3.67 \(\int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=30 \[ \frac {\tan ^4(c+d x) (a \cot (c+d x)+b)^4}{4 b d} \]

[Out]

1/4*(b+a*cot(d*x+c))^4*tan(d*x+c)^4/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} \[ \frac {\tan ^4(c+d x) (a \cot (c+d x)+b)^4}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

((b + a*Cot[c + d*x])^4*Tan[c + d*x]^4)/(4*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 ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^3}{x^5} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {(b+a \cot (c+d x))^4 \tan ^4(c+d x)}{4 b d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 57, normalized size = 1.90 \[ \frac {\tan (c+d x) \left (4 a^3+6 a^2 b \tan (c+d x)+4 a b^2 \tan ^2(c+d x)+b^3 \tan ^3(c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

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fricas [B]  time = 0.76, size = 78, normalized size = 2.60 \[ \frac {b^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (a b^{2} \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.60, size = 57, normalized size = 1.90 \[ \frac {b^{3} \tan \left (d x + c\right )^{4} + 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b \tan \left (d x + c\right )^{2} + 4 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/4*(b^3*tan(d*x + c)^4 + 4*a*b^2*tan(d*x + c)^3 + 6*a^2*b*tan(d*x + c)^2 + 4*a^3*tan(d*x + c))/d

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maple [B]  time = 11.42, size = 72, normalized size = 2.40 \[ \frac {a^{3} \tan \left (d x +c \right )+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+\frac {b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

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

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maxima [B]  time = 0.34, size = 87, normalized size = 2.90 \[ \frac {4 \, a b^{2} \tan \left (d x + c\right )^{3} + 4 \, a^{3} \tan \left (d x + c\right ) + \frac {{\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} b^{3}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \frac {6 \, a^{2} b}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*(4*a*b^2*tan(d*x + c)^3 + 4*a^3*tan(d*x + c) + (2*sin(d*x + c)^2 - 1)*b^3/(sin(d*x + c)^4 - 2*sin(d*x + c)
^2 + 1) - 6*a^2*b/(sin(d*x + c)^2 - 1))/d

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mupad [B]  time = 0.64, size = 88, normalized size = 2.93 \[ \frac {{\cos \left (c+d\,x\right )}^3\,\left (a^3\,\sin \left (c+d\,x\right )-a\,b^2\,\sin \left (c+d\,x\right )\right )+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )+\frac {b^3}{4}+a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^5,x)

[Out]

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

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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)**5*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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