3.530 \(\int \cos ^7(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=138 \[ -\frac {\left (6 a^2+b^2\right ) \sin ^7(c+d x)}{42 d}+\frac {\left (6 a^2+b^2\right ) \sin ^5(c+d x)}{10 d}-\frac {\left (6 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac {\left (6 a^2+b^2\right ) \sin (c+d x)}{6 d}-\frac {5 a b \cos ^7(c+d x)}{42 d}-\frac {b \cos ^7(c+d x) (a+b \tan (c+d x))}{6 d} \]

[Out]

-5/42*a*b*cos(d*x+c)^7/d+1/6*(6*a^2+b^2)*sin(d*x+c)/d-1/6*(6*a^2+b^2)*sin(d*x+c)^3/d+1/10*(6*a^2+b^2)*sin(d*x+
c)^5/d-1/42*(6*a^2+b^2)*sin(d*x+c)^7/d-1/6*b*cos(d*x+c)^7*(a+b*tan(d*x+c))/d

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Rubi [A]  time = 0.11, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3508, 3486, 2633} \[ -\frac {\left (6 a^2+b^2\right ) \sin ^7(c+d x)}{42 d}+\frac {\left (6 a^2+b^2\right ) \sin ^5(c+d x)}{10 d}-\frac {\left (6 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac {\left (6 a^2+b^2\right ) \sin (c+d x)}{6 d}-\frac {5 a b \cos ^7(c+d x)}{42 d}-\frac {b \cos ^7(c+d x) (a+b \tan (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + b*Tan[c + d*x])^2,x]

[Out]

(-5*a*b*Cos[c + d*x]^7)/(42*d) + ((6*a^2 + b^2)*Sin[c + d*x])/(6*d) - ((6*a^2 + b^2)*Sin[c + d*x]^3)/(6*d) + (
(6*a^2 + b^2)*Sin[c + d*x]^5)/(10*d) - ((6*a^2 + b^2)*Sin[c + d*x]^7)/(42*d) - (b*Cos[c + d*x]^7*(a + b*Tan[c
+ d*x]))/(6*d)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3508

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

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac {b \cos ^7(c+d x) (a+b \tan (c+d x))}{6 d}-\frac {1}{6} \int \cos ^7(c+d x) \left (-6 a^2-b^2-5 a b \tan (c+d x)\right ) \, dx\\ &=-\frac {5 a b \cos ^7(c+d x)}{42 d}-\frac {b \cos ^7(c+d x) (a+b \tan (c+d x))}{6 d}-\frac {1}{6} \left (-6 a^2-b^2\right ) \int \cos ^7(c+d x) \, dx\\ &=-\frac {5 a b \cos ^7(c+d x)}{42 d}-\frac {b \cos ^7(c+d x) (a+b \tan (c+d x))}{6 d}-\frac {\left (6 a^2+b^2\right ) \operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{6 d}\\ &=-\frac {5 a b \cos ^7(c+d x)}{42 d}+\frac {\left (6 a^2+b^2\right ) \sin (c+d x)}{6 d}-\frac {\left (6 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac {\left (6 a^2+b^2\right ) \sin ^5(c+d x)}{10 d}-\frac {\left (6 a^2+b^2\right ) \sin ^7(c+d x)}{42 d}-\frac {b \cos ^7(c+d x) (a+b \tan (c+d x))}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 154, normalized size = 1.12 \[ -\frac {-3675 a^2 \sin (c+d x)-735 a^2 \sin (3 (c+d x))-147 a^2 \sin (5 (c+d x))-15 a^2 \sin (7 (c+d x))+1050 a b \cos (c+d x)+630 a b \cos (3 (c+d x))+210 a b \cos (5 (c+d x))+30 a b \cos (7 (c+d x))-525 b^2 \sin (c+d x)+35 b^2 \sin (3 (c+d x))+63 b^2 \sin (5 (c+d x))+15 b^2 \sin (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + b*Tan[c + d*x])^2,x]

[Out]

-1/6720*(1050*a*b*Cos[c + d*x] + 630*a*b*Cos[3*(c + d*x)] + 210*a*b*Cos[5*(c + d*x)] + 30*a*b*Cos[7*(c + d*x)]
- 3675*a^2*Sin[c + d*x] - 525*b^2*Sin[c + d*x] - 735*a^2*Sin[3*(c + d*x)] + 35*b^2*Sin[3*(c + d*x)] - 147*a^2
*Sin[5*(c + d*x)] + 63*b^2*Sin[5*(c + d*x)] - 15*a^2*Sin[7*(c + d*x)] + 15*b^2*Sin[7*(c + d*x)])/d

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fricas [A]  time = 0.62, size = 94, normalized size = 0.68 \[ -\frac {30 \, a b \cos \left (d x + c\right )^{7} - {\left (15 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, a^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(30*a*b*cos(d*x + c)^7 - (15*(a^2 - b^2)*cos(d*x + c)^6 + 3*(6*a^2 + b^2)*cos(d*x + c)^4 + 4*(6*a^2 + b
^2)*cos(d*x + c)^2 + 48*a^2 + 8*b^2)*sin(d*x + c))/d

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giac [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(cos(d*x+c)^7*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.50, size = 108, normalized size = 0.78 \[ \frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {2 a b \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{2} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+b*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.33, size = 98, normalized size = 0.71 \[ -\frac {30 \, a b \cos \left (d x + c\right )^{7} + 3 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{2} - {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(30*a*b*cos(d*x + c)^7 + 3*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))
*a^2 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*b^2)/d

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mupad [B]  time = 3.84, size = 176, normalized size = 1.28 \[ \frac {16\,a^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {8\,b^2\,\sin \left (c+d\,x\right )}{105\,d}+\frac {8\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {6\,a^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}+\frac {4\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{105\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}-\frac {b^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}-\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^7}{7\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(a + b*tan(c + d*x))^2,x)

[Out]

(16*a^2*sin(c + d*x))/(35*d) + (8*b^2*sin(c + d*x))/(105*d) + (8*a^2*cos(c + d*x)^2*sin(c + d*x))/(35*d) + (6*
a^2*cos(c + d*x)^4*sin(c + d*x))/(35*d) + (a^2*cos(c + d*x)^6*sin(c + d*x))/(7*d) + (4*b^2*cos(c + d*x)^2*sin(
c + d*x))/(105*d) + (b^2*cos(c + d*x)^4*sin(c + d*x))/(35*d) - (b^2*cos(c + d*x)^6*sin(c + d*x))/(7*d) - (2*a*
b*cos(c + d*x)^7)/(7*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \cos ^{7}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*cos(c + d*x)**7, x)

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