3.2.81 \(\int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx\) [181]

Optimal. Leaf size=77 \[ \frac {a^2 x}{2}-b^2 x+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3957, 2801, 2715, 8, 2672, 327, 212, 3554} \begin {gather*} -\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a^2 x}{2}-\frac {2 a b \sin (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}-b^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2715

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] && Integ
erQ[2*n]

Rule 2801

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx &=\int (-b-a \cos (c+d x))^2 \tan ^2(c+d x) \, dx\\ &=\int \left (a^2 \sin ^2(c+d x)+2 a b \sin (c+d x) \tan (c+d x)+b^2 \tan ^2(c+d x)\right ) \, dx\\ &=a^2 \int \sin ^2(c+d x) \, dx+(2 a b) \int \sin (c+d x) \tan (c+d x) \, dx+b^2 \int \tan ^2(c+d x) \, dx\\ &=-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {1}{2} a^2 \int 1 \, dx-b^2 \int 1 \, dx+\frac {(2 a b) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {a^2 x}{2}-b^2 x-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {(2 a b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {a^2 x}{2}-b^2 x+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.40, size = 121, normalized size = 1.57 \begin {gather*} -\frac {-2 a^2 c+4 b^2 c-2 a^2 d x+4 b^2 d x+8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \sin (c+d x)+a^2 \sin (2 (c+d x))-4 b^2 \tan (c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(-2*a^2*c + 4*b^2*c - 2*a^2*d*x + 4*b^2*d*x + 8*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 8*a*b*Log[
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*a*b*Sin[c + d*x] + a^2*Sin[2*(c + d*x)] - 4*b^2*Tan[c + d*x])/d

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 77, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 b a \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(77\)
default	\(\frac {b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 b a \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(77\)
risch	\(\frac {a^{2} x}{2}-b^{2} x +\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i b a \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i b a \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(146\)
norman	\(\frac {\left (-\frac {a^{2}}{2}+b^{2}\right ) x +\left (-\frac {a^{2}}{2}+b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {a^{2}}{2}-b^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {a^{2}}{2}-b^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (a^{2}-4 b a -2 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a^{2}+4 b a -2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2*sin(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 80, normalized size = 1.04 \begin {gather*} \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 )} b^{2} + 4 \, a b {\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [A]
time = 2.89, size = 108, normalized size = 1.40 \begin {gather*} \frac {{\left (a^{2} - 2 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, a b \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 b \cos \left (d x + c\right ) - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (73) = 146\).
time = 0.52, size = 159, normalized size = 2.06 \begin {gather*} \frac {4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, b^{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 (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 1.17, size = 143, normalized size = 1.86 \begin {gather*} \frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (2*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(b^2*sin(c + d*x))/(d*cos(c + d*x)) - (2*a*b*sin(c + d*x))/d + (4*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)
/2)))/d - (a^2*cos(c + d*x)*sin(c + d*x))/(2*d)

________________________________________________________________________________________