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

Optimal. Leaf size=99 \[ \frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+a^2 (-x)-\frac{a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a b \tan (c+d x) \sec (c+d x)}{3 d}+\frac{\tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.466797, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4397, 2889, 3048, 3031, 3021, 2735, 3770} \[ \frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+a^2 (-x)-\frac{a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a b \tan (c+d x) \sec (c+d x)}{3 d}+\frac{\tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) - (a*b*ArcTanh[Sin[c + d*x]])/d + ((2*a^2 - b^2)*Tan[c + d*x])/(3*d) + (a*b*Sec[c + d*x]*Tan[c + d*x]
)/(3*d) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 3021

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

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 ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\int (b+a \cos (c+d x))^2 \sec ^2(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 ^4(c+d x) \, dx\\ &=\frac{(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac{(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-2 \left (2 a^2-b^2\right )+6 a b \cos (c+d x)+6 a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac{(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (6 a b+6 a^2 \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-a^2 x+\frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac{(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-(a b) \int \sec (c+d x) \, dx\\ &=-a^2 x-\frac{a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac{(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 1.14578, size = 201, normalized size = 2.03 \[ \frac{\sec ^3(c+d x) \left (2 \sin (c+d x) \left (\left (3 a^2-b^2\right ) \cos (2 (c+d x))+3 a^2+6 a b \cos (c+d x)+b^2\right )-9 a \cos (c+d x) \left (a (c+d x)-b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-3 a \cos (3 (c+d x)) \left (a (c+d x)-b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(-9*a*Cos[c + d*x]*(a*(c + d*x) - b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + b*Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]]) - 3*a*Cos[3*(c + d*x)]*(a*(c + d*x) - b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*(3*a^2 + b^2 + 6*a*b*Cos[c + d*x] + (3*a^2 - b^2)*Cos[2*(c +
d*x)])*Sin[c + d*x]))/(12*d)

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Maple [A]  time = 0.073, size = 109, normalized size = 1.1 \begin{align*} -{a}^{2}x+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}c}{d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab\sin \left ( dx+c \right ) }{d}}-{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.69924, size = 111, normalized size = 1.12 \begin{align*} \frac{2 \, b^{2} \tan \left (d x + c\right )^{3} - 6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 3 \, a b{\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 )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^2*tan(d*x + c)^3 - 6*(d*x + c - tan(d*x + c))*a^2 - 3*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(
sin(d*x + c) + 1) - log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 0.523383, size = 294, normalized size = 2.97 \begin{align*} -\frac{6 \, a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (3 \, a b \cos \left (d x + c\right ) +{\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a^2*d*x*cos(d*x + c)^3 + 3*a*b*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a*b*cos(d*x + c)^3*log(-sin(d*
x + c) + 1) - 2*(3*a*b*cos(d*x + c) + (3*a^2 - b^2)*cos(d*x + c)^2 + b^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

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Giac [A]  time = 2.90522, size = 213, normalized size = 2.15 \begin{align*} -\frac{3 \,{\left (d x + c\right )} a^{2} + 3 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a b \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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(d*x + c)*a^2 + 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) +
2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*b^2*tan(1/2*
d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d