3.180 \(\int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx\)

Optimal. Leaf size=178 \[ -\frac{b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac{\left (12 a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{12 a b d}-\frac{\left (39 a^2+2 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{3}{8} x \left (a^2-4 b^2\right )+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\sin (c+d x) (a \cos (c+d x)+b)^3}{4 a d}+\frac{\tan (c+d x) (a \cos (c+d x)+b)^3}{b d} \]

[Out]

(3*(a^2 - 4*b^2)*x)/8 + (2*a*b*ArcTanh[Sin[c + d*x]])/d - (b*(28*a^2 + b^2)*Sin[c + d*x])/(6*a*d) - ((39*a^2 +
2*b^2)*Cos[c + d*x]*Sin[c + d*x])/(24*d) - ((12*a^2 + b^2)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(12*a*b*d) +
((b + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*a*d) + ((b + a*Cos[c + d*x])^3*Tan[c + d*x])/(b*d)

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Rubi [A]  time = 0.556468, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2894, 3049, 3033, 3023, 2735, 3770} \[ -\frac{b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac{\left (12 a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{12 a b d}-\frac{\left (39 a^2+2 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{3}{8} x \left (a^2-4 b^2\right )+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\sin (c+d x) (a \cos (c+d x)+b)^3}{4 a d}+\frac{\tan (c+d x) (a \cos (c+d x)+b)^3}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^4,x]

[Out]

(3*(a^2 - 4*b^2)*x)/8 + (2*a*b*ArcTanh[Sin[c + d*x]])/d - (b*(28*a^2 + b^2)*Sin[c + d*x])/(6*a*d) - ((39*a^2 +
2*b^2)*Cos[c + d*x]*Sin[c + d*x])/(24*d) - ((12*a^2 + b^2)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(12*a*b*d) +
((b + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*a*d) + ((b + a*Cos[c + d*x])^3*Tan[c + d*x])/(b*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2894

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a*b*d*(n + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n
+ 2) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n
+ 4))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/(
b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m
, 2*n]) &&  !m < -1 && LtQ[n, -1] && NeQ[m + n + 4, 0]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[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 (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx &=\int (-b-a \cos (c+d x))^2 \sin ^2(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac{\int (-b-a \cos (c+d x))^2 \left (-8 a^2+5 a b \cos (c+d x)+\left (12 a^2+b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{4 a b}\\ &=-\frac{\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac{(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac{\int (-b-a \cos (c+d x)) \left (24 a^2 b-17 a b^2 \cos (c+d x)-b \left (39 a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a b}\\ &=-\frac{\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac{(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac{\int \left (-48 a^2 b^2-9 a b \left (a^2-4 b^2\right ) \cos (c+d x)+4 b^2 \left (28 a^2+b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a b}\\ &=-\frac{b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac{\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac{(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac{\int \left (-48 a^2 b^2-9 a b \left (a^2-4 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a b}\\ &=\frac{3}{8} \left (a^2-4 b^2\right ) x-\frac{b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac{\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac{(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}+(2 a b) \int \sec (c+d x) \, dx\\ &=\frac{3}{8} \left (a^2-4 b^2\right ) x+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac{\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac{(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}\\ \end{align*}

Mathematica [A]  time = 1.01218, size = 157, normalized size = 0.88 \[ \frac{\tan (c+d x) \left (-6 \left (3 a^2-4 b^2\right ) \cos (2 (c+d x))+3 \left (a^2 \cos (4 (c+d x))-7 a^2+40 b^2\right )+16 a b \cos (3 (c+d x))\right )+12 \left (3 \left (a^2-4 b^2\right ) (c+d x)-16 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+16 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-208 a b \sin (c+d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^4,x]

[Out]

(12*(3*(a^2 - 4*b^2)*(c + d*x) - 16*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 16*a*b*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) - 208*a*b*Sin[c + d*x] + (-6*(3*a^2 - 4*b^2)*Cos[2*(c + d*x)] + 16*a*b*Cos[3*(c + d*x)]
+ 3*(-7*a^2 + 40*b^2 + a^2*Cos[4*(c + d*x)]))*Tan[c + d*x])/(96*d)

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Maple [A]  time = 0.041, size = 187, normalized size = 1.1 \begin{align*} -{\frac{{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}x}{8}}+{\frac{3\,{a}^{2}c}{8\,d}}-{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-2\,{\frac{ab\sin \left ( dx+c \right ) }{d}}+2\,{\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 ) ^{5}}{d\cos \left ( dx+c \right ) }}+{\frac{{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{b}^{2}x}{2}}-{\frac{3\,{b}^{2}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2*sin(d*x+c)^4,x)

[Out]

-1/4/d*a^2*cos(d*x+c)*sin(d*x+c)^3-3/8*a^2*cos(d*x+c)*sin(d*x+c)/d+3/8*a^2*x+3/8/d*a^2*c-2/3*a*b*sin(d*x+c)^3/
d-2*a*b*sin(d*x+c)/d+2/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^2*sin(d*x+c)^5/cos(d*x+c)+1/d*b^2*cos(d*x+c)*sin(
d*x+c)^3+3/2/d*b^2*cos(d*x+c)*sin(d*x+c)-3/2*b^2*x-3/2/d*b^2*c

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Maxima [A]  time = 1.49883, size = 169, normalized size = 0.95 \begin{align*} \frac{3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 32 \,{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b - 48 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{2}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^4,x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a^2 - 32*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c
) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*a*b - 48*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) -
2*tan(d*x + c))*b^2)/d

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Fricas [A]  time = 1.88299, size = 371, normalized size = 2.08 \begin{align*} \frac{9 \,{\left (a^{2} - 4 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 24 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 24 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a b \cos \left (d x + c\right )^{3} - 64 \, a b \cos \left (d x + c\right ) - 3 \,{\left (5 \, a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, b^{2}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^4,x, algorithm="fricas")

[Out]

1/24*(9*(a^2 - 4*b^2)*d*x*cos(d*x + c) + 24*a*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 24*a*b*cos(d*x + c)*log(-
sin(d*x + c) + 1) + (6*a^2*cos(d*x + c)^4 + 16*a*b*cos(d*x + c)^3 - 64*a*b*cos(d*x + c) - 3*(5*a^2 - 4*b^2)*co
s(d*x + c)^2 + 24*b^2)*sin(d*x + c))/(d*cos(d*x + c))

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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((a+b*sec(d*x+c))**2*sin(d*x+c)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.31842, size = 385, normalized size = 2.16 \begin{align*} \frac{48 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 9 \,{\left (a^{2} - 4 \, b^{2}\right )}{\left (d x + c\right )} - \frac{48 \, b^{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 (9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 33 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 208 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 33 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 208 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^4,x, algorithm="giac")

[Out]

1/24*(48*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 48*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 9*(a^2 - 4*b^2)*
(d*x + c) - 48*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(9*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*
b*tan(1/2*d*x + 1/2*c)^7 - 12*b^2*tan(1/2*d*x + 1/2*c)^7 + 33*a^2*tan(1/2*d*x + 1/2*c)^5 - 208*a*b*tan(1/2*d*x
+ 1/2*c)^5 - 12*b^2*tan(1/2*d*x + 1/2*c)^5 - 33*a^2*tan(1/2*d*x + 1/2*c)^3 - 208*a*b*tan(1/2*d*x + 1/2*c)^3 +
12*b^2*tan(1/2*d*x + 1/2*c)^3 - 9*a^2*tan(1/2*d*x + 1/2*c) - 48*a*b*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d