∫ cos4(c + dx)(a + b tan(c + dx))2 dx

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

Optimal. Leaf size=88 \[ -\frac {\cos ^2(c+d x) \left (2 a b-\left (3 a^2+b^2\right ) \tan (c+d x)\right )}{8 d}+\frac {1}{8} x \left (3 a^2+b^2\right )-\frac {\cos ^4(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{4 d} \]

[Out]

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

(d*x+c))/d

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Rubi [A] time = 0.08, antiderivative size = 88, 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 = {3506, 739, 639, 203} \[ -\frac {\cos ^2(c+d x) \left (2 a b-\left (3 a^2+b^2\right ) \tan (c+d x)\right )}{8 d}+\frac {1}{8} x \left (3 a^2+b^2\right )-\frac {\cos ^4(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{4 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

b - (3*a^2 + b^2)*Tan[c + d*x]))/(8*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 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a

*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]

&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a

+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -

c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, 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, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst

[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b

^2, 0] && IntegerQ[m/2]

Rubi steps

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

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Mathematica [B] time = 2.98, size = 216, normalized size = 2.45 \[ \frac {4 \left (a^2+b^2\right ) \cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^3+\frac {\left (3 a^2+b^2\right ) \left (-\sqrt {-b^2} \left (b^4-a^4\right ) \sin (2 (c+d x))-2 a b \sqrt {-b^2} \left (a^2+b^2\right ) \cos (2 (c+d x))+b \left (a^2+b^2\right )^2 \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-b \left (a^2+b^2\right )^2 \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+2 a b \sqrt {-b^2} \left (2 a^2+b^2\right )\right )}{\sqrt {-b^2}}}{16 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

)^2*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - b*(a^2 + b^2)^2*Log[Sqrt[-b^2] + b*Tan[c + d*x]] - Sqrt[-b^2]*(-a^4 + b

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

(16*(a^2 + b^2)^2*d)

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fricas [A] time = 0.55, size = 75, normalized size = 0.85 \[ -\frac {4 \, a b \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} + b^{2}\right )} d x - {\left (2 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

in(d*x + c))/d

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/