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

Optimal. Leaf size=98 \[ -\frac{3 a^3 \sin (c+d x)}{d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac{5 a^3 x}{2} \]

[Out]

(-5*a^3*x)/2 + (5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^3*Sin[c + d*x])/d - (a^3*Cos[c + d*x]*Sin[c + d*x])/
(2*d) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.183177, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2872, 2637, 2635, 8, 3770, 3767, 3768} \[ -\frac{3 a^3 \sin (c+d x)}{d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac{5 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^3*x)/2 + (5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^3*Sin[c + d*x])/d - (a^3*Cos[c + d*x]*Sin[c + d*x])/
(2*d) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rubi steps

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

Mathematica [B]  time = 2.49477, size = 300, normalized size = 3.06 \[ \frac{1}{32} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{12 \sin (c) \cos (d x)}{d}-\frac{\sin (2 c) \cos (2 d x)}{d}-\frac{12 \cos (c) \sin (d x)}{d}-\frac{\cos (2 c) \sin (2 d x)}{d}+\frac{12 \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{12 \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{10 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{10 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}-10 x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(-10*x - (10*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (10*Lo
g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d - (12*Cos[d*x]*Sin[c])/d - (Cos[2*d*x]*Sin[2*c])/d - (12*Cos[c]*Sin[
d*x])/d - (Cos[2*c]*Sin[2*d*x])/d + 1/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (12*Sin[(d*x)/2])/(d*(Cos[
c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 1/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (12*
Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/32

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Maple [A]  time = 0.038, size = 111, normalized size = 1.1 \begin{align*} -{\frac{{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{a}^{3}x}{2}}-{\frac{5\,{a}^{3}c}{2\,d}}+{\frac{5\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{5\,{a}^{3}\sin \left ( dx+c \right ) }{2\,d}}+3\,{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5123, size = 171, normalized size = 1.74 \begin{align*} \frac{{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - a^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a^3 - 12*(d*x + c - tan(d*x + c))*a^3 - a^3*(2*sin(d*x + c)/(sin(d*x + c
)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) -
1) - 2*sin(d*x + c)))/d

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Fricas [A]  time = 1.83919, size = 313, normalized size = 3.19 \begin{align*} -\frac{10 \, a^{3} d x \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 5 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{3} \cos \left (d x + c\right )^{2} - 6 \, a^{3} \cos \left (d x + c\right ) - a^{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((a+a*sec(d*x+c))^3*sin(d*x+c)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.28844, size = 138, normalized size = 1.41 \begin{align*} -\frac{5 \,{\left (d x + c\right )} a^{3} - 5 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 5 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{4 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(5*(d*x + c)*a^3 - 5*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 5*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) +
4*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 9*a^3*tan(1/2*d*x + 1/2*c)^3)/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d