3.1456 \(\int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=104 \[ \frac{\left (a^2+b^2\right ) \tan (c+d x)}{d}-\frac{\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{3 a b \sec (c+d x)}{d}-\frac{3 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a b \csc ^2(c+d x) \sec (c+d x)}{d} \]

[Out]

(-3*a*b*ArcTanh[Cos[c + d*x]])/d - ((2*a^2 + b^2)*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) + (3*a*b*Sec[c
+ d*x])/d - (a*b*Csc[c + d*x]^2*Sec[c + d*x])/d + ((a^2 + b^2)*Tan[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.23094, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2911, 2622, 288, 321, 207, 3200, 448} \[ \frac{\left (a^2+b^2\right ) \tan (c+d x)}{d}-\frac{\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{3 a b \sec (c+d x)}{d}-\frac{3 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a b \csc ^2(c+d x) \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

(-3*a*b*ArcTanh[Cos[c + d*x]])/d - ((2*a^2 + b^2)*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) + (3*a*b*Sec[c
+ d*x])/d - (a*b*Csc[c + d*x]^2*Sec[c + d*x])/d + ((a^2 + b^2)*Tan[c + d*x])/d

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

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 207

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

Rule 3200

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.),
x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff^(n + 1)/f, Subst[Int[(x^n*(a + (a + b)*ff^2*x^2
)^p)/(1 + ff^2*x^2)^((m + n)/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/
2] && IntegerQ[n/2] && IntegerQ[p]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \csc ^4(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx+\int \csc ^4(c+d x) \sec ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right ) \left (a^2+\left (a^2+b^2\right ) x^2\right )}{x^4} \, dx,x,\tan (c+d x)\right )}{d}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \left (a^2 \left (1+\frac{b^2}{a^2}\right )+\frac{a^2}{x^4}+\frac{2 a^2+b^2}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{(3 a b) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{3 a b \sec (c+d x)}{d}-\frac{a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac{\left (a^2+b^2\right ) \tan (c+d x)}{d}+\frac{(3 a b) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{3 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac{\left (2 a^2+b^2\right ) \cot (c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{3 a b \sec (c+d x)}{d}-\frac{a b \csc ^2(c+d x) \sec (c+d x)}{d}+\frac{\left (a^2+b^2\right ) \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.992715, size = 196, normalized size = 1.88 \[ \frac{\csc ^5\left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (-4 \left (4 a^2+3 b^2\right ) \cos (2 (c+d x))+\left (8 a^2+6 b^2\right ) \cos (4 (c+d x))+3 b \left (10 a \sin (c+d x)-6 a \sin (3 (c+d x))-3 a \sin (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 a \sin (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+3 a \sin (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 b\right )\right )}{192 d \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

(Csc[(c + d*x)/2]^5*Sec[(c + d*x)/2]^3*(-4*(4*a^2 + 3*b^2)*Cos[2*(c + d*x)] + (8*a^2 + 6*b^2)*Cos[4*(c + d*x)]
+ 3*b*(2*b + 10*a*Sin[c + d*x] - 6*a*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[2*(c + d*x)] - 6*a*S
in[3*(c + d*x)] + 3*a*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] - 3*a*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)])))/(
192*d*(-1 + Cot[(c + d*x)/2]^2))

________________________________________________________________________________________

Maple [A]  time = 0.082, size = 162, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{4\,{a}^{2}}{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{8\,{a}^{2}\cot \left ( dx+c \right ) }{3\,d}}-{\frac{ab}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+3\,{\frac{ab}{d\cos \left ( dx+c \right ) }}+3\,{\frac{ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-2\,{\frac{{b}^{2}\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x)

[Out]

-1/3/d*a^2/sin(d*x+c)^3/cos(d*x+c)+4/3/d*a^2/sin(d*x+c)/cos(d*x+c)-8/3*a^2*cot(d*x+c)/d-1/d*a*b/sin(d*x+c)^2/c
os(d*x+c)+3/d*a*b/cos(d*x+c)+3/d*a*b*ln(csc(d*x+c)-cot(d*x+c))+1/d*b^2/sin(d*x+c)/cos(d*x+c)-2*b^2*cot(d*x+c)/
d

________________________________________________________________________________________

Maxima [A]  time = 1.01463, size = 166, normalized size = 1.6 \begin{align*} \frac{3 \, a b{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, b^{2}{\left (\frac{1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} - 2 \, a^{2}{\left (\frac{6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.75016, size = 489, normalized size = 4.7 \begin{align*} -\frac{4 \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 6 \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 9 \,{\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 9 \,{\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 6 \, a^{2} + 6 \, b^{2} - 6 \,{\left (3 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{6 \,{\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(4*(4*a^2 + 3*b^2)*cos(d*x + c)^4 - 6*(4*a^2 + 3*b^2)*cos(d*x + c)^2 + 9*(a*b*cos(d*x + c)^3 - a*b*cos(d*
x + c))*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*(a*b*cos(d*x + c)^3 - a*b*cos(d*x + c))*log(-1/2*cos(d*x
+ c) + 1/2)*sin(d*x + c) + 6*a^2 + 6*b^2 - 6*(3*a*b*cos(d*x + c)^2 - 2*a*b)*sin(d*x + c))/((d*cos(d*x + c)^3 -
d*cos(d*x + c))*sin(d*x + c))

________________________________________________________________________________________

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(csc(d*x+c)**4*sec(d*x+c)**2*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.23387, size = 275, normalized size = 2.64 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 72 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 21 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{48 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a b\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \frac{132 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a*b*tan(1/2*d*x + 1/2*c)^2 + 72*a*b*log(abs(tan(1/2*d*x + 1/2*c))) + 21*a
^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c) - 48*(a^2*tan(1/2*d*x + 1/2*c) + b^2*tan(1/2*d*x + 1/2*c
) + 2*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (132*a*b*tan(1/2*d*x + 1/2*c)^3 + 21*a^2*tan(1/2*d*x + 1/2*c)^2 + 12
*b^2*tan(1/2*d*x + 1/2*c)^2 + 6*a*b*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^3)/d