\(\int x (a+b \tan (c+d x^2)) \, dx\) [3]

Optimal result

Integrand size = 14, antiderivative size = 26 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {a x^2}{2}-\frac {b \log \left (\cos \left (c+d x^2\right )\right )}{2 d} \]

[Out]

1/2*a*x^2-1/2*b*ln(cos(d*x^2+c))/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {14, 3832, 3556} \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {a x^2}{2}-\frac {b \log \left (\cos \left (c+d x^2\right )\right )}{2 d} \]

[In]

Int[x*(a + b*Tan[c + d*x^2]),x]

[Out]

(a*x^2)/2 - (b*Log[Cos[c + d*x^2]])/(2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3832

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x+b x \tan \left (c+d x^2\right )\right ) \, dx \\ & = \frac {a x^2}{2}+b \int x \tan \left (c+d x^2\right ) \, dx \\ & = \frac {a x^2}{2}+\frac {1}{2} b \text {Subst}\left (\int \tan (c+d x) \, dx,x,x^2\right ) \\ & = \frac {a x^2}{2}-\frac {b \log \left (\cos \left (c+d x^2\right )\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {a x^2}{2}-\frac {b \log \left (\cos \left (c+d x^2\right )\right )}{2 d} \]

[In]

Integrate[x*(a + b*Tan[c + d*x^2]),x]

[Out]

(a*x^2)/2 - (b*Log[Cos[c + d*x^2]])/(2*d)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
parts	\(\frac {a \,x^{2}}{2}-\frac {b \ln \left (\cos \left (d \,x^{2}+c \right )\right )}{2 d}\)	\(23\)
norman	\(\frac {a \,x^{2}}{2}+\frac {b \ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right )}{4 d}\)	\(27\)
derivativedivides	\(\frac {\left (d \,x^{2}+c \right ) a -b \ln \left (\cos \left (d \,x^{2}+c \right )\right )}{2 d}\)	\(28\)
default	\(\frac {\left (d \,x^{2}+c \right ) a -b \ln \left (\cos \left (d \,x^{2}+c \right )\right )}{2 d}\)	\(28\)
parallelrisch	\(\frac {2 a d \,x^{2}+b \ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right )}{4 d}\)	\(29\)
risch	\(\frac {i b \,x^{2}}{2}+\frac {a \,x^{2}}{2}+\frac {i b c}{d}-\frac {b \ln \left (1+{\mathrm e}^{2 i \left (d \,x^{2}+c \right )}\right )}{2 d}\)	\(43\)

[In]

int(x*(a+b*tan(d*x^2+c)),x,method=_RETURNVERBOSE)

[Out]

1/2*a*x^2-1/2*b*ln(cos(d*x^2+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {2 \, a d x^{2} - b \log \left (\frac {1}{\tan \left (d x^{2} + c\right )^{2} + 1}\right )}{4 \, d} \]

[In]

integrate(x*(a+b*tan(d*x^2+c)),x, algorithm="fricas")

[Out]

1/4*(2*a*d*x^2 - b*log(1/(tan(d*x^2 + c)^2 + 1)))/d

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{2}}{2} + \frac {b \log {\left (\tan ^{2}{\left (c + d x^{2} \right )} + 1 \right )}}{4 d} & \text {for}\: d \neq 0 \\\frac {x^{2} \left (a + b \tan {\left (c \right )}\right )}{2} & \text {otherwise} \end {cases} \]

[In]

integrate(x*(a+b*tan(d*x**2+c)),x)

[Out]

Piecewise((a*x**2/2 + b*log(tan(c + d*x**2)**2 + 1)/(4*d), Ne(d, 0)), (x**2*(a + b*tan(c))/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {b \log \left (\sec \left (d x^{2} + c\right )\right )}{2 \, d} \]

[In]

integrate(x*(a+b*tan(d*x^2+c)),x, algorithm="maxima")

[Out]

1/2*a*x^2 + 1/2*b*log(sec(d*x^2 + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {{\left (d x^{2} + c\right )} a - b \log \left ({\left | \cos \left (d x^{2} + c\right ) \right |}\right )}{2 \, d} \]

[In]

integrate(x*(a+b*tan(d*x^2+c)),x, algorithm="giac")

[Out]

1/2*((d*x^2 + c)*a - b*log(abs(cos(d*x^2 + c))))/d

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \tan \left (c+d x^2\right )\right ) \, dx=\frac {a\,x^2}{2}+\frac {b\,\ln \left ({\mathrm {tan}\left (d\,x^2+c\right )}^2+1\right )}{4\,d} \]

[In]

int(x*(a + b*tan(c + d*x^2)),x)

[Out]

(a*x^2)/2 + (b*log(tan(c + d*x^2)^2 + 1))/(4*d)