3.85 \(\int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=30 \[ \frac{\tan ^5(c+d x) (a \cot (c+d x)+b)^5}{5 b d} \]

[Out]

((b + a*Cot[c + d*x])^5*Tan[c + d*x]^5)/(5*b*d)

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Rubi [A]  time = 0.0476237, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3088, 37} \[ \frac{\tan ^5(c+d x) (a \cot (c+d x)+b)^5}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

((b + a*Cot[c + d*x])^5*Tan[c + d*x]^5)/(5*b*d)

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^4}{x^6} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{(b+a \cot (c+d x))^5 \tan ^5(c+d x)}{5 b d}\\ \end{align*}

Mathematica [B]  time = 0.308616, size = 73, normalized size = 2.43 \[ \frac{\tan (c+d x) \left (10 a^2 b^2 \tan ^2(c+d x)+10 a^3 b \tan (c+d x)+5 a^4+5 a b^3 \tan ^3(c+d x)+b^4 \tan ^4(c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(Tan[c + d*x]*(5*a^4 + 10*a^3*b*Tan[c + d*x] + 10*a^2*b^2*Tan[c + d*x]^2 + 5*a*b^3*Tan[c + d*x]^3 + b^4*Tan[c
+ d*x]^4))/(5*d)

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Maple [B]  time = 0.146, size = 96, normalized size = 3.2 \begin{align*}{\frac{1}{d} \left ({a}^{4}\tan \left ( dx+c \right ) +2\,{\frac{{a}^{3}b}{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{{a}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/d*(a^4*tan(d*x+c)+2*a^3*b/cos(d*x+c)^2+2*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^3+a*b^3*sin(d*x+c)^4/cos(d*x+c)^4+1
/5*b^4*sin(d*x+c)^5/cos(d*x+c)^5)

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Maxima [B]  time = 1.23157, size = 139, normalized size = 4.63 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{5} + 10 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 5 \, a^{4} \tan \left (d x + c\right ) + \frac{5 \,{\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \frac{10 \, a^{3} b}{\sin \left (d x + c\right )^{2} - 1}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/5*(b^4*tan(d*x + c)^5 + 10*a^2*b^2*tan(d*x + c)^3 + 5*a^4*tan(d*x + c) + 5*(2*sin(d*x + c)^2 - 1)*a*b^3/(sin
(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 10*a^3*b/(sin(d*x + c)^2 - 1))/d

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Fricas [B]  time = 0.484743, size = 250, normalized size = 8.33 \begin{align*} \frac{5 \, a b^{3} \cos \left (d x + c\right ) + 10 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left ({\left (5 \, a^{4} - 10 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + b^{4} + 2 \,{\left (5 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/5*(5*a*b^3*cos(d*x + c) + 10*(a^3*b - a*b^3)*cos(d*x + c)^3 + ((5*a^4 - 10*a^2*b^2 + b^4)*cos(d*x + c)^4 + b
^4 + 2*(5*a^2*b^2 - b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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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(sec(d*x+c)**6*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.2078, size = 99, normalized size = 3.3 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{5} + 5 \, a b^{3} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b \tan \left (d x + c\right )^{2} + 5 \, a^{4} \tan \left (d x + c\right )}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/5*(b^4*tan(d*x + c)^5 + 5*a*b^3*tan(d*x + c)^4 + 10*a^2*b^2*tan(d*x + c)^3 + 10*a^3*b*tan(d*x + c)^2 + 5*a^4
*tan(d*x + c))/d