3.448 \(\int \cos ^4(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=87 \[ \frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2+2 a b+3 b^2\right )+\frac {(a-b) \sin (c+d x) \cos ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3675, 413, 385, 203} \[ \frac {3 \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2+2 a b+3 b^2\right )+\frac {(a-b) \sin (c+d x) \cos ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^2}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {(a-b) \cos ^3(c+d x) \sin (c+d x) \left (a+b \tan ^2(c+d x)\right )}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {a (3 a+b)+b (a+3 b) x^2}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}\\ &=\frac {3 \left (a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(a-b) \cos ^3(c+d x) \sin (c+d x) \left (a+b \tan ^2(c+d x)\right )}{4 d}+\frac {\left (3 a^2+2 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {1}{8} \left (3 a^2+2 a b+3 b^2\right ) x+\frac {3 \left (a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(a-b) \cos ^3(c+d x) \sin (c+d x) \left (a+b \tan ^2(c+d x)\right )}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 65, normalized size = 0.75 \[ \frac {4 \left (3 a^2+2 a b+3 b^2\right ) (c+d x)+8 \left (a^2-b^2\right ) \sin (2 (c+d x))+(a-b)^2 \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(4*(3*a^2 + 2*a*b + 3*b^2)*(c + d*x) + 8*(a^2 - b^2)*Sin[2*(c + d*x)] + (a - b)^2*Sin[4*(c + d*x)])/(32*d)

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fricas [A]  time = 0.53, size = 75, normalized size = 0.86 \[ \frac {{\left (3 \, a^{2} + 2 \, a b + 3 \, b^{2}\right )} d x + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} + 2 \, a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.79, size = 122, normalized size = 1.40 \[ \frac {b^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.76, size = 97, normalized size = 1.11 \[ \frac {{\left (3 \, a^{2} + 2 \, a b + 3 \, b^{2}\right )} {\left (d x + c\right )} + \frac {{\left (3 \, a^{2} + 2 \, a b - 5 \, b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (5 \, a^{2} - 2 \, a b - 3 \, b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.24, size = 93, normalized size = 1.07 \[ x\,\left (\frac {3\,a^2}{8}+\frac {a\,b}{4}+\frac {3\,b^2}{8}\right )-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {5\,a^2}{8}+\frac {a\,b}{4}+\frac {3\,b^2}{8}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {3\,a^2}{8}+\frac {a\,b}{4}-\frac {5\,b^2}{8}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + b*tan(c + d*x)^2)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \cos ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tan(c + d*x)**2)**2*cos(c + d*x)**4, x)

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