3.542 \(\int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=105 \[ -\frac {2 \left (4 a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{15 d}-\frac {\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac {\sin (c+d x) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{5 d} \]

[Out]

-2/15*(4*a^2+b^2)*cos(d*x+c)*(b-a*tan(d*x+c))/d-1/15*cos(d*x+c)^3*(b-4*a*tan(d*x+c))*(a+b*tan(d*x+c))^2/d+1/5*
cos(d*x+c)^4*sin(d*x+c)*(a+b*tan(d*x+c))^3/d

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Rubi [A]  time = 0.09, antiderivative size = 105, 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 = {3512, 737, 805, 637} \[ -\frac {2 \left (4 a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{15 d}-\frac {\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac {\sin (c+d x) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(-2*(4*a^2 + b^2)*Cos[c + d*x]*(b - a*Tan[c + d*x]))/(15*d) - (Cos[c + d*x]^3*(b - 4*a*Tan[c + d*x])*(a + b*Ta
n[c + d*x])^2)/(15*d) + (Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(5*d)

Rule 637

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[(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 805

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[(m*(c*d*f + a*e*g))/(2*a*c*(p + 1)), Int[(d + e*
x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplif
y[m + 2*p + 3], 0] && LtQ[p, -1]

Rule 3512

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 ^5(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{7/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {\cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{5 d}-\frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {(-4 a-x) (a+x)^2}{\left (1+\frac {x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{5 b d}\\ &=-\frac {\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac {\cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{5 d}+\frac {\left (2 \left (4 a^2+b^2\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {a+x}{\left (1+\frac {x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{15 b d}\\ &=-\frac {2 \left (4 a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{15 d}-\frac {\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac {\cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 150, normalized size = 1.43 \[ \frac {150 a^3 \sin (c+d x)+25 a^3 \sin (3 (c+d x))+3 a^3 \sin (5 (c+d x))-5 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-30 b \left (3 a^2+b^2\right ) \cos (c+d x)-9 a^2 b \cos (5 (c+d x))+90 a b^2 \sin (c+d x)-15 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))+3 b^3 \cos (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(-30*b*(3*a^2 + b^2)*Cos[c + d*x] - 5*(9*a^2*b + b^3)*Cos[3*(c + d*x)] - 9*a^2*b*Cos[5*(c + d*x)] + 3*b^3*Cos[
5*(c + d*x)] + 150*a^3*Sin[c + d*x] + 90*a*b^2*Sin[c + d*x] + 25*a^3*Sin[3*(c + d*x)] - 15*a*b^2*Sin[3*(c + d*
x)] + 3*a^3*Sin[5*(c + d*x)] - 9*a*b^2*Sin[5*(c + d*x)])/(240*d)

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fricas [A]  time = 0.66, size = 102, normalized size = 0.97 \[ -\frac {5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} + {\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/15*(5*b^3*cos(d*x + c)^3 + 3*(3*a^2*b - b^3)*cos(d*x + c)^5 - (3*(a^3 - 3*a*b^2)*cos(d*x + c)^4 + 8*a^3 + 6
*a*b^2 + (4*a^3 + 3*a*b^2)*cos(d*x + c)^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)^5*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.56, size = 125, normalized size = 1.19 \[ \frac {b^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x)

[Out]

1/d*(b^3*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3)+3*b^2*a*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(
d*x+c)^2)*sin(d*x+c))-3/5*a^2*b*cos(d*x+c)^5+1/5*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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maxima [A]  time = 0.36, size = 107, normalized size = 1.02 \[ -\frac {9 \, a^{2} b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} + 3 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/15*(9*a^2*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 + 3*(3*sin(d*x +
c)^5 - 5*sin(d*x + c)^3)*a*b^2 - (3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*b^3)/d

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mupad [B]  time = 3.81, size = 147, normalized size = 1.40 \[ \frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^3-\frac {9\,a^2\,b\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {9\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {3\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+3\,\sin \left (c+d\,x\right )\,a\,b^2+\frac {3\,b^3\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {5\,b^3\,{\cos \left (c+d\,x\right )}^3}{2}\right )}{15\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + b*tan(c + d*x))^3,x)

[Out]

(2*(4*a^3*sin(c + d*x) - (5*b^3*cos(c + d*x)^3)/2 + (3*b^3*cos(c + d*x)^5)/2 - (9*a^2*b*cos(c + d*x)^5)/2 + 2*
a^3*cos(c + d*x)^2*sin(c + d*x) + (3*a^3*cos(c + d*x)^4*sin(c + d*x))/2 + 3*a*b^2*sin(c + d*x) + (3*a*b^2*cos(
c + d*x)^2*sin(c + d*x))/2 - (9*a*b^2*cos(c + d*x)^4*sin(c + d*x))/2))/(15*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*cos(c + d*x)**5, x)

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