3.1490 \(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=135 \[ \frac{a \tan ^4(c+d x)}{4 d}+\frac{3 a \tan ^2(c+d x)}{2 d}-\frac{a \cot ^2(c+d x)}{2 d}+\frac{3 a \log (\tan (c+d x))}{d}-\frac{15 b \csc (c+d x)}{8 d}+\frac{15 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{5 b \csc (c+d x) \sec ^2(c+d x)}{8 d} \]

[Out]

(15*b*ArcTanh[Sin[c + d*x]])/(8*d) - (a*Cot[c + d*x]^2)/(2*d) - (15*b*Csc[c + d*x])/(8*d) + (3*a*Log[Tan[c + d
*x]])/d + (5*b*Csc[c + d*x]*Sec[c + d*x]^2)/(8*d) + (b*Csc[c + d*x]*Sec[c + d*x]^4)/(4*d) + (3*a*Tan[c + d*x]^
2)/(2*d) + (a*Tan[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.156585, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2834, 2620, 266, 43, 2621, 288, 321, 207} \[ \frac{a \tan ^4(c+d x)}{4 d}+\frac{3 a \tan ^2(c+d x)}{2 d}-\frac{a \cot ^2(c+d x)}{2 d}+\frac{3 a \log (\tan (c+d x))}{d}-\frac{15 b \csc (c+d x)}{8 d}+\frac{15 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{5 b \csc (c+d x) \sec ^2(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(15*b*ArcTanh[Sin[c + d*x]])/(8*d) - (a*Cot[c + d*x]^2)/(2*d) - (15*b*Csc[c + d*x])/(8*d) + (3*a*Log[Tan[c + d
*x]])/d + (5*b*Csc[c + d*x]*Sec[c + d*x]^2)/(8*d) + (b*Csc[c + d*x]*Sec[c + d*x]^4)/(4*d) + (3*a*Tan[c + d*x]^
2)/(2*d) + (a*Tan[c + d*x]^4)/(4*d)

Rule 2834

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

Rule 2620

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2621

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

Rubi steps

\begin{align*} \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \csc ^3(c+d x) \sec ^5(c+d x) \, dx+b \int \csc ^2(c+d x) \sec ^5(c+d x) \, dx\\ &=\frac{a \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{x^3} \, dx,x,\tan (c+d x)\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac{b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{a \operatorname{Subst}\left (\int \frac{(1+x)^3}{x^2} \, dx,x,\tan ^2(c+d x)\right )}{2 d}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d}\\ &=\frac{5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{a \operatorname{Subst}\left (\int \left (3+\frac{1}{x^2}+\frac{3}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=-\frac{a \cot ^2(c+d x)}{2 d}-\frac{15 b \csc (c+d x)}{8 d}+\frac{3 a \log (\tan (c+d x))}{d}+\frac{5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{3 a \tan ^2(c+d x)}{2 d}+\frac{a \tan ^4(c+d x)}{4 d}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=\frac{15 b \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{a \cot ^2(c+d x)}{2 d}-\frac{15 b \csc (c+d x)}{8 d}+\frac{3 a \log (\tan (c+d x))}{d}+\frac{5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{3 a \tan ^2(c+d x)}{2 d}+\frac{a \tan ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.605704, size = 86, normalized size = 0.64 \[ -\frac{a \left (2 \csc ^2(c+d x)-\sec ^4(c+d x)-4 \sec ^2(c+d x)-12 \log (\sin (c+d x))+12 \log (\cos (c+d x))\right )}{4 d}-\frac{b \csc (c+d x) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\sin ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

-((b*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[c + d*x]^2])/d) - (a*(2*Csc[c + d*x]^2 + 12*Log[Cos[c +
d*x]] - 12*Log[Sin[c + d*x]] - 4*Sec[c + d*x]^2 - Sec[c + d*x]^4))/(4*d)

