\(\int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\) [536]

Optimal result

Integrand size = 21, antiderivative size = 84 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3}{8} a \left (a^2+b^2\right ) x-\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3587, 743, 737, 209} \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3}{8} a x \left (a^2+b^2\right )-\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d} \]

[In]

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

[Out]

(3*a*(a^2 + b^2)*x)/8 - (3*a*Cos[c + d*x]^2*(b - a*Tan[c + d*x])*(a + b*Tan[c + d*x]))/(8*d) + (Cos[c + d*x]^3
*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(4*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[(2*p + 3)*((c*d^2 + a*e^2)/(2*a*c*(p + 1))), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^m)*(2*c*x)*((a + c*x^2)
^(p + 1)/(4*a*c*(p + 1))), x] - Dist[m*((2*c*d)/(4*a*c*(p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x
], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

Rule 3587

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^3} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{4 d}+\frac {(3 a) \text {Subst}\left (\int \frac {(a+x)^2}{\left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 b d} \\ & = -\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{4 d}+\frac {\left (3 a \left (a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{8 b d} \\ & = \frac {3}{8} a \left (a^2+b^2\right ) x-\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{4 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(488\) vs. \(2(84)=168\).

Time = 1.22 (sec) , antiderivative size = 488, normalized size of antiderivative = 5.81 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {39 a^6 b^2+41 a^4 b^4+21 a^2 b^6+3 b^8-4 b^2 \left (a^2+b^2\right )^2 \left (3 a^2+b^2\right ) \cos (2 (c+d x))+b^2 \left (-3 a^2+b^2\right ) \left (a^2+b^2\right )^2 \cos (4 (c+d x))-6 a^7 \sqrt {-b^2} \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+18 a^5 \left (-b^2\right )^{3/2} \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+6 a b^4 \left (-b^2\right )^{3/2} \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-18 a^3 \left (-b^2\right )^{5/2} \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+6 a^7 \sqrt {-b^2} \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+18 a^3 b^4 \sqrt {-b^2} \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+6 a b^6 \sqrt {-b^2} \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )-18 a^5 \left (-b^2\right )^{3/2} \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+8 a^7 b \sin (2 (c+d x))+16 a^5 b^3 \sin (2 (c+d x))+8 a^3 b^5 \sin (2 (c+d x))+a^7 b \sin (4 (c+d x))-a^5 b^3 \sin (4 (c+d x))-5 a^3 b^5 \sin (4 (c+d x))-3 a b^7 \sin (4 (c+d x))}{32 b \left (a^2+b^2\right )^2 d} \]

[In]

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

[Out]

(39*a^6*b^2 + 41*a^4*b^4 + 21*a^2*b^6 + 3*b^8 - 4*b^2*(a^2 + b^2)^2*(3*a^2 + b^2)*Cos[2*(c + d*x)] + b^2*(-3*a
^2 + b^2)*(a^2 + b^2)^2*Cos[4*(c + d*x)] - 6*a^7*Sqrt[-b^2]*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 18*a^5*(-b^2)^(
3/2)*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 6*a*b^4*(-b^2)^(3/2)*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 18*a^3*(-b^2)^
(5/2)*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 6*a^7*Sqrt[-b^2]*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + 18*a^3*b^4*Sqrt[-
b^2]*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + 6*a*b^6*Sqrt[-b^2]*Log[Sqrt[-b^2] + b*Tan[c + d*x]] - 18*a^5*(-b^2)^(3
/2)*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + 8*a^7*b*Sin[2*(c + d*x)] + 16*a^5*b^3*Sin[2*(c + d*x)] + 8*a^3*b^5*Sin[
2*(c + d*x)] + a^7*b*Sin[4*(c + d*x)] - a^5*b^3*Sin[4*(c + d*x)] - 5*a^3*b^5*Sin[4*(c + d*x)] - 3*a*b^7*Sin[4*
(c + d*x)])/(32*b*(a^2 + b^2)^2*d)

Maple [A] (verified)

Time = 20.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+3 a \,b^{2} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {3 a^{2} b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(114\)
default	\(\frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+3 a \,b^{2} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {3 a^{2} b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(114\)
risch	\(\frac {3 a^{3} x}{8}+\frac {3 x a \,b^{2}}{8}-\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}+\frac {b^{3} \cos \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}-\frac {3 a \sin \left (4 d x +4 c \right ) b^{2}}{32 d}-\frac {3 b \cos \left (2 d x +2 c \right ) a^{2}}{8 d}-\frac {b^{3} \cos \left (2 d x +2 c \right )}{8 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d}\)	\(137\)

