3.248 \(\int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=111 \[ \frac{b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac{a \left (a^2-3 b^2\right ) \sec (c+d x)}{d}+\frac{3 a^2 b \log (\cos (c+d x))}{d}+\frac{a^3 \cos (c+d x)}{d}+\frac{a b^2 \sec ^3(c+d x)}{d}+\frac{b^3 \sec ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.254024, antiderivative size = 111, 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} \[ \frac{b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac{a \left (a^2-3 b^2\right ) \sec (c+d x)}{d}+\frac{3 a^2 b \log (\cos (c+d x))}{d}+\frac{a^3 \cos (c+d x)}{d}+\frac{a b^2 \sec ^3(c+d x)}{d}+\frac{b^3 \sec ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

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

Mathematica [A]  time = 1.97822, size = 97, normalized size = 0.87 \[ \frac{\left (6 a^2 b-2 b^3\right ) \sec ^2(c+d x)+4 a \left (a^2-3 b^2\right ) \sec (c+d x)+12 a^2 b \log (\cos (c+d x))+4 a^3 \cos (c+d x)+4 a b^2 \sec ^3(c+d x)+b^3 \sec ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

(4*a^3*Cos[c + d*x] + 12*a^2*b*Log[Cos[c + d*x]] + 4*a*(a^2 - 3*b^2)*Sec[c + d*x] + (6*a^2*b - 2*b^3)*Sec[c +
d*x]^2 + 4*a*b^2*Sec[c + d*x]^3 + b^3*Sec[c + d*x]^4)/(4*d)

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Maple [A]  time = 0.09, size = 204, normalized size = 1.8 \begin{align*}{\frac{{a}^{3} \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}^{3}}{d}}+2\,{\frac{{a}^{3}\cos \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{2}b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{{a}^{2}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{a{b}^{2} \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{b}^{2}}{d}}-2\,{\frac{a{b}^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.0863, size = 130, normalized size = 1.17 \begin{align*} \frac{b^{3} \tan \left (d x + c\right )^{4} - 6 \, a^{2} b{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} + 4 \, a^{3}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac{4 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{2}}{\cos \left (d x + c\right )^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.542339, size = 258, normalized size = 2.32 \begin{align*} \frac{4 \, a^{3} \cos \left (d x + c\right )^{5} + 12 \, a^{2} b \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 4 \, a b^{2} \cos \left (d x + c\right ) + 4 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} + 2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{4 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 105.274, size = 567, normalized size = 5.11 \begin{align*} -\frac{12 \, a^{2} b \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 12 \, a^{2} b \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{4 \,{\left (2 \, a^{3} + 3 \, a^{2} b - \frac{3 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1} - \frac{8 \, a^{3} - 25 \, a^{2} b - 16 \, a b^{2} + \frac{24 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{124 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{64 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{24 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{198 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{48 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{16 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{8 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{124 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{25 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{4}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/4*(12*a^2*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 12*a^2*b*log(abs(-(cos(d*x + c) - 1)/(co
s(d*x + c) + 1) - 1)) + 4*(2*a^3 + 3*a^2*b - 3*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - 1) - (8*a^3 - 25*a^2*b - 16*a*b^2 + 24*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 124*
a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 64*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 24*a^3*(cos(d*x
+ c) - 1)^2/(cos(d*x + c) + 1)^2 - 198*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 48*a*b^2*(cos(d*x +
c) - 1)^2/(cos(d*x + c) + 1)^2 + 16*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 8*a^3*(cos(d*x + c) - 1)^3
/(cos(d*x + c) + 1)^3 - 124*a^2*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 25*a^2*b*(cos(d*x + c) - 1)^4/(c
os(d*x + c) + 1)^4)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^4)/d