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

Optimal. Leaf size=91 \[ \frac{\sin ^2(c+d x) \left (a \left (a^2-3 b^2\right ) \cot (c+d x)+b \left (3 a^2-b^2\right )\right )}{2 d}+\frac{1}{2} a x \left (a^2+3 b^2\right )-\frac{b^3 \log (\sin (c+d x))}{d}+\frac{b^3 \log (\tan (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.119388, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3088, 1805, 801, 635, 203, 260} \[ \frac{\sin ^2(c+d x) \left (a \left (a^2-3 b^2\right ) \cot (c+d x)+b \left (3 a^2-b^2\right )\right )}{2 d}+\frac{1}{2} a x \left (a^2+3 b^2\right )-\frac{b^3 \log (\sin (c+d x))}{d}+\frac{b^3 \log (\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 + 3*b^2)*x)/2 - (b^3*Log[Sin[c + d*x]])/d + (b^3*Log[Tan[c + d*x]])/d + ((b*(3*a^2 - b^2) + a*(a^2 - 3
*b^2)*Cot[c + d*x])*Sin[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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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

Mathematica [B]  time = 0.734223, size = 401, normalized size = 4.41 \[ \frac{a b \left (-2 a^2 b^2+a^4-3 b^4\right ) \sin (2 (c+d x))+\left (-2 a^2 b^4-3 a^4 b^2+b^6\right ) \cos (2 (c+d x))+2 a^2 b^4 \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+2 a^2 b^4 \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+4 a^3 \left (-b^2\right )^{3/2} \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )-a^5 \sqrt{-b^2} \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )-4 a^3 \left (-b^2\right )^{3/2} \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+a^5 \sqrt{-b^2} \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+2 a^2 b^4+5 a^4 b^2+3 a \sqrt{-b^2} b^4 \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )-3 a \left (-b^2\right )^{5/2} \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+2 b^6 \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+2 b^6 \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )-b^6}{4 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.109, size = 123, normalized size = 1.4 \begin{align*}{\frac{{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}x}{2}}+{\frac{{a}^{3}c}{2\,d}}-{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{3\,a{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,a{b}^{2}x}{2}}+{\frac{3\,a{b}^{2}c}{2\,d}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{2\,d}}-{\frac{{b}^{3}\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)*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.24556, size = 123, normalized size = 1.35 \begin{align*} \frac{6 \, a^{2} b \sin \left (d x + c\right )^{2} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 3 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} - 2 \,{\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} b^{3}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.515797, size = 181, normalized size = 1.99 \begin{align*} -\frac{2 \, b^{3} \log \left (-\cos \left (d x + c\right )\right ) -{\left (a^{3} + 3 \, a b^{2}\right )} d x +{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} -{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.20132, size = 126, normalized size = 1.38 \begin{align*} \frac{b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) +{\left (a^{3} + 3 \, a b^{2}\right )}{\left (d x + c\right )} - \frac{b^{3} \tan \left (d x + c\right )^{2} - a^{3} \tan \left (d x + c\right ) + 3 \, a b^{2} \tan \left (d x + c\right ) + 3 \, a^{2} b}{\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)*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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