3.54 \(\int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=85 \[ \frac{\left (a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \tan ^4(c+d x)}{2 d}+\frac{a b \tan ^2(c+d x)}{d}+\frac{b^2 \tan ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0766075, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3088, 894} \[ \frac{\left (a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \tan ^4(c+d x)}{2 d}+\frac{a b \tan ^2(c+d x)}{d}+\frac{b^2 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(a^2*Tan[c + d*x])/d + (a*b*Tan[c + d*x]^2)/d + ((a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + (a*b*Tan[c + d*x]^4)/(2*d
) + (b^2*Tan[c + d*x]^5)/(5*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 894

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

Rubi steps

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

Mathematica [A]  time = 0.184587, size = 54, normalized size = 0.64 \[ \frac{(a+b \tan (c+d x))^3 \left (a^2-3 a b \tan (c+d x)+6 b^2 \tan ^2(c+d x)+10 b^2\right )}{30 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

((a + b*Tan[c + d*x])^3*(a^2 + 10*b^2 - 3*a*b*Tan[c + d*x] + 6*b^2*Tan[c + d*x]^2))/(30*b^3*d)

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Maple [A]  time = 0.111, size = 82, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) +{\frac{ab}{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) \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))^2,x)

[Out]

1/d*(-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+1/2*a*b/cos(d*x+c)^4+b^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(
d*x+c)^3/cos(d*x+c)^3))

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Maxima [A]  time = 1.14459, size = 95, normalized size = 1.12 \begin{align*} \frac{10 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} b^{2} + \frac{15 \, a b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{30 \, 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))^2,x, algorithm="maxima")

[Out]

1/30*(10*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 + 2*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*b^2 + 15*a*b/(sin(d*x
+ c)^2 - 1)^2)/d

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Fricas [A]  time = 0.474388, size = 184, normalized size = 2.16 \begin{align*} \frac{15 \, a b \cos \left (d x + c\right ) + 2 \,{\left (2 \,{\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} +{\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, 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))^2,x, algorithm="fricas")

[Out]

1/30*(15*a*b*cos(d*x + c) + 2*(2*(5*a^2 - b^2)*cos(d*x + c)^4 + (5*a^2 - b^2)*cos(d*x + c)^2 + 3*b^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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15193, size = 108, normalized size = 1.27 \begin{align*} \frac{6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{3} + 10 \, b^{2} \tan \left (d x + c\right )^{3} + 30 \, a b \tan \left (d x + c\right )^{2} + 30 \, a^{2} \tan \left (d x + c\right )}{30 \, 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))^2,x, algorithm="giac")

[Out]

1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 + 10*a^2*tan(d*x + c)^3 + 10*b^2*tan(d*x + c)^3 + 30*a*b*ta
n(d*x + c)^2 + 30*a^2*tan(d*x + c))/d