[In]

int(cos(d*x+c)^4*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {4 \, b^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (a^{3} + a b^{2}\right )} d x - {\left (2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**4*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*cos(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.31 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} + a b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, b^{3} \tan \left (d x + c\right )^{2} - 3 \, {\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b + 2 \, b^{3} - {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]

[In]

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

[Out]

1/8*(3*(a^3 + a*b^2)*(d*x + c) - (4*b^3*tan(d*x + c)^2 - 3*(a^3 + a*b^2)*tan(d*x + c)^3 + 6*a^2*b + 2*b^3 - (5
*a^3 - 3*a*b^2)*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2496 vs. \(2 (79) = 158\).

Time = 12.25 (sec) , antiderivative size = 2496, normalized size of antiderivative = 29.71 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/64*(9*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c)^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) -
2*tan(c))*tan(d*x)^4*tan(c)^4 + 24*a^3*d*x*tan(d*x)^4*tan(c)^4 + 24*a*b^2*d*x*tan(d*x)^4*tan(c)^4 + 9*pi*a*b^2
*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^4*tan(c)^4 + 18*pi*a*b^2*sgn
(2*tan(d*x)^2*tan(c)^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^4
*tan(c)^2 + 18*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c)^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(
d*x) - 2*tan(c))*tan(d*x)^2*tan(c)^4 + 18*a*b^2*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^4*t
an(c)^4 - 18*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)*tan(c) + 1))*tan(d*x)^4*tan(c)^4 + 48*a^3*d*x*tan(d*x
)^4*tan(c)^2 + 48*a*b^2*d*x*tan(d*x)^4*tan(c)^2 + 18*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 +
2*tan(d*x) - 2*tan(c))*tan(d*x)^4*tan(c)^2 + 48*a^3*d*x*tan(d*x)^2*tan(c)^4 + 48*a*b^2*d*x*tan(d*x)^2*tan(c)^
4 + 18*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^2*tan(c)^4 -
30*a^2*b*tan(d*x)^4*tan(c)^4 - 6*b^3*tan(d*x)^4*tan(c)^4 + 9*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c)^2 - 2)*sgn(-2*ta
n(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^4 + 36*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c
)^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^2*tan(c)^2 + 36*a*b^
2*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^4*tan(c)^2 - 36*a*b^2*arctan(-(tan(d*x) - tan(c))
/(tan(d*x)*tan(c) + 1))*tan(d*x)^4*tan(c)^2 - 40*a^3*tan(d*x)^4*tan(c)^3 + 24*a*b^2*tan(d*x)^4*tan(c)^3 + 9*pi
*a*b^2*sgn(2*tan(d*x)^2*tan(c)^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*
tan(c)^4 + 36*a*b^2*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^2*tan(c)^4 - 36*a*b^2*arctan(-(
tan(d*x) - tan(c))/(tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^4 - 40*a^3*tan(d*x)^3*tan(c)^4 + 24*a*b^2*tan(d*x)
^3*tan(c)^4 + 24*a^3*d*x*tan(d*x)^4 + 24*a*b^2*d*x*tan(d*x)^4 + 9*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*
x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^4 + 96*a^3*d*x*tan(d*x)^2*tan(c)^2 + 96*a*b^2*d*x*tan(d*x)^2*tan
(c)^2 + 36*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^2*tan(c)^
2 + 36*a^2*b*tan(d*x)^4*tan(c)^2 - 12*b^3*tan(d*x)^4*tan(c)^2 + 192*a^2*b*tan(d*x)^3*tan(c)^3 + 24*a^3*d*x*tan
(c)^4 + 24*a*b^2*d*x*tan(c)^4 + 9*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan
(c))*tan(c)^4 + 36*a^2*b*tan(d*x)^2*tan(c)^4 - 12*b^3*tan(d*x)^2*tan(c)^4 + 18*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c
