3.441 \(\int \cos ^5(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=57 \[ \frac {a^2 \sin (c+d x)}{d}+\frac {(a-b)^2 \sin ^5(c+d x)}{5 d}-\frac {2 a (a-b) \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3676, 194} \[ \frac {a^2 \sin (c+d x)}{d}+\frac {(a-b)^2 \sin ^5(c+d x)}{5 d}-\frac {2 a (a-b) \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

Rule 194

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

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 52, normalized size = 0.91 \[ \frac {15 a^2 \sin (c+d x)+3 (a-b)^2 \sin ^5(c+d x)-10 a (a-b) \sin ^3(c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(15*a^2*Sin[c + d*x] - 10*a*(a - b)*Sin[c + d*x]^3 + 3*(a - b)^2*Sin[c + d*x]^5)/(15*d)

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fricas [A]  time = 0.49, size = 71, normalized size = 1.25 \[ \frac {{\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, a^{2} + a b - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.85, size = 89, normalized size = 1.56 \[ \frac {\frac {b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.38, size = 56, normalized size = 0.98 \[ \frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{5} - 10 \, {\left (a^{2} - a b\right )} \sin \left (d x + c\right )^{3} + 15 \, a^{2} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.20, size = 119, normalized size = 2.09 \[ \frac {\frac {5\,a^2\,\sin \left (c+d\,x\right )}{8}+\frac {b^2\,\sin \left (c+d\,x\right )}{8}+\frac {5\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{80}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{16}+\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{80}+\frac {a\,b\,\sin \left (c+d\,x\right )}{4}-\frac {a\,b\,\sin \left (3\,c+3\,d\,x\right )}{24}-\frac {a\,b\,\sin \left (5\,c+5\,d\,x\right )}{40}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + b*tan(c + d*x)^2)^2,x)

[Out]

((5*a^2*sin(c + d*x))/8 + (b^2*sin(c + d*x))/8 + (5*a^2*sin(3*c + 3*d*x))/48 + (a^2*sin(5*c + 5*d*x))/80 - (b^
2*sin(3*c + 3*d*x))/16 + (b^2*sin(5*c + 5*d*x))/80 + (a*b*sin(c + d*x))/4 - (a*b*sin(3*c + 3*d*x))/24 - (a*b*s
in(5*c + 5*d*x))/40)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \cos ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tan(c + d*x)**2)**2*cos(c + d*x)**5, x)

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