∫ cos4(c + dx)(a + b tan(c + dx))3 dx

3.536 \(\int \cos ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=84 \[ \frac {3}{8} a x \left (a^2+b^2\right )-\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d} \]

[Out]

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

d*x+c))^3/d

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Rubi [A] time = 0.07, antiderivative size = 84, 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, 729, 723, 203} \[ \frac {3}{8} a x \left (a^2+b^2\right )-\frac {3 a \cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Cos[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]

[Out]

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

*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(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 723

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[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -

2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt

Q[p, -1]

Rule 729

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

p + 1))/(4*a*c*(p + 1)), x] - Dist[(m*(2*c*d))/(4*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x],

x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

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

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Mathematica [B] time = 3.58, size = 257, normalized size = 3.06 \[ \frac {-24 a^4 b^4 \tan ^2(c+d x)-3 a \sqrt {-b^2} \left (a^2+b^2\right )^3 \left (\log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-\log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right )+8 a^2 b^2 \cos ^2(c+d x) (a+b \tan (c+d x))^4+2 a b \left (3 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^4+4 b \left (a^2+b^2\right ) (a \tan (c+d x)+b) (a \cos (c+d x)+b \sin (c+d x))^4+2 a b^5 \left (b^2-3 a^2\right ) \tan ^3(c+d x)-6 a b^3 \left (6 a^4+3 a^2 b^2+b^4\right ) \tan (c+d x)}{16 b d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]

[Out]

(-3*a*Sqrt[-b^2]*(a^2 + b^2)^3*(Log[Sqrt[-b^2] - b*Tan[c + d*x]] - Log[Sqrt[-b^2] + b*Tan[c + d*x]]) - 6*a*b^3

*(6*a^4 + 3*a^2*b^2 + b^4)*Tan[c + d*x] - 24*a^4*b^4*Tan[c + d*x]^2 + 2*a*b^5*(-3*a^2 + b^2)*Tan[c + d*x]^3 +

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

[c + d*x])^4 + 2*a*b*(3*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan[c + d*x])^4)/(16*b*(a^2 + b^2)^2*d)

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fricas [A] time = 0.69, size = 100, normalized size = 1.19 \[ -\frac {4 \, b^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (a^{3} + a b^{2}\right )} d x - {\left (2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

*x + c)^3 + 3*(a^3 + a*b^2)*cos(d*x + c))*sin(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))^3,x, algorithm="giac")

[Out]

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