3.1489 \(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=115 \[ -\frac {15 a \csc (c+d x)}{8 d}+\frac {15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \log (\tan (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 115, normalized size of antiderivative = 1.00, 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, 2621, 288, 321, 207, 2620, 266, 43} \[ -\frac {15 a \csc (c+d x)}{8 d}+\frac {15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \log (\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

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

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 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 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 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 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 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 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])

Rubi steps

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

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Mathematica [C]  time = 0.39, size = 76, normalized size = 0.66 \[ -\frac {a \csc (c+d x) \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\sin ^2(c+d x)\right )}{d}-\frac {b \left (-\sec ^4(c+d x)-2 \sec ^2(c+d x)-4 \log (\sin (c+d x))+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.44, size = 159, normalized size = 1.38 \[ \frac {16 \, b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (15 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (15 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 30 \, a \cos \left (d x + c\right )^{4} + 10 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(16*b*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) + (15*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) +
1)*sin(d*x + c) - (15*a + 8*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d*x + c) - 30*a*cos(d*x + c)^4 + 10*a
*cos(d*x + c)^2 + 4*(2*b*cos(d*x + c)^2 + b)*sin(d*x + c) + 4*a)/(d*cos(d*x + c)^4*sin(d*x + c))

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giac [A]  time = 0.29, size = 134, normalized size = 1.17 \[ \frac {{\left (15 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (15 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {16 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (6 \, b \sin \left (d x + c\right )^{4} - 7 \, a \sin \left (d x + c\right )^{3} - 16 \, b \sin \left (d x + c\right )^{2} + 9 \, a \sin \left (d x + c\right ) + 12 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*((15*a - 8*b)*log(abs(sin(d*x + c) + 1)) - (15*a + 8*b)*log(abs(sin(d*x + c) - 1)) + 16*b*log(abs(sin(d*x
+ c))) - 16*(b*sin(d*x + c) + a)/sin(d*x + c) + 2*(6*b*sin(d*x + c)^4 - 7*a*sin(d*x + c)^3 - 16*b*sin(d*x + c
)^2 + 9*a*sin(d*x + c) + 12*b)/(sin(d*x + c)^2 - 1)^2)/d

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maple [A]  time = 0.30, size = 120, normalized size = 1.04 \[ \frac {a}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 a}{8 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 a}{8 d \sin \left (d x +c \right )}+\frac {15 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b}{4 d \cos \left (d x +c \right )^{4}}+\frac {b}{2 d \cos \left (d x +c \right )^{2}}+\frac {b \ln \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.31, size = 126, normalized size = 1.10 \[ \frac {{\left (15 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{4} + 4 \, b \sin \left (d x + c\right )^{3} - 25 \, a \sin \left (d x + c\right )^{2} - 6 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 11.89, size = 130, normalized size = 1.13 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {15\,a}{16}-\frac {b}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {15\,a}{16}+\frac {b}{2}\right )}{d}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^4}{8}+\frac {b\,{\sin \left (c+d\,x\right )}^3}{2}-\frac {25\,a\,{\sin \left (c+d\,x\right )}^2}{8}-\frac {3\,b\,\sin \left (c+d\,x\right )}{4}+a}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)^2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**5*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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