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]

1/5*(b+a*cot(d*x+c))^5*tan(d*x+c)^5/b/d

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Rubi [A]  time = 0.05, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [B]  time = 0.32, size = 73, normalized size = 2.43 \[ \frac {\tan (c+d x) \left (5 a^4+10 a^3 b \tan (c+d x)+10 a^2 b^2 \tan ^2(c+d x)+5 a b^3 \tan ^3(c+d x)+b^4 \tan ^4(c+d x)\right )}{5 d} \]

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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fricas [B]  time = 0.53, size = 109, normalized size = 3.63 \[ \frac {5 \, a b^{3} \cos \left (d x + c\right ) + 10 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (5 \, a^{4} - 10 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + b^{4} + 2 \, {\left (5 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{5}} \]

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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giac [B]  time = 0.66, size = 73, normalized size = 2.43 \[ \frac {b^{4} \tan \left (d x + c\right )^{5} + 5 \, a b^{3} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b \tan \left (d x + c\right )^{2} + 5 \, a^{4} \tan \left (d x + c\right )}{5 \, d} \]

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

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maple [B]  time = 27.47, size = 96, normalized size = 3.20 \[ \frac {a^{4} \tan \left (d x +c \right )+\frac {2 a^{3} b}{\cos \left (d x +c \right )^{2}}+\frac {2 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{4}}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )^{5}}}{d} \]

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 = 0.33, size = 103, normalized size = 3.43 \[ \frac {b^{4} \tan \left (d x + c\right )^{5} + 10 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 5 \, a^{4} \tan \left (d x + c\right ) + \frac {5 \, {\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \frac {10 \, a^{3} b}{\sin \left (d x + c\right )^{2} - 1}}{5 \, d} \]

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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mupad [B]  time = 0.80, size = 139, normalized size = 4.63 \[ \frac {\frac {b^4\,\sin \left (c+d\,x\right )}{5}-{\cos \left (c+d\,x\right )}^3\,\left (2\,a\,b^3-2\,a^3\,b\right )-{\cos \left (c+d\,x\right )}^2\,\left (\frac {2\,b^4\,\sin \left (c+d\,x\right )}{5}-2\,a^2\,b^2\,\sin \left (c+d\,x\right )\right )+{\cos \left (c+d\,x\right )}^4\,\left (\sin \left (c+d\,x\right )\,a^4-2\,\sin \left (c+d\,x\right )\,a^2\,b^2+\frac {\sin \left (c+d\,x\right )\,b^4}{5}\right )+a\,b^3\,\cos \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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