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]

-1/2*cos(b*x+a)^2/b+2*ln(cos(b*x+a))/b+1/2*sec(b*x+a)^2/b

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.50, size = 54, normalized size = 1.26 \[ -\frac {2 \, \cos \left (b x + a\right )^{4} - 8 \, \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right )\right ) - \cos \left (b x + a\right )^{2} - 2}{4 \, b \cos \left (b x + a\right )^{2}} \]

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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giac [B]  time = 0.30, size = 182, normalized size = 4.23 \[ -\frac {\frac {4 \, {\left (\frac {\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} + \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}}{{\left (\frac {\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} + \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}^{2} - 4} + \log \left ({\left | -\frac {\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 2 \right |}\right ) - \log \left ({\left | -\frac {\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2 \right |}\right )}{b} \]

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

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maple [A]  time = 0.03, size = 60, normalized size = 1.40 \[ \frac {\sin ^{6}\left (b x +a \right )}{2 b \cos \left (b x +a \right )^{2}}+\frac {\sin ^{4}\left (b x +a \right )}{2 b}+\frac {\sin ^{2}\left (b x +a \right )}{b}+\frac {2 \ln \left (\cos \left (b x +a \right )\right )}{b} \]

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 = 0.32, size = 41, normalized size = 0.95 \[ \frac {\sin \left (b x + a\right )^{2} - \frac {1}{\sin \left (b x + a\right )^{2} - 1} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right )}{2 \, b} \]

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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mupad [B]  time = 0.40, size = 37, normalized size = 0.86 \[ -\frac {\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )+\frac {{\cos \left (a+b\,x\right )}^2}{2}-\frac {{\mathrm {tan}\left (a+b\,x\right )}^2}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^5/cos(a + b*x)^3,x)

[Out]

-(log(tan(a + b*x)^2 + 1) + cos(a + b*x)^2/2 - tan(a + b*x)^2/2)/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)**5,x)

[Out]

Timed out

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