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]

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

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Rubi [A]  time = 0.08, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 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 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 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 ^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*}

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Mathematica [A]  time = 0.42, size = 55, normalized size = 1.00 \[ \frac {2 \left (a^2+2 a b-3 b^2\right ) (c+d x)+(a-b)^2 \sin (2 (c+d x))+4 b^2 \tan (c+d x)}{4 d} \]

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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fricas [A]  time = 0.62, size = 69, normalized size = 1.25 \[ \frac {{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} d x \cos \left (d x + c\right ) + {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

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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giac [B]  time = 3.36, size = 594, normalized size = 10.80 \[ \frac {a^{2} d x \tan \left (d x\right )^{3} \tan \relax (c)^{3} + 2 \, a b d x \tan \left (d x\right )^{3} \tan \relax (c)^{3} - 3 \, b^{2} d x \tan \left (d x\right )^{3} \tan \relax (c)^{3} + a^{2} d x \tan \left (d x\right )^{3} \tan \relax (c) + 2 \, a b d x \tan \left (d x\right )^{3} \tan \relax (c) - 3 \, b^{2} d x \tan \left (d x\right )^{3} \tan \relax (c) - a^{2} d x \tan \left (d x\right )^{2} \tan \relax (c)^{2} - 2 \, a b d x \tan \left (d x\right )^{2} \tan \relax (c)^{2} + 3 \, b^{2} d x \tan \left (d x\right )^{2} \tan \relax (c)^{2} + a^{2} d x \tan \left (d x\right ) \tan \relax (c)^{3} + 2 \, a b d x \tan \left (d x\right ) \tan \relax (c)^{3} - 3 \, b^{2} d x \tan \left (d x\right ) \tan \relax (c)^{3} - a^{2} \tan \left (d x\right )^{3} \tan \relax (c)^{2} + 2 \, a b \tan \left (d x\right )^{3} \tan \relax (c)^{2} - 3 \, b^{2} \tan \left (d x\right )^{3} \tan \relax (c)^{2} - a^{2} \tan \left (d x\right )^{2} \tan \relax (c)^{3} + 2 \, a b \tan \left (d x\right )^{2} \tan \relax (c)^{3} - 3 \, b^{2} \tan \left (d x\right )^{2} \tan \relax (c)^{3} - a^{2} d x \tan \left (d x\right )^{2} - 2 \, a b d x \tan \left (d x\right )^{2} + 3 \, b^{2} d x \tan \left (d x\right )^{2} + a^{2} d x \tan \left (d x\right ) \tan \relax (c) + 2 \, a b d x \tan \left (d x\right ) \tan \relax (c) - 3 \, b^{2} d x \tan \left (d x\right ) \tan \relax (c) - a^{2} d x \tan \relax (c)^{2} - 2 \, a b d x \tan \relax (c)^{2} + 3 \, b^{2} d x \tan \relax (c)^{2} - 2 \, b^{2} \tan \left (d x\right )^{3} + 2 \, a^{2} \tan \left (d x\right )^{2} \tan \relax (c) - 4 \, a b \tan \left (d x\right )^{2} \tan \relax (c) + 2 \, a^{2} \tan \left (d x\right ) \tan \relax (c)^{2} - 4 \, a b \tan \left (d x\right ) \tan \relax (c)^{2} - 2 \, b^{2} \tan \relax (c)^{3} - a^{2} d x - 2 \, a b d x + 3 \, b^{2} d x - a^{2} \tan \left (d x\right ) + 2 \, a b \tan \left (d x\right ) - 3 \, b^{2} \tan \left (d x\right ) - a^{2} \tan \relax (c) + 2 \, a b \tan \relax (c) - 3 \, b^{2} \tan \relax (c)}{2 \, {\left (d \tan \left (d x\right )^{3} \tan \relax (c)^{3} + d \tan \left (d x\right )^{3} \tan \relax (c) - d \tan \left (d x\right )^{2} \tan \relax (c)^{2} + d \tan \left (d x\right ) \tan \relax (c)^{3} - d \tan \left (d x\right )^{2} + d \tan \left (d x\right ) \tan \relax (c) - d \tan \relax (c)^{2} - d\right )}} \]

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)

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maple [B]  time = 0.56, size = 111, normalized size = 2.02 \[ \frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d} \]

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

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maxima [A]  time = 0.76, size = 66, normalized size = 1.20 \[ \frac {2 \, b^{2} \tan \left (d x + c\right ) + {\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} {\left (d x + c\right )} + \frac {{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

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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mupad [B]  time = 12.25, size = 91, normalized size = 1.65 \[ \frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {\sin \left (2\,c+2\,d\,x\right )\,\left (\frac {a^2}{2}-a\,b+\frac {b^2}{2}\right )}{2\,d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\left (a+3\,b\right )}{2\,\left (\frac {a^2}{2}+a\,b-\frac {3\,b^2}{2}\right )}\right )\,\left (a-b\right )\,\left (a+3\,b\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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