3.6.43 \(\int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx\) [543]

Optimal. Leaf size=142 \[ \frac {8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac {3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac {\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac {2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d} \]

[Out]

8/35*a*(2*a^2+b^2)*sin(d*x+c)/d-3/35*cos(d*x+c)^5*(b-2*a*tan(d*x+c))*(a+b*tan(d*x+c))^2/d+1/7*cos(d*x+c)^6*sin
(d*x+c)*(a+b*tan(d*x+c))^3/d-2/35*cos(d*x+c)^3*(b*(6*a^2+b^2)-a*(4*a^2-b^2)*tan(d*x+c))/d

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3593, 751, 835, 792, 197} \begin {gather*} \frac {8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac {2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}-\frac {3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac {\sin (c+d x) \cos ^6(c+d x) (a+b \tan (c+d x))^3}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + b*Tan[c + d*x])^3,x]

[Out]

(8*a*(2*a^2 + b^2)*Sin[c + d*x])/(35*d) - (3*Cos[c + d*x]^5*(b - 2*a*Tan[c + d*x])*(a + b*Tan[c + d*x])^2)/(35
*d) + (Cos[c + d*x]^6*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(7*d) - (2*Cos[c + d*x]^3*(b*(6*a^2 + b^2) - a*(4*a
^2 - b^2)*Tan[c + d*x]))/(35*d)

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 751

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

Rule 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 835

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

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 ^7(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 )^{9/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {(-6 a-3 x) (a+x)^2}{\left (1+\frac {x^2}{b^2}\right )^{7/2}} \, dx,x,b \tan (c+d x)\right )}{7 b d}\\ &=-\frac {3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac {\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac {\left (b \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {(a+x) \left (-6 \left (1+\frac {4 a^2}{b^2}\right )-\frac {12 a x}{b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{35 d}\\ &=-\frac {3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac {\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac {2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}+\frac {\left (8 a \left (1+\frac {2 a^2}{b^2}\right ) b \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{35 d}\\ &=\frac {8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac {3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac {\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac {2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.15, size = 204, normalized size = 1.44 \begin {gather*} \frac {-105 b \left (5 a^2+b^2\right ) \cos (c+d x)-35 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-105 a^2 b \cos (5 (c+d x))+7 b^3 \cos (5 (c+d x))-15 a^2 b \cos (7 (c+d x))+5 b^3 \cos (7 (c+d x))+1225 a^3 \sin (c+d x)+525 a b^2 \sin (c+d x)+245 a^3 \sin (3 (c+d x))-35 a b^2 \sin (3 (c+d x))+49 a^3 \sin (5 (c+d x))-63 a b^2 \sin (5 (c+d x))+5 a^3 \sin (7 (c+d x))-15 a b^2 \sin (7 (c+d x))}{2240 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + b*Tan[c + d*x])^3,x]

[Out]

(-105*b*(5*a^2 + b^2)*Cos[c + d*x] - 35*(9*a^2*b + b^3)*Cos[3*(c + d*x)] - 105*a^2*b*Cos[5*(c + d*x)] + 7*b^3*
Cos[5*(c + d*x)] - 15*a^2*b*Cos[7*(c + d*x)] + 5*b^3*Cos[7*(c + d*x)] + 1225*a^3*Sin[c + d*x] + 525*a*b^2*Sin[
c + d*x] + 245*a^3*Sin[3*(c + d*x)] - 35*a*b^2*Sin[3*(c + d*x)] + 49*a^3*Sin[5*(c + d*x)] - 63*a*b^2*Sin[5*(c
+ d*x)] + 5*a^3*Sin[7*(c + d*x)] - 15*a*b^2*Sin[7*(c + d*x)])/(2240*d)

________________________________________________________________________________________

Maple [A]
time = 0.24, size = 145, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(145\)
default	\(\frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(145\)
risch	\(-\frac {15 b \cos \left (d x +c \right ) a^{2}}{64 d}-\frac {3 b^{3} \cos \left (d x +c \right )}{64 d}+\frac {35 a^{3} \sin \left (d x +c \right )}{64 d}+\frac {15 a \,b^{2} \sin \left (d x +c \right )}{64 d}-\frac {3 b \cos \left (7 d x +7 c \right ) a^{2}}{448 d}+\frac {b^{3} \cos \left (7 d x +7 c \right )}{448 d}+\frac {a^{3} \sin \left (7 d x +7 c \right )}{448 d}-\frac {3 a \sin \left (7 d x +7 c \right ) b^{2}}{448 d}-\frac {3 b \cos \left (5 d x +5 c \right ) a^{2}}{64 d}+\frac {b^{3} \cos \left (5 d x +5 c \right )}{320 d}+\frac {7 a^{3} \sin \left (5 d x +5 c \right )}{320 d}-\frac {9 a \sin \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {9 b \cos \left (3 d x +3 c \right ) a^{2}}{64 d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{64 d}+\frac {7 a^{3} \sin \left (3 d x +3 c \right )}{64 d}-\frac {a \sin \left (3 d x +3 c \right ) b^{2}}{64 d}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(b^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+3*b^2*a*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+co
s(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-3/7*a^2*b*cos(d*x+c)^7+1/7*a^3*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/
5*cos(d*x+c)^2)*sin(d*x+c))

