{\mathrm e}^{n \ln \left (x^{3} d +b x \right )}}{1+n}+\frac {d \,x^{3} {\mathrm e}^{n \ln \left (x^{3} d +b x \right )}}{1+n}\)	\(46\)

[In]

int((3*d*x^2+b)*(d*x^3+b*x)^n,x,method=_RETURNVERBOSE)

[Out]

(d*x^3+b*x)^(1+n)/(1+n)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + b x\right )} {\left (d x^{3} + b x\right )}^{n}}{n + 1} \]

[In]

integrate((3*d*x^2+b)*(d*x^3+b*x)^n,x, algorithm="fricas")

[Out]

(d*x^3 + b*x)*(d*x^3 + b*x)^n/(n + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (14) = 28\).

Time = 5.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.32 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\begin {cases} \frac {b x \left (b x + d x^{3}\right )^{n}}{n + 1} + \frac {d x^{3} \left (b x + d x^{3}\right )^{n}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (x \right )} + \log {\left (x - \sqrt {- \frac {b}{d}} \right )} + \log {\left (x + \sqrt {- \frac {b}{d}} \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((3*d*x**2+b)*(d*x**3+b*x)**n,x)

[Out]

Piecewise((b*x*(b*x + d*x**3)**n/(n + 1) + d*x**3*(b*x + d*x**3)**n/(n + 1), Ne(n, -1)), (log(x) + log(x - sqr
t(-b/d)) + log(x + sqrt(-b/d)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + b x\right )}^{n + 1}}{n + 1} \]

[In]

integrate((3*d*x^2+b)*(d*x^3+b*x)^n,x, algorithm="maxima")

[Out]

(d*x^3 + b*x)^(n + 1)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + b x\right )}^{n + 1}}{n + 1} \]

[In]

integrate((3*d*x^2+b)*(d*x^3+b*x)^n,x, algorithm="giac")

[Out]

(d*x^3 + b*x)^(n + 1)/(n + 1)

Mupad [B] (verification not implemented)

Time = 9.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^n \, dx=\frac {x\,{\left (d\,x^3+b\,x\right )}^n\,\left (d\,x^2+b\right )}{n+1} \]

[In]

int((b*x + d*x^3)^n*(b + 3*d*x^2),x)

[Out]

(x*(b*x + d*x^3)^n*(b + d*x^2))/(n + 1)