3.1491 \(\int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=155 \[ -\frac{35 a \csc ^3(c+d x)}{24 d}-\frac{35 a \csc (c+d x)}{8 d}+\frac{35 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac{7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d}+\frac{3 b \tan ^2(c+d x)}{2 d}-\frac{b \cot ^2(c+d x)}{2 d}+\frac{3 b \log (\tan (c+d x))}{d} \]

[Out]

(35*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Cot[c + d*x]^2)/(2*d) - (35*a*Csc[c + d*x])/(8*d) - (35*a*Csc[c + d*x]
^3)/(24*d) + (3*b*Log[Tan[c + d*x]])/d + (7*a*Csc[c + d*x]^3*Sec[c + d*x]^2)/(8*d) + (a*Csc[c + d*x]^3*Sec[c +
d*x]^4)/(4*d) + (3*b*Tan[c + d*x]^2)/(2*d) + (b*Tan[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.160565, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2834, 2621, 288, 302, 207, 2620, 266, 43} \[ -\frac{35 a \csc ^3(c+d x)}{24 d}-\frac{35 a \csc (c+d x)}{8 d}+\frac{35 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac{7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d}+\frac{3 b \tan ^2(c+d x)}{2 d}-\frac{b \cot ^2(c+d x)}{2 d}+\frac{3 b \log (\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(35*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Cot[c + d*x]^2)/(2*d) - (35*a*Csc[c + d*x])/(8*d) - (35*a*Csc[c + d*x]
^3)/(24*d) + (3*b*Log[Tan[c + d*x]])/d + (7*a*Csc[c + d*x]^3*Sec[c + d*x]^2)/(8*d) + (a*Csc[c + d*x]^3*Sec[c +
d*x]^4)/(4*d) + (3*b*Tan[c + d*x]^2)/(2*d) + (b*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 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 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])

Rubi steps

\begin{align*} \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \csc ^4(c+d x) \sec ^5(c+d x) \, dx+b \int \csc ^3(c+d x) \sec ^5(c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{x^8}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d}+\frac{b \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{x^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}-\frac{(7 a) \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{(1+x)^3}{x^2} \, dx,x,\tan ^2(c+d x)\right )}{2 d}\\ &=\frac{7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}-\frac{(35 a) \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}+\frac{b \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{b \cot ^2(c+d x)}{2 d}+\frac{3 b \log (\tan (c+d x))}{d}+\frac{7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac{3 b \tan ^2(c+d x)}{2 d}+\frac{b \tan ^4(c+d x)}{4 d}-\frac{(35 a) \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=-\frac{b \cot ^2(c+d x)}{2 d}-\frac{35 a \csc (c+d x)}{8 d}-\frac{35 a \csc ^3(c+d x)}{24 d}+\frac{3 b \log (\tan (c+d x))}{d}+\frac{7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac{3 b \tan ^2(c+d x)}{2 d}+\frac{b \tan ^4(c+d x)}{4 d}-\frac{(35 a) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=\frac{35 a \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{b \cot ^2(c+d x)}{2 d}-\frac{35 a \csc (c+d x)}{8 d}-\frac{35 a \csc ^3(c+d x)}{24 d}+\frac{3 b \log (\tan (c+d x))}{d}+\frac{7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac{3 b \tan ^2(c+d x)}{2 d}+\frac{b \tan ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.859305, size = 90, normalized size = 0.58 \[ -\frac{a \csc ^3(c+d x) \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};\sin ^2(c+d x)\right )}{3 d}-\frac{b \left (2 \csc ^2(c+d x)-\sec ^4(c+d x)-4 \sec ^2(c+d x)-12 \log (\sin (c+d x))+12 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

-(a*Csc[c + d*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[c + d*x]^2])/(3*d) - (b*(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 = 0.068, size = 173, normalized size = 1.1 \begin{align*}{\frac{a}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{7\,a}{12\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{35\,a}{24\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{35\,a}{8\,d\sin \left ( dx+c \right ) }}+{\frac{35\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{b}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,b}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,b}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b\ln \left ( \tan \left ( dx+c \right ) \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)^5*(a+b*sin(d*x+c)),x)

[Out]

