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

Optimal. Leaf size=21 \[ \frac{\cos (a+b x)}{b}+\frac{\sec (a+b x)}{b} \]

[Out]

Cos[a + b*x]/b + Sec[a + b*x]/b

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Rubi [A]  time = 0.0209529, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2590, 14} \[ \frac{\cos (a+b x)}{b}+\frac{\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[a + b*x]^2,x]

[Out]

Cos[a + b*x]/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0221629, size = 21, normalized size = 1. \[ \frac{\cos (a+b x)}{b}+\frac{\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[a + b*x]/b + Sec[a + b*x]/b

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Maple [A]  time = 0.014, size = 40, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{\cos \left ( bx+a \right ) }}+ \left ( 2+ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \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)^3,x)

[Out]

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

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Maxima [A]  time = 0.988198, size = 26, normalized size = 1.24 \begin{align*} \frac{\frac{1}{\cos \left (b x + a\right )} + \cos \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

(1/cos(b*x + a) + cos(b*x + a))/b

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Fricas [A]  time = 1.57206, size = 53, normalized size = 2.52 \begin{align*} \frac{\cos \left (b x + a\right )^{2} + 1}{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)^3,x, algorithm="fricas")

[Out]

(cos(b*x + a)^2 + 1)/(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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1924, size = 31, normalized size = 1.48 \begin{align*} \frac{\cos \left (b x + a\right )}{b} + \frac{1}{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)^3,x, algorithm="giac")

[Out]

cos(b*x + a)/b + 1/(b*cos(b*x + a))