3.6.36 \(\int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\) [536]
Optimal. Leaf size=84 \[ \frac {3}{8} a \left (a^2+b^2\right ) x-\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{4 d} \]
[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]
time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3587, 743, 737,
209} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
[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*} \int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx &=\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] Leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(84)=168\).
time = 3.81, size = 257, normalized size = 3.06 \begin {gather*} \frac {-3 a \sqrt {-b^2} \left (a^2+b^2\right )^3 \left (\log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-\log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right )-6 a b^3 \left (6 a^4+3 a^2 b^2+b^4\right ) \tan (c+d x)-24 a^4 b^4 \tan ^2(c+d x)+2 a b^5 \left (-3 a^2+b^2\right ) \tan ^3(c+d x)+4 b \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^4 (b+a \tan (c+d x))+8 a^2 b^2 \cos ^2(c+d x) (a+b \tan (c+d x))^4+2 a b \left (3 a^2-b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{16 b \left (a^2+b^2\right )^2 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]
[Out]
(-3*a*Sqrt[-b^2]*(a^2 + b^2)^3*(Log[Sqrt[-b^2] - b*Tan[c + d*x]] - Log[Sqrt[-b^2] + b*Tan[c + d*x]]) - 6*a*b^3
*(6*a^4 + 3*a^2*b^2 + b^4)*Tan[c + d*x] - 24*a^4*b^4*Tan[c + d*x]^2 + 2*a*b^5*(-3*a^2 + b^2)*Tan[c + d*x]^3 +
4*b*(a^2 + b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^4*(b + a*Tan[c + d*x]) + 8*a^2*b^2*Cos[c + d*x]^2*(a + b*Tan
[c + d*x])^4 + 2*a*b*(3*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan[c + d*x])^4)/(16*b*(a^2 + b^2)^2*d)
________________________________________________________________________________________
Maple [A]
time = 0.22, size = 114, normalized size = 1.36
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {3 a^{2} b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) |
\(114\) |
| default |
\(\frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {3 a^{2} b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) |
\(114\) |
| risch |
\(\frac {3 a^{3} x}{8}+\frac {3 a \,b^{2} x}{8}-\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}+\frac {b^{3} \cos \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}-\frac {3 a \sin \left (4 d x +4 c \right ) b^{2}}{32 d}-\frac {3 b \cos \left (2 d x +2 c \right ) a^{2}}{8 d}-\frac {b^{3} \cos \left (2 d x +2 c \right )}{8 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d}\) |
\(137\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[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*b^2*a*(-1/4*sin(d*x+c)*cos(d*x+c)^3+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))
________________________________________________________________________________________
Maxima [A]
time = 0.48, size = 110, normalized size = 1.31 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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
________________________________________________________________________________________
Fricas [A]
time = 0.44, size = 100, normalized size = 1.19 \begin {gather*} -\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{4}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 2496 vs.
\(2 (79) = 158\).
time = 12.21, size = 2496, normalized size = 29.71 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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(...
________________________________________________________________________________________
Mupad [B]
time = 3.80, size = 109, normalized size = 1.30 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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
________________________________________________________________________________________