3.803 \(\int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=110 \[ \frac {a \tan ^3(c+d x)}{3 d}+\frac {2 a \tan (c+d x)}{d}-\frac {a \cot (c+d x)}{d}+\frac {5 a \sec ^3(c+d x)}{6 d}+\frac {5 a \sec (c+d x)}{2 d}-\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a \csc ^2(c+d x) \sec ^3(c+d x)}{2 d} \]

[Out]

-5/2*a*arctanh(cos(d*x+c))/d-a*cot(d*x+c)/d+5/2*a*sec(d*x+c)/d+5/6*a*sec(d*x+c)^3/d-1/2*a*csc(d*x+c)^2*sec(d*x
+c)^3/d+2*a*tan(d*x+c)/d+1/3*a*tan(d*x+c)^3/d

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Rubi [A]  time = 0.14, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2838, 2622, 288, 302, 207, 2620, 270} \[ \frac {a \tan ^3(c+d x)}{3 d}+\frac {2 a \tan (c+d x)}{d}-\frac {a \cot (c+d x)}{d}+\frac {5 a \sec ^3(c+d x)}{6 d}+\frac {5 a \sec (c+d x)}{2 d}-\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a \csc ^2(c+d x) \sec ^3(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(-5*a*ArcTanh[Cos[c + d*x]])/(2*d) - (a*Cot[c + d*x])/d + (5*a*Sec[c + d*x])/(2*d) + (5*a*Sec[c + d*x]^3)/(6*d
) - (a*Csc[c + d*x]^2*Sec[c + d*x]^3)/(2*d) + (2*a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*d)

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 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 2838

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

Rubi steps

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

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Mathematica [B]  time = 6.10, size = 359, normalized size = 3.26 \[ \frac {5 a \tan (c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {5 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {5 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {13 a \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {a \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {13 a \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {a}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {a}{12 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {a \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

-((a*Cot[c + d*x])/d) - (a*Csc[(c + d*x)/2]^2)/(8*d) - (5*a*Log[Cos[(c + d*x)/2]])/(2*d) + (5*a*Log[Sin[(c + d
*x)/2]])/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d) + a/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (a*Sin[(c +
d*x)/2])/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (13*a*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] - Sin
[(c + d*x)/2])) - (a*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + a/(12*d*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2])^2) - (13*a*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (5*a*Tan[c +
d*x])/(3*d) + (a*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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fricas [B]  time = 0.48, size = 222, normalized size = 2.02 \[ \frac {32 \, a \cos \left (d x + c\right )^{4} - 18 \, a \cos \left (d x + c\right )^{2} - 15 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) - 8 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(32*a*cos(d*x + c)^4 - 18*a*cos(d*x + c)^2 - 15*(a*cos(d*x + c)^3 - a*cos(d*x + c) - (a*cos(d*x + c)^3 -
a*cos(d*x + c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 15*(a*cos(d*x + c)^3 - a*cos(d*x + c) - (a*cos(d*x
+ c)^3 - a*cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(a*cos(d*x + c)^2 + 2*a)*sin(d*x + c)
- 8*a)/(d*cos(d*x + c)^3 - d*cos(d*x + c) - (d*cos(d*x + c)^3 - d*cos(d*x + c))*sin(d*x + c))

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giac [A]  time = 0.23, size = 148, normalized size = 1.35 \[ \frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 60 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {12 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} - \frac {3 \, {\left (30 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {4 \, {\left (27 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*a*tan(1/2*d*x + 1/2*c)^2 + 60*a*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a*tan(1/2*d*x + 1/2*c) + 12*a/(tan
(1/2*d*x + 1/2*c) + 1) - 3*(30*a*tan(1/2*d*x + 1/2*c)^2 + 4*a*tan(1/2*d*x + 1/2*c) + a)/tan(1/2*d*x + 1/2*c)^2
- 4*(27*a*tan(1/2*d*x + 1/2*c)^2 - 48*a*tan(1/2*d*x + 1/2*c) + 25*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [A]  time = 0.51, size = 138, normalized size = 1.25 \[ \frac {a}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4 a}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 a \cot \left (d x +c \right )}{3 d}+\frac {a}{3 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}-\frac {5 a}{6 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {5 a}{2 d \cos \left (d x +c \right )}+\frac {5 a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c)),x)

[Out]

1/3/d*a/sin(d*x+c)/cos(d*x+c)^3+4/3/d*a/sin(d*x+c)/cos(d*x+c)-8/3*a*cot(d*x+c)/d+1/3/d*a/sin(d*x+c)^2/cos(d*x+
c)^3-5/6/d*a/sin(d*x+c)^2/cos(d*x+c)+5/2/d*a/cos(d*x+c)+5/2/d*a*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.32, size = 106, normalized size = 0.96 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a + a {\left (\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a + a*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 2
)/(cos(d*x + c)^5 - cos(d*x + c)^3) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 8.98, size = 180, normalized size = 1.64 \[ \frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {-18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {23\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+\frac {67\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {68\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {5\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))/(cos(c + d*x)^4*sin(c + d*x)^3),x)

[Out]

(a*tan(c/2 + (d*x)/2))/(2*d) - (a/2 + a*tan(c/2 + (d*x)/2) - (68*a*tan(c/2 + (d*x)/2)^2)/3 + (67*a*tan(c/2 + (
d*x)/2)^3)/3 + (23*a*tan(c/2 + (d*x)/2)^4)/2 - 18*a*tan(c/2 + (d*x)/2)^5)/(d*(4*tan(c/2 + (d*x)/2)^2 - 8*tan(c
/2 + (d*x)/2)^3 + 8*tan(c/2 + (d*x)/2)^5 - 4*tan(c/2 + (d*x)/2)^6)) + (a*tan(c/2 + (d*x)/2)^2)/(8*d) + (5*a*lo
g(tan(c/2 + (d*x)/2)))/(2*d)

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

[Out]

Timed out

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