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

Optimal. Leaf size=119 \[ \frac{b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 d}+\frac{a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 d}-\frac{3 a^2 b \sec (c+d x)}{d}+\frac{a^3 \log (\cos (c+d x))}{d}+\frac{3 a b^2 \sec ^4(c+d x)}{4 d}+\frac{b^3 \sec ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.219906, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4397, 2837, 12, 894} \[ \frac{b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 d}+\frac{a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 d}-\frac{3 a^2 b \sec (c+d x)}{d}+\frac{a^3 \log (\cos (c+d x))}{d}+\frac{3 a b^2 \sec ^4(c+d x)}{4 d}+\frac{b^3 \sec ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.3156, size = 99, normalized size = 0.83 \[ \frac{-20 b \left (b^2-3 a^2\right ) \sec ^3(c+d x)+30 a \left (a^2-3 b^2\right ) \sec ^2(c+d x)-180 a^2 b \sec (c+d x)+60 a^3 \log (\cos (c+d x))+45 a b^2 \sec ^4(c+d x)+12 b^3 \sec ^5(c+d x)}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a^3*Log[Cos[c + d*x]] - 180*a^2*b*Sec[c + d*x] + 30*a*(a^2 - 3*b^2)*Sec[c + d*x]^2 - 20*b*(-3*a^2 + b^2)*S
ec[c + d*x]^3 + 45*a*b^2*Sec[c + d*x]^4 + 12*b^3*Sec[c + d*x]^5)/(60*d)

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Maple [B]  time = 0.096, size = 252, normalized size = 2.1 \begin{align*}{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}b}{d}}-2\,{\frac{{a}^{2}b\cos \left ( dx+c \right ) }{d}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,d\cos \left ( dx+c \right ) }}-{\frac{{b}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{2\,{b}^{3}\cos \left ( dx+c \right ) }{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12428, size = 173, normalized size = 1.45 \begin{align*} -\frac{30 \, a^{3}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - \frac{45 \,{\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{2}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + \frac{60 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2} b}{\cos \left (d x + c\right )^{3}} + \frac{4 \,{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(30*a^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1)) - 45*(2*sin(d*x + c)^2 - 1)*a*b^2/(sin(d*x +
c)^4 - 2*sin(d*x + c)^2 + 1) + 60*(3*cos(d*x + c)^2 - 1)*a^2*b/cos(d*x + c)^3 + 4*(5*cos(d*x + c)^2 - 3)*b^3/c
os(d*x + c)^5)/d

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Fricas [A]  time = 0.535732, size = 270, normalized size = 2.27 \begin{align*} \frac{60 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 180 \, a^{2} b \cos \left (d x + c\right )^{4} + 45 \, a b^{2} \cos \left (d x + c\right ) + 30 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 12 \, b^{3} + 20 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{60 \, 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)^3*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(60*a^3*cos(d*x + c)^5*log(-cos(d*x + c)) - 180*a^2*b*cos(d*x + c)^4 + 45*a*b^2*cos(d*x + c) + 30*(a^3 -
3*a*b^2)*cos(d*x + c)^3 + 12*b^3 + 20*(3*a^2*b - b^3)*cos(d*x + c)^2)/(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)**3*(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 104.493, size = 579, normalized size = 4.87 \begin{align*} -\frac{60 \, a^{3} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{3} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{137 \, a^{3} + 240 \, a^{2} b + 16 \, b^{3} + \frac{805 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{1200 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{80 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{1730 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{1680 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{720 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{80 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{1730 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{720 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{720 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{240 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{805 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{137 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d
*x + c) + 1) - 1)) + (137*a^3 + 240*a^2*b + 16*b^3 + 805*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1200*a^2*
b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 80*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1730*a^3*(cos(d*x + c
) - 1)^2/(cos(d*x + c) + 1)^2 + 1680*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 720*a*b^2*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 - 80*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1730*a^3*(cos(d*x + c) - 1)^
3/(cos(d*x + c) + 1)^3 + 720*a^2*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 720*a*b^2*(cos(d*x + c) - 1)^3/
(cos(d*x + c) + 1)^3 + 240*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*a^3*(cos(d*x + c) - 1)^4/(cos(d
*x + c) + 1)^4 + 137*a^3*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1
)^5)/d