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

Optimal. Leaf size=169 \[ -\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\sin (c+d x))}{d}+\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac{\sin ^2(c+d x) \left (a \left (-10 a^2 b^2+a^4+5 b^4\right ) \cot (c+d x)+b \left (-10 a^2 b^2+5 a^4+b^4\right )\right )}{2 d}+\frac{1}{2} a x \left (10 a^2 b^2+a^4-15 b^4\right )+\frac{5 a b^4 \tan (c+d x)}{d}+\frac{b^5 \tan ^2(c+d x)}{2 d} \]

[Out]

(a*(a^4 + 10*a^2*b^2 - 15*b^4)*x)/2 - (2*b^3*(5*a^2 - b^2)*Log[Sin[c + d*x]])/d + (2*b^3*(5*a^2 - b^2)*Log[Tan
[c + d*x]])/d + ((b*(5*a^4 - 10*a^2*b^2 + b^4) + a*(a^4 - 10*a^2*b^2 + 5*b^4)*Cot[c + d*x])*Sin[c + d*x]^2)/(2
*d) + (5*a*b^4*Tan[c + d*x])/d + (b^5*Tan[c + d*x]^2)/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.231263, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3088, 1805, 1802, 635, 203, 260} \[ -\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\sin (c+d x))}{d}+\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac{\sin ^2(c+d x) \left (a \left (-10 a^2 b^2+a^4+5 b^4\right ) \cot (c+d x)+b \left (-10 a^2 b^2+5 a^4+b^4\right )\right )}{2 d}+\frac{1}{2} a x \left (10 a^2 b^2+a^4-15 b^4\right )+\frac{5 a b^4 \tan (c+d x)}{d}+\frac{b^5 \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^4 + 10*a^2*b^2 - 15*b^4)*x)/2 - (2*b^3*(5*a^2 - b^2)*Log[Sin[c + d*x]])/d + (2*b^3*(5*a^2 - b^2)*Log[Tan
[c + d*x]])/d + ((b*(5*a^4 - 10*a^2*b^2 + b^4) + a*(a^4 - 10*a^2*b^2 + 5*b^4)*Cot[c + d*x])*Sin[c + d*x]^2)/(2
*d) + (5*a*b^4*Tan[c + d*x])/d + (b^5*Tan[c + d*x]^2)/(2*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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^5}{x^3 \left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{-2 b^5-10 a b^4 x-2 b^3 \left (10 a^2-b^2\right ) x^2-a \left (a^4+10 a^2 b^2-5 b^4\right ) x^3}{x^3 \left (1+x^2\right )} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac{\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 d}+\frac{\operatorname{Subst}\left (\int \left (-\frac{2 b^5}{x^3}-\frac{10 a b^4}{x^2}-\frac{4 \left (5 a^2 b^3-b^5\right )}{x}+\frac{-a \left (a^4+10 a^2 b^2-15 b^4\right )+4 b^3 \left (5 a^2-b^2\right ) x}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac{\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 d}+\frac{5 a b^4 \tan (c+d x)}{d}+\frac{b^5 \tan ^2(c+d x)}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{-a \left (a^4+10 a^2 b^2-15 b^4\right )+4 b^3 \left (5 a^2-b^2\right ) x}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac{\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 d}+\frac{5 a b^4 \tan (c+d x)}{d}+\frac{b^5 \tan ^2(c+d x)}{2 d}+\frac{\left (2 b^3 \left (5 