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Maple [A]  time = 0.066, size = 151, normalized size = 1.1 \begin{align*}{\frac{a}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,a}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{a\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{b}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,b}{8\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15\,b}{8\,d\sin \left ( dx+c \right ) }}+{\frac{15\,b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x)

[Out]

1/4/d*a/sin(d*x+c)^2/cos(d*x+c)^4+3/4/d*a/sin(d*x+c)^2/cos(d*x+c)^2-3/2/d*a/sin(d*x+c)^2+3*a*ln(tan(d*x+c))/d+
1/4/d*b/sin(d*x+c)/cos(d*x+c)^4+5/8/d*b/sin(d*x+c)/cos(d*x+c)^2-15/8/d*b/sin(d*x+c)+15/8/d*b*ln(sec(d*x+c)+tan
(d*x+c))

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Maxima [A]  time = 1.00901, size = 189, normalized size = 1.4 \begin{align*} -\frac{3 \,{\left (8 \, a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (8 \, a + 5 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac{2 \,{\left (15 \, b \sin \left (d x + c\right )^{5} + 12 \, a \sin \left (d x + c\right )^{4} - 25 \, b \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 8 \, b \sin \left (d x + c\right ) + 4 \, a\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(3*(8*a - 5*b)*log(sin(d*x + c) + 1) + 3*(8*a + 5*b)*log(sin(d*x + c) - 1) - 48*a*log(sin(d*x + c)) + 2*
(15*b*sin(d*x + c)^5 + 12*a*sin(d*x + c)^4 - 25*b*sin(d*x + c)^3 - 18*a*sin(d*x + c)^2 + 8*b*sin(d*x + c) + 4*
a)/(sin(d*x + c)^6 - 2*sin(d*x + c)^4 + sin(d*x + c)^2))/d

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Fricas [A]  time = 2.17341, size = 532, normalized size = 3.94 \begin{align*} \frac{24 \, a \cos \left (d x + c\right )^{4} - 12 \, a \cos \left (d x + c\right )^{2} + 48 \,{\left (a \cos \left (d x + c\right )^{6} - a \cos \left (d x + c\right )^{4}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 3 \,{\left ({\left (8 \, a - 5 \, b\right )} \cos \left (d x + c\right )^{6} -{\left (8 \, a - 5 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (8 \, a + 5 \, b\right )} \cos \left (d x + c\right )^{6} -{\left (8 \, a + 5 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (15 \, b \cos \left (d x + c\right )^{4} - 5 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \,{\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(24*a*cos(d*x + c)^4 - 12*a*cos(d*x + c)^2 + 48*(a*cos(d*x + c)^6 - a*cos(d*x + c)^4)*log(1/2*sin(d*x + c
)) - 3*((8*a - 5*b)*cos(d*x + c)^6 - (8*a - 5*b)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 3*((8*a + 5*b)*cos(d*
x + c)^6 - (8*a + 5*b)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(15*b*cos(d*x + c)^4 - 5*b*cos(d*x + c)^2 -
2*b)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^6 - 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(csc(d*x+c)**3*sec(d*x+c)**5*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.27938, size = 180, normalized size = 1.33 \begin{align*} -\frac{3 \,{\left (8 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \,{\left (8 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 48 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac{2 \,{\left (15 \, b \sin \left (d x + c\right )^{5} + 12 \, a \sin \left (d x + c\right )^{4} - 25 \, b \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 8 \, b \sin \left (d x + c\right ) + 4 \, a\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(3*(8*a - 5*b)*log(abs(sin(d*x + c) + 1)) + 3*(8*a + 5*b)*log(abs(sin(d*x + c) - 1)) - 48*a*log(abs(sin(
d*x + c))) + 2*(15*b*sin(d*x + c)^5 + 12*a*sin(d*x + c)^4 - 25*b*sin(d*x + c)^3 - 18*a*sin(d*x + c)^2 + 8*b*si
n(d*x + c) + 4*a)/(sin(d*x + c)^3 - sin(d*x + c))^2)/d