3.240 \(\int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=90 \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]

[Out]

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

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Rubi [A]  time = 0.42642, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {4397, 2889, 3048, 3031, 3023, 2735, 3770} \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.192681, size = 75, normalized size = 0.83 \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))-2 a^2 \sin (c+d x)-4 a b \tan ^{-1}(\tan (c+d x))+4 a b \tan (c+d x)+b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

(-4*a*b*ArcTan[Tan[c + d*x]] + (2*a^2 - b^2)*ArcTanh[Sin[c + d*x]] - 2*a^2*Sin[c + d*x] + 4*a*b*Tan[c + d*x] +
b^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Maple [A]  time = 0.063, size = 123, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-2\,abx+2\,{\frac{ab\tan \left ( dx+c \right ) }{d}}-2\,{\frac{abc}{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.71679, size = 138, normalized size = 1.53 \begin{align*} -\frac{8 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a b + b^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*(d*x + c - tan(d*x + c))*a*b + b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) - 2*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)))/d

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Fricas [A]  time = 0.545411, size = 305, normalized size = 3.39 \begin{align*} -\frac{8 \, a b d x \cos \left (d x + c\right )^{2} -{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) - b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right )^{2} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))**2,x)

[Out]

Integral((a*sin(c + d*x) + b*tan(c + d*x))**2*sec(c + d*x), x)

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Giac [B]  time = 2.75911, size = 231, normalized size = 2.57 \begin{align*} -\frac{4 \,{\left (d x + c\right )} a b -{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) +{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{4 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{2 \,{\left (4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(d*x + c)*a*b - (2*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + (2*a^2 - b^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) + 4*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(4*a*b*tan(1/2*d*x + 1/2*c)^3 - b^
2*tan(1/2*d*x + 1/2*c)^3 - 4*a*b*tan(1/2*d*x + 1/2*c) - b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)
^2)/d