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}-\frac{\left (a \left (a^4+10 a^2 b^2-15 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac{1}{2} a \left (a^4+10 a^2 b^2-15 b^4\right ) x-\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\sin (c+d x))}{d}+\frac{2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac{\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 d}+\frac{5 a b^4 \tan (c+d x)}{d}+\frac{b^5 \tan ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 6.37759, size = 571, normalized size = 3.38 \[ \frac{b^3 \left (\frac{\cos ^2(c+d x) (a+b \tan (c+d x))^6 \left (a b \tan (c+d x)+b^2\right )}{2 b^4 \left (a^2+b^2\right )}-\frac{\left (4 b^2-6 a^2\right ) \left (\frac{1}{2} b^2 \left (10 a^2-b^2\right ) \tan ^2(c+d x)+5 a b \left (2 a^2-b^2\right ) \tan (c+d x)+\frac{1}{2} \left (-10 a^2 b^2+\frac{-10 a^3 b^2+a^5+5 a b^4}{\sqrt{-b^2}}+5 a^4+b^4\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+\frac{1}{2} \left (-10 a^2 b^2-\frac{-10 a^3 b^2+a^5+5 a b^4}{\sqrt{-b^2}}+5 a^4+b^4\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+\frac{5}{3} a b^3 \tan ^3(c+d x)+\frac{1}{4} b^4 \tan ^4(c+d x)\right )+5 a \left (\frac{1}{3} b^3 \left (15 a^2-b^2\right ) \tan ^3(c+d x)+a b^2 \left (10 a^2-3 b^2\right ) \tan ^2(c+d x)+b \left (-15 a^2 b^2+15 a^4+b^4\right ) \tan (c+d x)+\frac{1}{2} \left (-20 a^3 b^2+\frac{-15 a^4 b^2+15 a^2 b^4+a^6-b^6}{\sqrt{-b^2}}+6 a^5+6 a b^4\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+\frac{1}{2} \left (-20 a^3 b^2-\frac{-15 a^4 b^2+15 a^2 b^4+a^6-b^6}{\sqrt{-b^2}}+6 a^5+6 a b^4\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+\frac{3}{2} a b^4 \tan ^4(c+d x)+\frac{1}{5} b^5 \tan ^5(c+d x)\right )}{2 b^2 \left (a^2+b^2\right )}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b^3*((Cos[c + d*x]^2*(a + b*Tan[c + d*x])^6*(b^2 + a*b*Tan[c + d*x]))/(2*b^4*(a^2 + b^2)) - ((-6*a^2 + 4*b^2)
*(((5*a^4 - 10*a^2*b^2 + b^4 + (a^5 - 10*a^3*b^2 + 5*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/2 +
((5*a^4 - 10*a^2*b^2 + b^4 - (a^5 - 10*a^3*b^2 + 5*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/2 + 5*
a*b*(2*a^2 - b^2)*Tan[c + d*x] + (b^2*(10*a^2 - b^2)*Tan[c + d*x]^2)/2 + (5*a*b^3*Tan[c + d*x]^3)/3 + (b^4*Tan
[c + d*x]^4)/4) + 5*a*(((6*a^5 - 20*a^3*b^2 + 6*a*b^4 + (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)/Sqrt[-b^2])*Log[
Sqrt[-b^2] - b*Tan[c + d*x]])/2 + ((6*a^5 - 20*a^3*b^2 + 6*a*b^4 - (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)/Sqrt[
-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/2 + b*(15*a^4 - 15*a^2*b^2 + b^4)*Tan[c + d*x] + a*b^2*(10*a^2 - 3*b^
2)*Tan[c + d*x]^2 + (b^3*(15*a^2 - b^2)*Tan[c + d*x]^3)/3 + (3*a*b^4*Tan[c + d*x]^4)/2 + (b^5*Tan[c + d*x]^5)/
5))/(2*b^2*(a^2 + b^2))))/d