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 126, normalized size = 0.89 \begin {gather*} -\frac {15 \, a^{2} b \cos \left (d x + c\right )^{7} + {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{3}}{35 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/35*(15*a^2*b*cos(d*x + c)^7 + (5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*
a^3 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*a*b^2 - (5*cos(d*x + c)^7 - 7*cos(d*x + c)^5
)*b^3)/d

________________________________________________________________________________________

Fricas [A]
time = 0.38, size = 123, normalized size = 0.87 \begin {gather*} -\frac {7 \, b^{3} \cos \left (d x + c\right )^{5} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{7} - {\left (5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4} + 16 \, a^{3} + 8 \, a b^{2} + 4 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{35 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/35*(7*b^3*cos(d*x + c)^5 + 5*(3*a^2*b - b^3)*cos(d*x + c)^7 - (5*(a^3 - 3*a*b^2)*cos(d*x + c)^6 + 3*(2*a^3
+ a*b^2)*cos(d*x + c)^4 + 16*a^3 + 8*a*b^2 + 4*(2*a^3 + a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 101962 vs. \(2 (136) = 272\).
time = 82.30, size = 101962, normalized size = 718.04 \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)^7*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/17920*(945*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)^14*tan(1/2*c)^14 + 210*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)^14*tan(1/2*c
)^14 + 945*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)^14*tan(1/2*c)^14 + 210*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)^14*tan(1/2*c)^
14 - 2205*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)^14*tan(1/2*c)^14 + 1995*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)^14*tan(1/2*c)^14 + 220
5*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*t
an(1/2*d*x) - 1)*tan(1/2*d*x)^14*tan(1/2*c)^14 - 1995*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)^14*tan(1/2*c)^14 - 4830*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)^14*tan(1/2*c)^14 - 420*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)^14*tan(1/2*c)^14 + 6615*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)^14*tan(1/2*c)^12 + 14
70*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*ta
n(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)^14*tan(1/2*c)^12 + 6615*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)^14*tan(1/2*c)^12 + 14
70*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*ta
n(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)^14*tan(1/2*c)^12 + 6615*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)^12*tan(1/2*c)^14 + 14
70*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*ta
n(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)^12*tan(1/2*c)^14 + 6615*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)^12*tan(1/2*c)^14 + 14
70*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*ta
n(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)^12*tan(1/2*c)^14 + 4830*pi*a^2*b*tan(1/2*d*x)^14*tan(1/2*c)^14 + 420*pi*b^3*t
an(1/2*d*x)^14*tan(1/2*c)^14 - 1890*a^2*b*arctan((tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1)/(ta
n(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1))*tan(1/2*d*x)^14*tan(1/2*c)^14 - 420*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)^14*tan(1/2*c)^14 - 1890*a^2*b*arctan((tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1)/(t
an(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1))*tan(1/2*d*x)^14*tan(1/2*c)^14 - 420*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)^14*tan(1/2*c)^14 + 7770*a^2*b*arc...

________________________________________________________________________________________

Mupad [B]
time = 3.94, size = 214, normalized size = 1.51 \begin {gather*} \frac {16\,a^3\,\sin \left (c+d\,x\right )}{35\,d}-\frac {b^3\,{\cos \left (c+d\,x\right )}^5}{5\,d}+\frac {b^3\,{\cos \left (c+d\,x\right )}^7}{7\,d}-\frac {3\,a^2\,b\,{\cos \left (c+d\,x\right )}^7}{7\,d}+\frac {8\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {6\,a^3\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}+\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {4\,a\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}-\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(a + b*tan(c + d*x))^3,x)

[Out]

(16*a^3*sin(c + d*x))/(35*d) - (b^3*cos(c + d*x)^5)/(5*d) + (b^3*cos(c + d*x)^7)/(7*d) - (3*a^2*b*cos(c + d*x)
^7)/(7*d) + (8*a^3*cos(c + d*x)^2*sin(c + d*x))/(35*d) + (6*a^3*cos(c + d*x)^4*sin(c + d*x))/(35*d) + (a^3*cos
(c + d*x)^6*sin(c + d*x))/(7*d) + (8*a*b^2*sin(c + d*x))/(35*d) + (4*a*b^2*cos(c + d*x)^2*sin(c + d*x))/(35*d)
+ (3*a*b^2*cos(c + d*x)^4*sin(c + d*x))/(35*d) - (3*a*b^2*cos(c + d*x)^6*sin(c + d*x))/(7*d)

________________________________________________________________________________________