3.191 \(\int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx\)

Optimal. Leaf size=236 \[ -\frac {b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}+\frac {3 b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{4 b^2 d}-\frac {\left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{4 b d}-\frac {a \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x \left (a^2-12 b^2\right )+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d} \]

[Out]

3/8*a*(a^2-12*b^2)*x+3/2*b*(2*a^2-b^2)*arctanh(sin(d*x+c))/d-1/2*b*(17*a^2-b^2)*sin(d*x+c)/d-1/8*a*(21*a^2-2*b
^2)*cos(d*x+c)*sin(d*x+c)/d-1/4*(6*a^2-b^2)*(b+a*cos(d*x+c))^2*sin(d*x+c)/b/d-1/4*(4*a^2-b^2)*(b+a*cos(d*x+c))
^3*sin(d*x+c)/b^2/d+a*(b+a*cos(d*x+c))^4*tan(d*x+c)/b^2/d+1/2*(b+a*cos(d*x+c))^4*sec(d*x+c)*tan(d*x+c)/b/d

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Rubi [A]  time = 0.75, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3872, 2893, 3049, 3033, 3023, 2735, 3770} \[ -\frac {b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}+\frac {3 b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{4 b^2 d}-\frac {\left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{4 b d}-\frac {a \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x \left (a^2-12 b^2\right )+\frac {a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^4,x]

[Out]

(3*a*(a^2 - 12*b^2)*x)/8 + (3*b*(2*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(2*d) - (b*(17*a^2 - b^2)*Sin[c + d*x])/(
2*d) - (a*(21*a^2 - 2*b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*d) - ((6*a^2 - b^2)*(b + a*Cos[c + d*x])^2*Sin[c + d*
x])/(4*b*d) - ((4*a^2 - b^2)*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*b^2*d) + (a*(b + a*Cos[c + d*x])^4*Tan[c
+ d*x])/(b^2*d) + ((b + a*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*b*d)

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 2893

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^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

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 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 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 3770

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

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]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx &=-\int (-b-a \cos (c+d x))^3 \sin (c+d x) \tan ^3(c+d x) \, dx\\ &=\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int (-b-a \cos (c+d x))^3 \left (-3 \left (2 a^2-b^2\right )+3 a b \cos (c+d x)+2 \left (4 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 b^2}\\ &=-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int (-b-a \cos (c+d x))^2 \left (12 b \left (2 a^2-b^2\right )-18 a b^2 \cos (c+d x)-6 b \left (6 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{8 b^2}\\ &=-\frac {\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int (-b-a \cos (c+d x)) \left (-36 b^2 \left (2 a^2-b^2\right )+78 a b^3 \cos (c+d x)+6 b^2 \left (21 a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 b^2}\\ &=-\frac {a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int \left (72 b^3 \left (2 a^2-b^2\right )+18 a b^2 \left (a^2-12 b^2\right ) \cos (c+d x)-24 b^3 \left (17 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{48 b^2}\\ &=-\frac {b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac {a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int \left (72 b^3 \left (2 a^2-b^2\right )+18 a b^2 \left (a^2-12 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx}{48 b^2}\\ &=\frac {3}{8} a \left (a^2-12 b^2\right ) x-\frac {b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac {a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac {1}{2} \left (3 b \left (2 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {3}{8} a \left (a^2-12 b^2\right ) x+\frac {3 b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac {a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac {\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac {a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac {(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}\\ \end {align*}

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Mathematica [B]  time = 6.19, size = 696, normalized size = 2.95 \[ \frac {a^3 \sin (4 (c+d x)) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{32 d (a \cos (c+d x)+b)^3}+\frac {3 \left (b^3-2 a^2 b\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b)^3}-\frac {3 \left (b^3-2 a^2 b\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b)^3}+\frac {3 a \left (a^2-12 b^2\right ) (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{8 d (a \cos (c+d x)+b)^3}+\frac {b \left (4 b^2-15 a^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (a \cos (c+d x)+b)^3}-\frac {a \left (a^2-3 b^2\right ) \sin (2 (c+d x)) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (a \cos (c+d x)+b)^3}+\frac {a^2 b \sin (3 (c+d x)) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (a \cos (c+d x)+b)^3}+\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b)^3}-\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b)^3}+\frac {3 a b^2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}+\frac {3 a b^2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^4,x]

[Out]

(3*a*(a^2 - 12*b^2)*(c + d*x)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(8*d*(b + a*Cos[c + d*x])^3) + (3*(-2*a^2
*b + b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)/(2*d*(b + a*Cos[c +
d*x])^3) - (3*(-2*a^2*b + b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)
/(2*d*(b + a*Cos[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b + a*Cos[c + d*x])^3*(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[(c + d*x)/2])/(d*(b +
a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b
+ a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3
*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (b*(-15*a^2 + 4*b^2)*Cos
[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d*(b + a*Cos[c + d*x])^3) - (a*(a^2 - 3*b^2)*Cos[c + d*x]^
3*(a + b*Sec[c + d*x])^3*Sin[2*(c + d*x)])/(4*d*(b + a*Cos[c + d*x])^3) + (a^2*b*Cos[c + d*x]^3*(a + b*Sec[c +
d*x])^3*Sin[3*(c + d*x)])/(4*d*(b + a*Cos[c + d*x])^3) + (a^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[4*(c
+ d*x)])/(32*d*(b + a*Cos[c + d*x])^3)

