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

Optimal. Leaf size=37 \[ -\frac{\cos ^3(a+b x)}{3 b}+\frac{2 \cos (a+b x)}{b}+\frac{\sec (a+b x)}{b} \]

[Out]

(2*Cos[a + b*x])/b - Cos[a + b*x]^3/(3*b) + Sec[a + b*x]/b

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Rubi [A]  time = 0.0342953, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2590, 270} \[ -\frac{\cos ^3(a+b x)}{3 b}+\frac{2 \cos (a+b x)}{b}+\frac{\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Tan[a + b*x]^2,x]

[Out]

(2*Cos[a + b*x])/b - Cos[a + b*x]^3/(3*b) + Sec[a + b*x]/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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0328598, size = 39, normalized size = 1.05 \[ \frac{7 \cos (a+b x)}{4 b}-\frac{\cos (3 (a+b x))}{12 b}+\frac{\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Cos[a + b*x])/(4*b) - Cos[3*(a + b*x)]/(12*b) + Sec[a + b*x]/b

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Maple [A]  time = 0.017, size = 50, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{6}}{\cos \left ( bx+a \right ) }}+ \left ({\frac{8}{3}}+ \left ( \sin \left ( bx+a \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) \cos \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2*sin(b*x+a)^5,x)

[Out]

1/b*(sin(b*x+a)^6/cos(b*x+a)+(8/3+sin(b*x+a)^4+4/3*sin(b*x+a)^2)*cos(b*x+a))

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Maxima [A]  time = 0.965872, size = 43, normalized size = 1.16 \begin{align*} -\frac{\cos \left (b x + a\right )^{3} - \frac{3}{\cos \left (b x + a\right )} - 6 \, \cos \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/3*(cos(b*x + a)^3 - 3/cos(b*x + a) - 6*cos(b*x + a))/b

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Fricas [A]  time = 1.57127, size = 85, normalized size = 2.3 \begin{align*} -\frac{\cos \left (b x + a\right )^{4} - 6 \, \cos \left (b x + a\right )^{2} - 3}{3 \, b \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/3*(cos(b*x + a)^4 - 6*cos(b*x + a)^2 - 3)/(b*cos(b*x + a))

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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)**2*sin(b*x+a)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.14049, size = 134, normalized size = 3.62 \begin{align*} \frac{2 \,{\left (\frac{3}{\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1} + \frac{\frac{12 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 5}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{3}}\right )}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a)^5,x, algorithm="giac")

[Out]

2/3*(3/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1) + (12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x +
a) - 1)^2/(cos(b*x + a) + 1)^2 - 5)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^3)/b