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

Optimal. Leaf size=55 \[ \frac{(a-b)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (a+3 b) (a-b)+\frac{b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0772035, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 203} \[ \frac{(a-b)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (a+3 b) (a-b)+\frac{b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.394028, size = 55, normalized size = 1. \[ \frac{2 \left (a^2+2 a b-3 b^2\right ) (c+d x)+(a-b)^2 \sin (2 (c+d x))+4 b^2 \tan (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.043, size = 111, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) \cos \left ( dx+c \right ) -{\frac{3\,dx}{2}}-{\frac{3\,c}{2}} \right ) +2\,ab \left ( -1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.67371, size = 89, normalized size = 1.62 \begin{align*} \frac{2 \, b^{2} \tan \left (d x + c\right ) +{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )}{\left (d x + c\right )} + \frac{{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.41154, size = 166, normalized size = 3.02 \begin{align*} \frac{{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} d x \cos \left (d x + c\right ) +{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \cos ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.95463, size = 802, normalized size = 14.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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