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

Optimal. Leaf size=73 \[ -\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^2 x}{2} \]

[Out]

-1/2*a^2*x+2*a^2*arctanh(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d-1/2*a^2*cos(d*x+c)*sin(d*x+c)/d+a^2*tan(d*x+c)/d

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Rubi [A]  time = 0.13, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3872, 2709, 2637, 2635, 8, 3770, 3767} \[ -\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2637

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

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

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 3770

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

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]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \tan ^2(c+d x) \, dx\\ &=\frac {\int \left (-2 a^4 \cos (c+d x)-a^4 \cos ^2(c+d x)+2 a^4 \sec (c+d x)+a^4 \sec ^2(c+d x)\right ) \, dx}{a^2}\\ &=-\left (a^2 \int \cos ^2(c+d x) \, dx\right )+a^2 \int \sec ^2(c+d x) \, dx-\left (2 a^2\right ) \int \cos (c+d x) \, dx+\left (2 a^2\right ) \int \sec (c+d x) \, dx\\ &=\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} a^2 \int 1 \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=-\frac {a^2 x}{2}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 1.16, size = 243, normalized size = 3.33 \[ \frac {1}{16} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {8 \sin (c) \cos (d x)}{d}-\frac {\sin (2 c) \cos (2 d x)}{d}-\frac {8 \cos (c) \sin (d x)}{d}-\frac {\cos (2 c) \sin (2 d x)}{d}+\frac {4 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-2 x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-2*x - (8*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (8*Log[C
os[(c + d*x)/2] + Sin[(c + d*x)/2]])/d - (8*Cos[d*x]*Sin[c])/d - (Cos[2*d*x]*Sin[2*c])/d - (8*Cos[c]*Sin[d*x])
/d - (Cos[2*c]*Sin[2*d*x])/d + (4*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
) + (4*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/16

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fricas [A]  time = 1.61, size = 104, normalized size = 1.42 \[ -\frac {a^{2} d x \cos \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 128, normalized size = 1.75 \[ -\frac {{\left (d x + c\right )} a^{2} - 4 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 4 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.40, size = 86, normalized size = 1.18 \[ -\frac {a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {a^{2} x}{2}-\frac {a^{2} c}{2 d}-\frac {2 a^{2} \sin \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 81, normalized size = 1.11 \[ \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} + 4 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a^2 - 4*(d*x + c - tan(d*x + c))*a^2 + 4*a^2*(log(sin(d*x + c) + 1) - lo
g(sin(d*x + c) - 1) - 2*sin(d*x + c)))/d

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mupad [B]  time = 1.16, size = 117, normalized size = 1.60 \[ \frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,x}{2}+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2*(a + a/cos(c + d*x))^2,x)

[Out]

(4*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (a^2*x)/2 + (6*a^2*tan(c/2 + (d*x)/2)^3 + 5*a^2*tan(c/2 + (d*x)/2)^5 - 3
*a^2*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^6 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(2*sin(c + d*x)**2*sec(c + d*x), x) + Integral(sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(si
n(c + d*x)**2, x))

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