3.1.94 \(\int \sin (a+b x) \tan ^3(a+b x) \, dx\) [94]

Optimal. Leaf size=49 \[ -\frac {3 \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {3 \sin (a+b x)}{2 b}+\frac {\sin (a+b x) \tan ^2(a+b x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2672, 294, 327, 212} \begin {gather*} \frac {3 \sin (a+b x)}{2 b}+\frac {\sin (a+b x) \tan ^2(a+b x)}{2 b}-\frac {3 \tanh ^{-1}(\sin (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[a + b*x]^3,x]

[Out]

(-3*ArcTanh[Sin[a + b*x]])/(2*b) + (3*Sin[a + b*x])/(2*b) + (Sin[a + b*x]*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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

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Mathematica [A]
time = 0.07, size = 40, normalized size = 0.82 \begin {gather*} \frac {-3 \tanh ^{-1}(\sin (a+b x))+(2+\cos (2 (a+b x))) \sec (a+b x) \tan (a+b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 58, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{5}\left (b x +a \right )}{2 \cos \left (b x +a \right )^{2}}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{2}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\)	\(58\)
default	\(\frac {\frac {\sin ^{5}\left (b x +a \right )}{2 \cos \left (b x +a \right )^{2}}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{2}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\)	\(58\)
risch	\(-\frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {i \left ({\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}\)	\(108\)
norman	\(\frac {\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}-\frac {2 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {3 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{2 b}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{2 b}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3*sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 56, normalized size = 1.14 \begin {gather*} -\frac {\frac {2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\sin \left (b x + a\right ) - 1\right ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 74, normalized size = 1.51 \begin {gather*} -\frac {3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \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 )^{2} + 1\right )} \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{4}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3*sin(b*x+a)**4,x)

[Out]

Integral(sin(a + b*x)**4*sec(a + b*x)**3, x)

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Giac [A]
time = 4.58, size = 58, normalized size = 1.18 \begin {gather*} -\frac {\frac {2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + 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 ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/4*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 3*log(abs(sin(b*x + a) + 1)) - 3*log(abs(sin(b*x + a) - 1)) - 4*si
n(b*x + a))/b

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Mupad [B]
time = 3.97, size = 98, normalized size = 2.00 \begin {gather*} -\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b}-\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+3\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{b\,\left (-{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^4/cos(a + b*x)^3,x)

[Out]

- (3*atanh(tan(a/2 + (b*x)/2)))/b - (3*tan(a/2 + (b*x)/2) - 2*tan(a/2 + (b*x)/2)^3 + 3*tan(a/2 + (b*x)/2)^5)/(
b*(tan(a/2 + (b*x)/2)^2 + tan(a/2 + (b*x)/2)^4 - tan(a/2 + (b*x)/2)^6 - 1))

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