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

Optimal. Leaf size=115 \[ \frac{b \left (3 a^2-b^2\right ) \sec (c+d x)}{d}-\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+\frac{3 a^2 b \cos (c+d x)}{d}+\frac{a^3 \cos ^2(c+d x)}{2 d}+\frac{3 a b^2 \sec ^2(c+d x)}{2 d}+\frac{b^3 \sec ^3(c+d x)}{3 d} \]

[Out]

(3*a^2*b*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^2)/(2*d) - (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/d + (b*(3*a^2 - b^
2)*Sec[c + d*x])/d + (3*a*b^2*Sec[c + d*x]^2)/(2*d) + (b^3*Sec[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.24159, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4397, 2837, 12, 894} \[ \frac{b \left (3 a^2-b^2\right ) \sec (c+d x)}{d}-\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+\frac{3 a^2 b \cos (c+d x)}{d}+\frac{a^3 \cos ^2(c+d x)}{2 d}+\frac{3 a b^2 \sec ^2(c+d x)}{2 d}+\frac{b^3 \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*b*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^2)/(2*d) - (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/d + (b*(3*a^2 - b^
2)*Sec[c + d*x])/d + (3*a*b^2*Sec[c + d*x]^2)/(2*d) + (b^3*Sec[c + d*x]^3)/(3*d)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.54442, size = 100, normalized size = 0.87 \[ \frac{2 \left (-6 b \left (b^2-3 a^2\right ) \sec (c+d x)-6 a \left (a^2-3 b^2\right ) \log (\cos (c+d x))+9 a b^2 \sec ^2(c+d x)+2 b^3 \sec ^3(c+d x)\right )+36 a^2 b \cos (c+d x)+3 a^3 \cos (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a^2*b*Cos[c + d*x] + 3*a^3*Cos[2*(c + d*x)] + 2*(-6*a*(a^2 - 3*b^2)*Log[Cos[c + d*x]] - 6*b*(-3*a^2 + b^2)
*Sec[c + d*x] + 9*a*b^2*Sec[c + d*x]^2 + 2*b^3*Sec[c + d*x]^3))/(12*d)

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Maple [A]  time = 0.078, size = 213, normalized size = 1.9 \begin{align*} -{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}b}{d}}+6\,{\frac{{a}^{2}b\cos \left ( dx+c \right ) }{d}}+{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{a{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{{b}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{2\,{b}^{3}\cos \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11744, size = 147, normalized size = 1.28 \begin{align*} -\frac{3 \,{\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a^{3} + 9 \, a b^{2}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 18 \, a^{2} b{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + \frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(sin(d*x + c)^2 + log(sin(d*x + c)^2 - 1))*a^3 + 9*a*b^2*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2
- 1)) - 18*a^2*b*(1/cos(d*x + c) + cos(d*x + c)) + 2*(3*cos(d*x + c)^2 - 1)*b^3/cos(d*x + c)^3)/d

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Fricas [A]  time = 0.549714, size = 297, normalized size = 2.58 \begin{align*} \frac{6 \, a^{3} \cos \left (d x + c\right )^{5} + 36 \, a^{2} b \cos \left (d x + c\right )^{4} - 3 \, a^{3} \cos \left (d x + c\right )^{3} - 12 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) + 18 \, a b^{2} \cos \left (d x + c\right ) + 4 \, b^{3} + 12 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a^3*cos(d*x + c)^5 + 36*a^2*b*cos(d*x + c)^4 - 3*a^3*cos(d*x + c)^3 - 12*(a^3 - 3*a*b^2)*cos(d*x + c)^
3*log(-cos(d*x + c)) + 18*a*b^2*cos(d*x + c) + 4*b^3 + 12*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^3)

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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*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 100.792, size = 691, normalized size = 6.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(a^3 - 3*a*b^2)*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 6*(a^3 - 3*a*b^2)*log(abs(-(cos(
d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) - 3*(3*a^3 - 12*a^2*b - 9*a*b^2 - 10*a^3*(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 12*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 18*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*
a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 9*a*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((cos(d*x +
c) - 1)/(cos(d*x + c) + 1) - 1)^2 + (11*a^3 + 36*a^2*b - 33*a*b^2 - 8*b^3 + 33*a^3*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) + 72*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 135*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
- 24*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 33*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 36*a^2*b*
(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 135*a*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 11*a^3*(cos(
d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 33*a*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/((cos(d*x + c) - 1)
/(cos(d*x + c) + 1) + 1)^3)/d