3.118 \(\int \sin ^3(a+b x) \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=66 \[ \frac {5 \sin ^3(a+b x)}{6 b}+\frac {5 \sin (a+b x)}{2 b}+\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b}-\frac {5 \tanh ^{-1}(\sin (a+b x))}{2 b} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2592, 288, 302, 206} \[ \frac {5 \sin ^3(a+b x)}{6 b}+\frac {5 \sin (a+b x)}{2 b}+\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{2 b}-\frac {5 \tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[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 206

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

Rule 288

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 302

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 2592

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*} \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx &=\frac {\operatorname {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 \operatorname {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 \operatorname {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 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b}\\ &=-\frac {5 \tanh ^{-1}(\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*}

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Mathematica [A]  time = 0.17, size = 52, normalized size = 0.79 \[ \frac {(24 \cos (2 (a+b x))-\cos (4 (a+b x))+37) \tan (a+b x) \sec (a+b x)-60 \tanh ^{-1}(\sin (a+b x))}{24 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*ArcTanh[Sin[a + b*x]] + (37 + 24*Cos[2*(a + b*x)] - Cos[4*(a + b*x)])*Sec[a + b*x]*Tan[a + b*x])/(24*b)

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fricas [A]  time = 0.45, size = 84, normalized size = 1.27 \[ -\frac {15 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 15 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (b x + a\right )^{4} - 14 \, \cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right )}{12 \, b \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.34, size = 68, normalized size = 1.03 \[ \frac {4 \, \sin \left (b x + a\right )^{3} - \frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right ) + 24 \, \sin \left (b x + a\right )}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 79, normalized size = 1.20 \[ \frac {\sin ^{7}\left (b x +a \right )}{2 b \cos \left (b x +a \right )^{2}}+\frac {\sin ^{5}\left (b x +a \right )}{2 b}+\frac {5 \left (\sin ^{3}\left (b x +a \right )\right )}{6 b}+\frac {5 \sin \left (b x +a \right )}{2 b}-\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 66, normalized size = 1.00 \[ \frac {4 \, \sin \left (b x + a\right )^{3} - \frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right ) + 24 \, \sin \left (b x + a\right )}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 7.24, size = 147, normalized size = 2.23 \[ \frac {5\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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