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]

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

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Rubi [A]  time = 0.16, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 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 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 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 ^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*}

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Mathematica [C]  time = 0.82, 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]

-1/3*(a*Csc[c + d*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[c + d*x]^2])/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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fricas [A]  time = 0.47, size = 248, normalized size = 1.60 \[ -\frac {210 \, a \cos \left (d x + c\right )^{6} - 280 \, a \cos \left (d x + c\right )^{4} + 42 \, a \cos \left (d x + c\right )^{2} - 144 \, {\left (b \cos \left (d x + c\right )^{6} - b \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 3 \, {\left ({\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 12 \, {\left (6 \, b \cos \left (d x + c\right )^{4} - 3 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) + 12 \, a}{48 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.30, size = 160, normalized size = 1.03 \[ \frac {3 \, {\left (35 \, a - 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (35 \, a + 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 144 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {6 \, {\left (18 \, b \sin \left (d x + c\right )^{4} - 11 \, a \sin \left (d x + c\right )^{3} - 44 \, b \sin \left (d x + c\right )^{2} + 13 \, a \sin \left (d x + c\right ) + 28 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}} - \frac {8 \, {\left (33 \, b \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} + 3 \, b \sin \left (d x + c\right ) + 2 \, a\right )}}{\sin \left (d x + c\right )^{3}}}{48 \, d} \]

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

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maple [A]  time = 0.37, size = 173, normalized size = 1.12 \[ \frac {a}{4 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7 a}{12 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {35 a}{24 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {35 a}{8 d \sin \left (d x +c \right )}+\frac {35 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3 b}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3 b}{2 d \sin \left (d x +c \right )^{2}}+\frac {3 b \ln \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^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 = 0.36, size = 151, normalized size = 0.97 \[ \frac {3 \, {\left (35 \, a - 24 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, a + 24 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (105 \, a \sin \left (d x + c\right )^{6} + 36 \, b \sin \left (d x + c\right )^{5} - 175 \, a \sin \left (d x + c\right )^{4} - 54 \, b \sin \left (d x + c\right )^{3} + 56 \, a \sin \left (d x + c\right )^{2} + 12 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{7} - 2 \, \sin \left (d x + c\right )^{5} + \sin \left (d x + c\right )^{3}}}{48 \, d} \]

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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mupad [B]  time = 11.91, size = 157, normalized size = 1.01 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {35\,a}{16}-\frac {3\,b}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {35\,a}{16}+\frac {3\,b}{2}\right )}{d}+\frac {3\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {35\,a\,{\sin \left (c+d\,x\right )}^6}{8}+\frac {3\,b\,{\sin \left (c+d\,x\right )}^5}{2}-\frac {175\,a\,{\sin \left (c+d\,x\right )}^4}{24}-\frac {9\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {7\,a\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {b\,\sin \left (c+d\,x\right )}{2}+\frac {a}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^7-2\,{\sin \left (c+d\,x\right )}^5+{\sin \left (c+d\,x\right )}^3\right )} \]

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)^4),x)

[Out]

(log(sin(c + d*x) + 1)*((35*a)/16 - (3*b)/2))/d - (log(sin(c + d*x) - 1)*((35*a)/16 + (3*b)/2))/d + (3*b*log(s
in(c + d*x)))/d - (a/3 + (b*sin(c + d*x))/2 + (7*a*sin(c + d*x)^2)/3 - (175*a*sin(c + d*x)^4)/24 + (35*a*sin(c
+ d*x)^6)/8 - (9*b*sin(c + d*x)^3)/4 + (3*b*sin(c + d*x)^5)/2)/(d*(sin(c + d*x)^3 - 2*sin(c + d*x)^5 + sin(c
+ d*x)^7))

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

[Out]

Timed out

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