\(\int \sin ^3(a+b x) \tan ^3(a+b x) \, dx\) [118]

Optimal result

Integrand size = 17, antiderivative size = 66 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {5 \sin (a+b x)}{2 b}+\frac {5 \sin ^3(a+b x)}{6 b}+\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b} \]

[Out]

-5/2*arctanh(sin(b*x+a))/b+5/2*sin(b*x+a)/b+5/6*sin(b*x+a)^3/b+1/2*sin(b*x+a)^3*tan(b*x+a)^2/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2672, 294, 308, 212} \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {5 \sin ^3(a+b x)}{6 b}+\frac {5 \sin (a+b x)}{2 b}+\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b} \]

[In]

Int[Sin[a + b*x]^3*Tan[a + b*x]^3,x]

[Out]

(-5*ArcTanh[Sin[a + b*x]])/(2*b) + (5*Sin[a + b*x])/(2*b) + (5*Sin[a + b*x]^3)/(6*b) + (Sin[a + b*x]^3*Tan[a +
b*x]^2)/(2*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 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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^6}{\left (1-x^2\right )^2} \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b}-\frac {5 \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b} \\ & = \frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b}-\frac {5 \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (a+b x)\right )}{2 b} \\ & = \frac {5 \sin (a+b x)}{2 b}+\frac {5 \sin ^3(a+b x)}{6 b}+\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b}-\frac {5 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b} \\ & = -\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {5 \sin (a+b x)}{2 b}+\frac {5 \sin ^3(a+b x)}{6 b}+\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {5 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {5 \sin (a+b x) \tan ^2(a+b x)}{3 b}-\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{3 b} \]

[In]

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

[Out]

(-5*ArcTanh[Sin[a + b*x]])/(2*b) + (5*Sec[a + b*x]*Tan[a + b*x])/(2*b) - (5*Sin[a + b*x]*Tan[a + b*x]^2)/(3*b)
- (Sin[a + b*x]^3*Tan[a + b*x]^2)/(3*b)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{7}\left (b x +a \right )}{2 \cos \left (b x +a \right )^{2}}+\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (b x +a \right )\right )}{6}+\frac {5 \sin \left (b x +a \right )}{2}-\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\)	\(68\)
default	\(\frac {\frac {\sin ^{7}\left (b x +a \right )}{2 \cos \left (b x +a \right )^{2}}+\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (b x +a \right )\right )}{6}+\frac {5 \sin \left (b x +a \right )}{2}-\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\)	\(68\)
parallelrisch	\(\frac {\left (60 \cos \left (2 b x +2 a \right )+60\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-60 \cos \left (2 b x +2 a \right )-60\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+50 \sin \left (b x +a \right )+25 \sin \left (3 b x +3 a \right )-\sin \left (5 b x +5 a \right )}{24 b \left (1+\cos \left (2 b x +2 a \right )\right )}\)	\(102\)
risch	\(\frac {i {\mathrm e}^{3 i \left (b x +a \right )}}{24 b}-\frac {9 i {\mathrm e}^{i \left (b x +a \right )}}{8 b}+\frac {9 i {\mathrm e}^{-i \left (b x +a \right )}}{8 b}-\frac {i {\mathrm e}^{-3 i \left (b x +a \right )}}{24 b}-\frac {i \left ({\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}\)	\(138\)
norman	\(\frac {\frac {5 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}+\frac {20 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {22 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {20 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {5 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{2}}+\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{2 b}-\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{2 b}\)	\(146\)

[In]

int(sec(b*x+a)^3*sin(b*x+a)^6,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/2*sin(b*x+a)^7/cos(b*x+a)^2+1/2*sin(b*x+a)^5+5/6*sin(b*x+a)^3+5/2*sin(b*x+a)-5/2*ln(sec(b*x+a)+tan(b*x+
a)))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {15 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 15 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (b x + a\right )^{4} - 14 \, \cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right )}{12 \, b \cos \left (b x + a\right )^{2}} \]

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^6,x, algorithm="fricas")

[Out]

-1/12*(15*cos(b*x + a)^2*log(sin(b*x + a) + 1) - 15*cos(b*x + a)^2*log(-sin(b*x + a) + 1) + 2*(2*cos(b*x + a)^
4 - 14*cos(b*x + a)^2 - 3)*sin(b*x + a))/(b*cos(b*x + a)^2)

Sympy [F(-1)]

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

[In]

integrate(sec(b*x+a)**3*sin(b*x+a)**6,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {4 \, \sin \left (b x + a\right )^{3} - \frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right ) + 24 \, \sin \left (b x + a\right )}{12 \, b} \]

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^6,x, algorithm="maxima")

[Out]

1/12*(4*sin(b*x + a)^3 - 6*sin(b*x + a)/(sin(b*x + a)^2 - 1) - 15*log(sin(b*x + a) + 1) + 15*log(sin(b*x + a)
- 1) + 24*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {4 \, \sin \left (b x + a\right )^{3} - \frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right ) + 24 \, \sin \left (b x + a\right )}{12 \, b} \]

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^6,x, algorithm="giac")

[Out]

1/12*(4*sin(b*x + a)^3 - 6*sin(b*x + a)/(sin(b*x + a)^2 - 1) - 15*log(abs(sin(b*x + a) + 1)) + 15*log(abs(sin(
b*x + a) - 1)) + 24*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 7.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.23 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {5\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b} \]

[In]

int(sin(a + b*x)^6/cos(a + b*x)^3,x)

[Out]

(5*tan(a/2 + (b*x)/2) + (20*tan(a/2 + (b*x)/2)^3)/3 - (22*tan(a/2 + (b*x)/2)^5)/3 + (20*tan(a/2 + (b*x)/2)^7)/
3 + 5*tan(a/2 + (b*x)/2)^9)/(b*(tan(a/2 + (b*x)/2)^2 - 2*tan(a/2 + (b*x)/2)^4 - 2*tan(a/2 + (b*x)/2)^6 + tan(a
/2 + (b*x)/2)^8 + tan(a/2 + (b*x)/2)^10 + 1)) - (5*atanh(tan(a/2 + (b*x)/2)))/b