3.6.41 \(\int \cos ^3(c+d x) (a+b \tan (c+d x))^3 \, dx\) [541]

Optimal. Leaf size=70 \[ -\frac {2 \left (a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{3 d}-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3593, 737, 651} \begin {gather*} -\frac {2 \left (a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{3 d}-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

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 3593

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

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d}+\frac {\left (2 \left (a^2+b^2\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {a+x}{\left (1+\frac {x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{3 b d}\\ &=-\frac {2 \left (a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{3 d}-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.41, size = 81, normalized size = 1.16 \begin {gather*} \frac {-9 b \left (a^2+b^2\right ) \cos (c+d x)+\left (-3 a^2 b+b^3\right ) \cos (3 (c+d x))+2 a \left (5 a^2+3 b^2+\left (a^2-3 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*b*(a^2 + b^2)*Cos[c + d*x] + (-3*a^2*b + b^3)*Cos[3*(c + d*x)] + 2*a*(5*a^2 + 3*b^2 + (a^2 - 3*b^2)*Cos[2*
(c + d*x)])*Sin[c + d*x])/(12*d)

________________________________________________________________________________________

Maple [A]
time = 0.18, size = 75, normalized size = 1.07

method	result	size
derivativedivides	\(\frac {-\frac {b^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )-a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )+\frac {a^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(75\)
default	\(\frac {-\frac {b^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )-a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )+\frac {a^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(75\)
risch	\(-\frac {3 b \cos \left (d x +c \right ) a^{2}}{4 d}-\frac {3 b^{3} \cos \left (d x +c \right )}{4 d}+\frac {3 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{4 d}-\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{4 d}+\frac {b^{3} \cos \left (3 d x +3 c \right )}{12 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}-\frac {a \sin \left (3 d x +3 c \right ) b^{2}}{4 d}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 77, normalized size = 1.10 \begin {gather*} -\frac {3 \, a^{2} b \cos \left (d x + c\right )^{3} - 3 \, a b^{2} \sin \left (d x + c\right )^{3} + {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b^{3}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.42, size = 77, normalized size = 1.10 \begin {gather*} -\frac {3 \, b^{3} \cos \left (d x + c\right ) + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{3} + 3 \, a b^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*b^3*cos(d*x + c) + (3*a^2*b - b^3)*cos(d*x + c)^3 - (2*a^3 + 3*a*b^2 + (a^3 - 3*a*b^2)*cos(d*x + c)^2)
*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 ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 24430 vs. \(2 (68) = 136\).
time = 126.46, size = 24430, normalized size = 349.00 \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)^3*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/768*(72*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)
^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/
2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 - 105*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(
1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*
tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 +
72*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*
tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2
- 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 - 105*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x
)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2
*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 - 72*pi*a
^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2
*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 + 105*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c
)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 + 72*pi*a^2*b*sgn(tan(1/
2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(
1/2*d*x)^6*tan(1/2*c)^6 - 105*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d
*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 - 144*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1
/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 + 210*pi*
b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d
*x)^6*tan(1/2*c)^6 + 216*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)
^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*
d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^4 - 315*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/
2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*ta
n(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*
tan(1/2*c)^4 + 216*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - t
an(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2
+ tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/2*c)^4 - 315*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2
- 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*
c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^6*tan(1/
2*c)^4 + 216*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2
*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan
(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^4*tan(1/2*c)^6 - 315*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*t
an(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 +
2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^4*tan(1/2*c)^6
+ 216*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2
- 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c
)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^4*tan(1/2*c)^6 - 315*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2
*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan
(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^4*tan(1/2*c)^6 + 144
*pi*a^2*b*tan(1/2*d*x)^6*tan(1/2*c)^6 - 210*pi*b^3*tan(1/2*d*x)^6*tan(1/2*c)^6 - 144*a^2*b*arctan((tan(1/2*d*x
)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1)/(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1))*tan(1
/2*d*x)^6*tan(1/2*c)^6 + 210*b^3*arctan((tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1)/(tan(1/2*d*x
)*tan(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1))*tan(1/2*d*x)^6*tan(1/2*c)^6 - 144*a^2*b*arctan((tan(1/2*d*x)*ta
n(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1)/(tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1))*tan(1/2*d
*x)^6*tan(1/2*c)^6 + 210*b^3*arctan((tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1)/(tan(1/2*d*x)*ta
n(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1))*tan(1/2*d*x)^6*tan(1/2*c)^6 + 144*a^2*b*arctan((tan(1/2*d*x)*tan(1/
2*c) + tan(1/2*d*x) + tan(1/2*c) - 1)/(tan(1/2*...

________________________________________________________________________________________

Mupad [B]
time = 3.71, size = 104, normalized size = 1.49 \begin {gather*} \frac {\frac {\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2}{3}+\frac {2\,\sin \left (c+d\,x\right )\,a^3}{3}-a^2\,b\,{\cos \left (c+d\,x\right )}^3-\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\,a\,b^2+\frac {b^3\,{\cos \left (c+d\,x\right )}^3}{3}-b^3\,\cos \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________