3.543 \(\int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=142 \[ \frac{8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac{2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}-\frac{3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac{\sin (c+d x) \cos ^6(c+d x) (a+b \tan (c+d x))^3}{7 d} \]

[Out]

(8*a*(2*a^2 + b^2)*Sin[c + d*x])/(35*d) - (3*Cos[c + d*x]^5*(b - 2*a*Tan[c + d*x])*(a + b*Tan[c + d*x])^2)/(35
*d) + (Cos[c + d*x]^6*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(7*d) - (2*Cos[c + d*x]^3*(b*(6*a^2 + b^2) - a*(4*a
^2 - b^2)*Tan[c + d*x]))/(35*d)

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Rubi [A]  time = 0.151976, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3512, 737, 821, 778, 191} \[ \frac{8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac{2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}-\frac{3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac{\sin (c+d x) \cos ^6(c+d x) (a+b \tan (c+d x))^3}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + b*Tan[c + d*x])^3,x]

[Out]

(8*a*(2*a^2 + b^2)*Sin[c + d*x])/(35*d) - (3*Cos[c + d*x]^5*(b - 2*a*Tan[c + d*x])*(a + b*Tan[c + d*x])^2)/(35
*d) + (Cos[c + d*x]^6*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(7*d) - (2*Cos[c + d*x]^3*(b*(6*a^2 + b^2) - a*(4*a
^2 - b^2)*Tan[c + d*x]))/(35*d)

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*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, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x)^3}{\left (1+\frac{x^2}{b^2}\right )^{9/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(-6 a-3 x) (a+x)^2}{\left (1+\frac{x^2}{b^2}\right )^{7/2}} \, dx,x,b \tan (c+d x)\right )}{7 b d}\\ &=-\frac{3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac{\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac{\left (b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x) \left (-6 \left (1+\frac{4 a^2}{b^2}\right )-\frac{12 a x}{b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{35 d}\\ &=-\frac{3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac{\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac{2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}+\frac{\left (8 a \left (1+\frac{2 a^2}{b^2}\right ) b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{35 d}\\ &=\frac{8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac{3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac{\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac{2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d}\\ \end{align*}

Mathematica [A]  time = 1.05369, size = 204, normalized size = 1.44 \[ \frac{-105 b \left (5 a^2+b^2\right ) \cos (c+d x)-35 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-105 a^2 b \cos (5 (c+d x))-15 a^2 b \cos (7 (c+d x))+1225 a^3 \sin (c+d x)+245 a^3 \sin (3 (c+d x))+49 a^3 \sin (5 (c+d x))+5 a^3 \sin (7 (c+d x))+525 a b^2 \sin (c+d x)-35 a b^2 \sin (3 (c+d x))-63 a b^2 \sin (5 (c+d x))-15 a b^2 \sin (7 (c+d x))+7 b^3 \cos (5 (c+d x))+5 b^3 \cos (7 (c+d x))}{2240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + b*Tan[c + d*x])^3,x]

[Out]

(-105*b*(5*a^2 + b^2)*Cos[c + d*x] - 35*(9*a^2*b + b^3)*Cos[3*(c + d*x)] - 105*a^2*b*Cos[5*(c + d*x)] + 7*b^3*
Cos[5*(c + d*x)] - 15*a^2*b*Cos[7*(c + d*x)] + 5*b^3*Cos[7*(c + d*x)] + 1225*a^3*Sin[c + d*x] + 525*a*b^2*Sin[
c + d*x] + 245*a^3*Sin[3*(c + d*x)] - 35*a*b^2*Sin[3*(c + d*x)] + 49*a^3*Sin[5*(c + d*x)] - 63*a*b^2*Sin[5*(c
+ d*x)] + 5*a^3*Sin[7*(c + d*x)] - 15*a*b^2*Sin[7*(c + d*x)])/(2240*d)

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Maple [A]  time = 0.069, size = 145, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{7}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) +3\,a{b}^{2} \left ( -1/7\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) +1/35\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{3\,b{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7}}+{\frac{{a}^{3}\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x)

[Out]

1/d*(b^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+3*a*b^2*(-1/7*cos(d*x+c)^6*sin(d*x+c)+1/35*(8/3+co
s(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-3/7*b*a^2*cos(d*x+c)^7+1/7*a^3*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/
5*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.15782, size = 170, normalized size = 1.2 \begin{align*} -\frac{15 \, a^{2} b \cos \left (d x + c\right )^{7} +{\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{3} -{\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} -{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{3}}{35 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/35*(15*a^2*b*cos(d*x + c)^7 + (5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*
a^3 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*a*b^2 - (5*cos(d*x + c)^7 - 7*cos(d*x + c)^5
)*b^3)/d

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Fricas [A]  time = 1.98461, size = 278, normalized size = 1.96 \begin{align*} -\frac{7 \, b^{3} \cos \left (d x + c\right )^{5} + 5 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{7} -{\left (5 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4} + 16 \, a^{3} + 8 \, a b^{2} + 4 \,{\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{35 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/35*(7*b^3*cos(d*x + c)^5 + 5*(3*a^2*b - b^3)*cos(d*x + c)^7 - (5*(a^3 - 3*a*b^2)*cos(d*x + c)^6 + 3*(2*a^3
+ a*b^2)*cos(d*x + c)^4 + 16*a^3 + 8*a*b^2 + 4*(2*a^3 + a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out