\(\int \sin ^3(a+b x) \tan (a+b x) \, dx\) [93]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \]

[Out]

arctanh(sin(b*x+a))/b-sin(b*x+a)/b-1/3*sin(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2672, 308, 212} \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin ^3(a+b x)}{3 b}-\frac {\sin (a+b x)}{b} \]

[In]

Int[Sin[a + b*x]^3*Tan[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b - Sin[a + b*x]/b - Sin[a + b*x]^3/(3*b)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = -\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \]

[In]

Integrate[Sin[a + b*x]^3*Tan[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b - Sin[a + b*x]/b - Sin[a + b*x]^3/(3*b)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{3}-\sin \left (b x +a \right )+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\)	\(38\)
default	\(\frac {-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{3}-\sin \left (b x +a \right )+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\)	\(38\)
parallelrisch	\(\frac {12 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )-12 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-15 \sin \left (b x +a \right )+\sin \left (3 b x +3 a \right )}{12 b}\)	\(52\)
risch	\(\frac {5 i {\mathrm e}^{i \left (b x +a \right )}}{8 b}-\frac {5 i {\mathrm e}^{-i \left (b x +a \right )}}{8 b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{b}+\frac {\sin \left (3 b x +3 a \right )}{12 b}\)	\(81\)
norman	\(\frac {-\frac {2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}-\frac {20 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {2 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}\)	\(98\)

[In]

int(sec(b*x+a)*sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/3*sin(b*x+a)^3-sin(b*x+a)+ln(sec(b*x+a)+tan(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {2 \, {\left (\cos \left (b x + a\right )^{2} - 4\right )} \sin \left (b x + a\right ) + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{6 \, b} \]

[In]

integrate(sec(b*x+a)*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/6*(2*(cos(b*x + a)^2 - 4)*sin(b*x + a) + 3*log(sin(b*x + a) + 1) - 3*log(-sin(b*x + a) + 1))/b

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)*sin(b*x+a)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=-\frac {2 \, \sin \left (b x + a\right )^{3} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right ) + 6 \, \sin \left (b x + a\right )}{6 \, b} \]

[In]

integrate(sec(b*x+a)*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/6*(2*sin(b*x + a)^3 - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1) + 6*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=-\frac {2 \, \sin \left (b x + a\right )^{3} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right ) + 6 \, \sin \left (b x + a\right )}{6 \, b} \]

[In]

integrate(sec(b*x+a)*sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/6*(2*sin(b*x + a)^3 - 3*log(abs(sin(b*x + a) + 1)) + 3*log(abs(sin(b*x + a) - 1)) + 6*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}{\cos \left (\frac {a}{2}+\frac {b\,x}{2}\right )}\right )}{b}-\frac {5\,\sin \left (a+b\,x\right )}{4\,b}+\frac {\sin \left (3\,a+3\,b\,x\right )}{12\,b} \]

[In]

int(sin(a + b*x)^4/cos(a + b*x),x)

[Out]

(2*atanh(sin(a/2 + (b*x)/2)/cos(a/2 + (b*x)/2)))/b - (5*sin(a + b*x))/(4*b) + sin(3*a + 3*b*x)/(12*b)