)^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^2 + 18*a*b^2*arctan(
(tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^4 - 18*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)*tan(c)
+ 1))*tan(d*x)^4 - 24*a^3*tan(d*x)^4*tan(c) - 24*a*b^2*tan(d*x)^4*tan(c) + 18*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c)
^2 - 2)*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(c)^2 + 72*a*b^2*arctan((ta
n(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^2*tan(c)^2 - 72*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)*t
an(c) + 1))*tan(d*x)^2*tan(c)^2 + 48*a^3*tan(d*x)^3*tan(c)^2 - 144*a*b^2*tan(d*x)^3*tan(c)^2 + 48*a^3*tan(d*x)
^2*tan(c)^3 - 144*a*b^2*tan(d*x)^2*tan(c)^3 + 18*a*b^2*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(c
)^4 - 18*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)*tan(c) + 1))*tan(c)^4 - 24*a^3*tan(d*x)*tan(c)^4 - 24*a*b
^2*tan(d*x)*tan(c)^4 + 48*a^3*d*x*tan(d*x)^2 + 48*a*b^2*d*x*tan(d*x)^2 + 18*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c)
+ 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(d*x)^2 + 18*a^2*b*tan(d*x)^4 + 10*b^3*tan(d*x)^4 + 64*b^3*t
an(d*x)^3*tan(c) + 48*a^3*d*x*tan(c)^2 + 48*a*b^2*d*x*tan(c)^2 + 18*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(
d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c))*tan(c)^2 - 216*a^2*b*tan(d*x)^2*tan(c)^2 + 72*b^3*tan(d*x)^2*tan(c)^2 +
64*b^3*tan(d*x)*tan(c)^3 + 18*a^2*b*tan(c)^4 + 10*b^3*tan(c)^4 + 9*pi*a*b^2*sgn(2*tan(d*x)^2*tan(c)^2 - 2)*sg
n(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c)) + 36*a*b^2*arctan((tan(d*x) + tan(c))/(t
an(d*x)*tan(c) - 1))*tan(d*x)^2 - 36*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)*tan(c) + 1))*tan(d*x)^2 + 24*
a^3*tan(d*x)^3 + 24*a*b^2*tan(d*x)^3 - 48*a^3*tan(d*x)^2*tan(c) + 144*a*b^2*tan(d*x)^2*tan(c) + 36*a*b^2*arcta
n((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(c)^2 - 36*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)*tan(c)
+ 1))*tan(c)^2 - 48*a^3*tan(d*x)*tan(c)^2 + 144*a*b^2*tan(d*x)*tan(c)^2 + 24*a^3*tan(c)^3 + 24*a*b^2*tan(c)^3
+ 24*a^3*d*x + 24*a*b^2*d*x + 9*pi*a*b^2*sgn(-2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 + 2*tan(d*x) - 2*tan(c
)) + 36*a^2*b*tan(d*x)^2 - 12*b^3*tan(d*x)^2 + 192*a^2*b*tan(d*x)*tan(c) + 36*a^2*b*tan(c)^2 - 12*b^3*tan(c)^2
+ 18*a*b^2*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1)) - 18*a*b^2*arctan(-(tan(d*x) - tan(c))/(tan(d*x)
*tan(c) + 1)) + 40*a^3*tan(d*x) - 24*a*b^2*tan(d*x) + 40*a^3*tan(c) - 24*a*b^2*tan(c) - 30*a^2*b - 6*b^3)/(d*t
an(d*x)^4*tan(c)^4 + 2*d*tan(d*x)^4*tan(c)^2 + 2*d*tan(d*x)^2*tan(c)^4 + d*tan(d*x)^4 + 4*d*tan(d*x)^2*tan(c)^
2 + d*tan(c)^4 + 2*d*tan(d*x)^2 + 2*d*tan(c)^2 + d)

Mupad [B] (verification not implemented)

Time = 4.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.30 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3\,a^3\,x}{8}-\frac {6\,a^2\,b-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^3+3\,a\,b^2\right )+2\,b^3+\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a\,b^2-5\,a^3\right )+4\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d\,\left (8\,{\mathrm {tan}\left (c+d\,x\right )}^4+16\,{\mathrm {tan}\left (c+d\,x\right )}^2+8\right )}+\frac {3\,a\,b^2\,x}{8} \]

[In]

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

[Out]

(3*a^3*x)/8 - (6*a^2*b - tan(c + d*x)^3*(3*a*b^2 + 3*a^3) + 2*b^3 + tan(c + d*x)*(3*a*b^2 - 5*a^3) + 4*b^3*tan
(c + d*x)^2)/(d*(16*tan(c + d*x)^2 + 8*tan(c + d*x)^4 + 8)) + (3*a*b^2*x)/8