3.107 \(\int \sin ^2(a+b x) \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=43 \[ -\frac{\cos ^2(a+b x)}{2 b}+\frac{\sec ^2(a+b x)}{2 b}+\frac{2 \log (\cos (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0379609, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2590, 266, 43} \[ -\frac{\cos ^2(a+b x)}{2 b}+\frac{\sec ^2(a+b x)}{2 b}+\frac{2 \log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2*Tan[a + b*x]^3,x]

[Out]

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

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sin ^2(a+b x) \tan ^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^3} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^2} \, dx,x,\cos ^2(a+b x)\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}-\frac{2}{x}\right ) \, dx,x,\cos ^2(a+b x)\right )}{2 b}\\ &=-\frac{\cos ^2(a+b x)}{2 b}+\frac{2 \log (\cos (a+b x))}{b}+\frac{\sec ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0403313, size = 33, normalized size = 0.77 \[ \frac{\sin ^2(a+b x)+\sec ^2(a+b x)+4 \log (\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2*Tan[a + b*x]^3,x]

[Out]

(4*Log[Cos[a + b*x]] + Sec[a + b*x]^2 + Sin[a + b*x]^2)/(2*b)

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Maple [A]  time = 0.021, size = 60, normalized size = 1.4 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{6}}{2\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{2\,b}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{b}}+2\,{\frac{\ln \left ( \cos \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3*sin(b*x+a)^5,x)

[Out]

1/2/b*sin(b*x+a)^6/cos(b*x+a)^2+1/2*sin(b*x+a)^4/b+sin(b*x+a)^2/b+2*ln(cos(b*x+a))/b

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Maxima [A]  time = 0.95793, size = 55, normalized size = 1.28 \begin{align*} \frac{\sin \left (b x + a\right )^{2} - \frac{1}{\sin \left (b x + a\right )^{2} - 1} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62585, size = 139, normalized size = 3.23 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{4} - 8 \, \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right )\right ) - \cos \left (b x + a\right )^{2} - 2}{4 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/4*(2*cos(b*x + a)^4 - 8*cos(b*x + a)^2*log(-cos(b*x + a)) - cos(b*x + a)^2 - 2)/(b*cos(b*x + a)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3*sin(b*x+a)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.20022, size = 246, normalized size = 5.72 \begin{align*} -\frac{\frac{4 \,{\left (\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} + \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}}{{\left (\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} + \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}^{2} - 4} + \log \left ({\left | -\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 2 \right |}\right ) - \log \left ({\left | -\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2 \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^5,x, algorithm="giac")

[Out]

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