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fricas [A]  time = 0.57, size = 196, normalized size = 0.83 \[ \frac {3 \, {\left (a^{3} - 12 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{2} b \cos \left (d x + c\right )^{4} + 24 \, a b^{2} \cos \left (d x + c\right ) - {\left (5 \, a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 4 \, b^{3} - 8 \, {\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="fricas")

[Out]

1/8*(3*(a^3 - 12*a*b^2)*d*x*cos(d*x + c)^2 + 6*(2*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 6*(2*a^2
*b - b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + (2*a^3*cos(d*x + c)^5 + 8*a^2*b*cos(d*x + c)^4 + 24*a*b^2*co
s(d*x + c) - (5*a^3 - 12*a*b^2)*cos(d*x + c)^3 + 4*b^3 - 8*(4*a^2*b - b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*co
s(d*x + c)^2)

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giac [A]  time = 0.40, size = 431, normalized size = 1.83 \[ \frac {3 \, {\left (a^{3} - 12 \, a b^{2}\right )} {\left (d x + c\right )} + 12 \, {\left (2 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (2 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {8 \, {\left (6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \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}} + \frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 11 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 104 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 11 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 104 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, b^{3} \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}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="giac")

[Out]

1/8*(3*(a^3 - 12*a*b^2)*(d*x + c) + 12*(2*a^2*b - b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*(2*a^2*b - b^3)
*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 8*(6*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2
*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(3*a^3*tan(1/2*d*x + 1/2*
c)^7 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 8*b^3*tan(1/2*d*x + 1/2*c)^7 + 11*a
^3*tan(1/2*d*x + 1/2*c)^5 - 104*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*b^3*tan(1/
2*d*x + 1/2*c)^5 - 11*a^3*tan(1/2*d*x + 1/2*c)^3 - 104*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 12*a*b^2*tan(1/2*d*x + 1
/2*c)^3 + 24*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^3*tan(1/2*d*x + 1/2*c) - 24*a^2*b*tan(1/2*d*x + 1/2*c) + 12*a*b^
2*tan(1/2*d*x + 1/2*c) + 8*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.62, size = 276, normalized size = 1.17 \[ -\frac {a^{3} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 a^{3} x}{8}+\frac {3 a^{3} c}{8 d}-\frac {a^{2} b \left (\sin ^{3}\left (d x +c \right )\right )}{d}-\frac {3 a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 b^{2} a \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {3 b^{2} a \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{d}+\frac {9 a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {9 a \,b^{2} x}{2}-\frac {9 a \,b^{2} c}{2 d}+\frac {b^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 b^{3} \sin \left (d x +c \right )}{2 d}-\frac {3 b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x)

[Out]

-1/4*a^3*cos(d*x+c)*sin(d*x+c)^3/d-3/8*a^3*cos(d*x+c)*sin(d*x+c)/d+3/8*a^3*x+3/8/d*a^3*c-a^2*b*sin(d*x+c)^3/d-
3*a^2*b*sin(d*x+c)/d+3/d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/d*b^2*a*sin(d*x+c)^5/cos(d*x+c)+3/d*b^2*a*cos(d*x+c
)*sin(d*x+c)^3+9/2*a*b^2*cos(d*x+c)*sin(d*x+c)/d-9/2*a*b^2*x-9/2/d*a*b^2*c+1/2/d*b^3*sin(d*x+c)^5/cos(d*x+c)^2
+1/2*b^3*sin(d*x+c)^3/d+3/2*b^3*sin(d*x+c)/d-3/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.73, size = 183, normalized size = 0.78 \[ \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 16 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{2} b - 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{2} - 8 \, b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="maxima")

[Out]

1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a^3 - 16*(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^2*b - 48*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) -
2*tan(d*x + c))*a*b^2 - 8*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x
+ c) - 1) - 4*sin(d*x + c)))/d

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mupad [B]  time = 1.47, size = 281, normalized size = 1.19 \[ \frac {b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}-\frac {3\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {5\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}-\frac {4\,a^2\,b\,\sin \left (c+d\,x\right )}{d}-\frac {9\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {a^2\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4*(a + b/cos(c + d*x))^3,x)

[Out]

(b^3*sin(c + d*x))/d + (3*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) - (3*b^3*atanh(sin(c/2 + (d*x
)/2)/cos(c/2 + (d*x)/2)))/d + (a^3*cos(c + d*x)^3*sin(c + d*x))/(4*d) + (b^3*sin(c + d*x))/(2*d*cos(c + d*x)^2
) - (5*a^3*cos(c + d*x)*sin(c + d*x))/(8*d) - (4*a^2*b*sin(c + d*x))/d - (9*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2)))/d + (6*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*a*b^2*cos(c + d*x)*sin(c +
d*x))/(2*d) + (3*a*b^2*sin(c + d*x))/(d*cos(c + d*x)) + (a^2*b*cos(c + d*x)^2*sin(c + d*x))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**3*sin(d*x+c)**4,x)

[Out]

Timed out

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