1/4/d*a/sin(d*x+c)^3/cos(d*x+c)^4-7/12/d*a/sin(d*x+c)^3/cos(d*x+c)^2+35/24/d*a/sin(d*x+c)/cos(d*x+c)^2-35/8/d*
a/sin(d*x+c)+35/8/d*a*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*b/sin(d*x+c)^2/cos(d*x+c)^4+3/4/d*b/sin(d*x+c)^2/cos(d*x
+c)^2-3/2/d*b/sin(d*x+c)^2+3*b*ln(tan(d*x+c))/d

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Maxima [A]  time = 0.978455, size = 204, normalized size = 1.32 \begin{align*} \frac{3 \,{\left (35 \, a - 24 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (35 \, a + 24 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac{2 \,{\left (105 \, a \sin \left (d x + c\right )^{6} + 36 \, b \sin \left (d x + c\right )^{5} - 175 \, a \sin \left (d x + c\right )^{4} - 54 \, b \sin \left (d x + c\right )^{3} + 56 \, a \sin \left (d x + c\right )^{2} + 12 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{7} - 2 \, \sin \left (d x + c\right )^{5} + \sin \left (d x + c\right )^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(3*(35*a - 24*b)*log(sin(d*x + c) + 1) - 3*(35*a + 24*b)*log(sin(d*x + c) - 1) + 144*b*log(sin(d*x + c))
- 2*(105*a*sin(d*x + c)^6 + 36*b*sin(d*x + c)^5 - 175*a*sin(d*x + c)^4 - 54*b*sin(d*x + c)^3 + 56*a*sin(d*x +
c)^2 + 12*b*sin(d*x + c) + 8*a)/(sin(d*x + c)^7 - 2*sin(d*x + c)^5 + sin(d*x + c)^3))/d

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Fricas [A]  time = 2.14723, size = 649, normalized size = 4.19 \begin{align*} -\frac{210 \, a \cos \left (d x + c\right )^{6} - 280 \, a \cos \left (d x + c\right )^{4} + 42 \, a \cos \left (d x + c\right )^{2} - 144 \,{\left (b \cos \left (d x + c\right )^{6} - b \cos \left (d x + c\right )^{4}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 3 \,{\left ({\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{6} -{\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \,{\left ({\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{6} -{\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 12 \,{\left (6 \, b \cos \left (d x + c\right )^{4} - 3 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) + 12 \, a}{48 \,{\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\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)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(210*a*cos(d*x + c)^6 - 280*a*cos(d*x + c)^4 + 42*a*cos(d*x + c)^2 - 144*(b*cos(d*x + c)^6 - b*cos(d*x +
c)^4)*log(1/2*sin(d*x + c))*sin(d*x + c) - 3*((35*a - 24*b)*cos(d*x + c)^6 - (35*a - 24*b)*cos(d*x + c)^4)*lo
g(sin(d*x + c) + 1)*sin(d*x + c) + 3*((35*a + 24*b)*cos(d*x + c)^6 - (35*a + 24*b)*cos(d*x + c)^4)*log(-sin(d*
x + c) + 1)*sin(d*x + c) - 12*(6*b*cos(d*x + c)^4 - 3*b*cos(d*x + c)^2 - b)*sin(d*x + c) + 12*a)/((d*cos(d*x +
c)^6 - d*cos(d*x + c)^4)*sin(d*x + c))

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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)**4*sec(d*x+c)**5*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.27312, size = 216, normalized size = 1.39 \begin{align*} \frac{3 \,{\left (35 \, a - 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (35 \, a + 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 144 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac{6 \,{\left (18 \, b \sin \left (d x + c\right )^{4} - 11 \, a \sin \left (d x + c\right )^{3} - 44 \, b \sin \left (d x + c\right )^{2} + 13 \, a \sin \left (d x + c\right ) + 28 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}} - \frac{8 \,{\left (33 \, b \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} + 3 \, b \sin \left (d x + c\right ) + 2 \, a\right )}}{\sin \left (d x + c\right )^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/48*(3*(35*a - 24*b)*log(abs(sin(d*x + c) + 1)) - 3*(35*a + 24*b)*log(abs(sin(d*x + c) - 1)) + 144*b*log(abs(
sin(d*x + c))) + 6*(18*b*sin(d*x + c)^4 - 11*a*sin(d*x + c)^3 - 44*b*sin(d*x + c)^2 + 13*a*sin(d*x + c) + 28*b
)/(sin(d*x + c)^2 - 1)^2 - 8*(33*b*sin(d*x + c)^3 + 18*a*sin(d*x + c)^2 + 3*b*sin(d*x + c) + 2*a)/sin(d*x + c)
^3)/d