________________________________________________________________________________________

Maple [A]  time = 0.253, size = 291, normalized size = 1.7 \begin{align*}{\frac{{a}^{5}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{5}x}{2}}+{\frac{{a}^{5}c}{2\,d}}-{\frac{5\,{a}^{4}b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-5\,{\frac{{a}^{3}{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+5\,{a}^{3}{b}^{2}x+5\,{\frac{{a}^{3}{b}^{2}c}{d}}-5\,{\frac{{a}^{2}{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-10\,{\frac{{a}^{2}{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+5\,{\frac{a{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d\cos \left ( dx+c \right ) }}+5\,{\frac{a{b}^{4}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{15\,a{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{15\,a{b}^{4}x}{2}}-{\frac{15\,a{b}^{4}c}{2\,d}}+{\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{5}}{d}}+2\,{\frac{{b}^{5}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^5*cos(d*x+c)*sin(d*x+c)/d+1/2*a^5*x+1/2/d*a^5*c-5/2/d*a^4*b*cos(d*x+c)^2-5*a^3*b^2*cos(d*x+c)*sin(d*x+c)
/d+5*a^3*b^2*x+5/d*a^3*b^2*c-5/d*a^2*b^3*sin(d*x+c)^2-10/d*a^2*b^3*ln(cos(d*x+c))+5/d*a*b^4*sin(d*x+c)^5/cos(d
*x+c)+5/d*a*b^4*cos(d*x+c)*sin(d*x+c)^3+15/2*a*b^4*cos(d*x+c)*sin(d*x+c)/d-15/2*a*b^4*x-15/2/d*a*b^4*c+1/2/d*b
^5*sin(d*x+c)^6/cos(d*x+c)^2+1/2/d*b^5*sin(d*x+c)^4+1/d*sin(d*x+c)^2*b^5+2/d*b^5*ln(cos(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.78528, size = 242, normalized size = 1.43 \begin{align*} \frac{10 \, a^{4} b \sin \left (d x + c\right )^{2} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{5} + 10 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b^{2} - 20 \,{\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a^{2} b^{3} - 10 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{4} + 2 \,{\left (\sin \left (d x + c\right )^{2} - \frac{1}{\sin \left (d x + c\right )^{2} - 1} + 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} b^{5}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(10*a^4*b*sin(d*x + c)^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*a^5 + 10*(2*d*x + 2*c - sin(2*d*x + 2*c))*a^3*
b^2 - 20*(sin(d*x + c)^2 + log(sin(d*x + c)^2 - 1))*a^2*b^3 - 10*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 +
1) - 2*tan(d*x + c))*a*b^4 + 2*(sin(d*x + c)^2 - 1/(sin(d*x + c)^2 - 1) + 2*log(sin(d*x + c)^2 - 1))*b^5)/d

________________________________________________________________________________________

Fricas [A]  time = 0.536343, size = 414, normalized size = 2.45 \begin{align*} \frac{2 \, b^{5} - 2 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 8 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) +{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5} + 2 \,{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (10 \, a b^{4} \cos \left (d x + c\right ) +{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b^5 - 2*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 - 8*(5*a^2*b^3 - b^5)*cos(d*x + c)^2*log(-cos(d*x +
c)) + (5*a^4*b - 10*a^2*b^3 + b^5 + 2*(a^5 + 10*a^3*b^2 - 15*a*b^4)*d*x)*cos(d*x + c)^2 + 2*(10*a*b^4*cos(d*x
+ c) + (a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.27347, size = 234, normalized size = 1.38 \begin{align*} \frac{b^{5} \tan \left (d x + c\right )^{2} + 10 \, a b^{4} \tan \left (d x + c\right ) +{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )}{\left (d x + c\right )} + 2 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac{10 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 2 \, b^{5} \tan \left (d x + c\right )^{2} - a^{5} \tan \left (d x + c\right ) + 10 \, a^{3} b^{2} \tan \left (d x + c\right ) - 5 \, a b^{4} \tan \left (d x + c\right ) + 5 \, a^{4} b - b^{5}}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^5*tan(d*x + c)^2 + 10*a*b^4*tan(d*x + c) + (a^5 + 10*a^3*b^2 - 15*a*b^4)*(d*x + c) + 2*(5*a^2*b^3 - b^5
)*log(tan(d*x + c)^2 + 1) - (10*a^2*b^3*tan(d*x + c)^2 - 2*b^5*tan(d*x + c)^2 - a^5*tan(d*x + c) + 10*a^3*b^2*
tan(d*x + c) - 5*a*b^4*tan(d*x + c) + 5*a^4*b - b^5)/(tan(d*x + c)^2 + 1))/d