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

Optimal. Leaf size=129 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {3 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \csc (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]

[Out]

-a*csc(d*x+c)/d-23/16*a*ln(1-sin(d*x+c))/d+a*ln(sin(d*x+c))/d+7/16*a*ln(1+sin(d*x+c))/d+1/8*a^3/d/(a-a*sin(d*x
+c))^2+3/4*a^2/d/(a-a*sin(d*x+c))-1/8*a^2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {3 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \csc (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Csc[c + d*x])/d) - (23*a*Log[1 - Sin[c + d*x]])/(16*d) + (a*Log[Sin[c + d*x]])/d + (7*a*Log[1 + Sin[c + d
*x]])/(16*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + (3*a^2)/(4*d*(a - a*Sin[c + d*x])) - a^2/(8*d*(a + a*Sin[c +
d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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

Rubi steps

\begin {align*} \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {a^2}{(a-x)^3 x^2 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 x^2 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {1}{4 a^4 (a-x)^3}+\frac {3}{4 a^5 (a-x)^2}+\frac {23}{16 a^6 (a-x)}+\frac {1}{a^5 x^2}+\frac {1}{a^6 x}+\frac {1}{8 a^5 (a+x)^2}+\frac {7}{16 a^6 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {a \csc (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {3 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 76, normalized size = 0.59 \[ -\frac {a \csc (c+d x) \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\sin ^2(c+d x)\right )}{d}-\frac {a \left (-\sec ^4(c+d x)-2 \sec ^2(c+d x)-4 \log (\sin (c+d x))+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[c + d*x]^2])/d) - (a*(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 = 0.47, size = 229, normalized size = 1.78 \[ -\frac {22 \, a \cos \left (d x + c\right )^{2} - 16 \, {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 7 \, {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 23 \, {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(22*a*cos(d*x + c)^2 - 16*(a*cos(d*x + c)^4 + a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(1/2*
sin(d*x + c)) - 7*(a*cos(d*x + c)^4 + a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1)
+ 23*(a*cos(d*x + c)^4 + a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(15*a*co
s(d*x + c)^2 - a)*sin(d*x + c) - 6*a)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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giac [A]  time = 0.28, size = 121, normalized size = 0.94 \[ \frac {14 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 46 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 32 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {23 \, a \sin \left (d x + c\right )^{2} + 59 \, a \sin \left (d x + c\right ) + 32 \, a}{\sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )} + \frac {69 \, a \sin \left (d x + c\right )^{2} - 162 \, a \sin \left (d x + c\right ) + 97 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(14*a*log(abs(sin(d*x + c) + 1)) - 46*a*log(abs(sin(d*x + c) - 1)) + 32*a*log(abs(sin(d*x + c))) - (23*a*
sin(d*x + c)^2 + 59*a*sin(d*x + c) + 32*a)/(sin(d*x + c)^2 + sin(d*x + c)) + (69*a*sin(d*x + c)^2 - 162*a*sin(
d*x + c) + 97*a)/(sin(d*x + c) - 1)^2)/d

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maple [A]  time = 0.35, size = 120, normalized size = 0.93 \[ \frac {a}{4 d \cos \left (d x +c \right )^{4}}+\frac {a}{2 d \cos \left (d x +c \right )^{2}}+\frac {a \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 a}{8 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 a}{8 d \sin \left (d x +c \right )}+\frac {15 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 114, normalized size = 0.88 \[ \frac {7 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{3} - 11 \, a \sin \left (d x + c\right )^{2} - 14 \, a \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{4} - \sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(7*a*log(sin(d*x + c) + 1) - 23*a*log(sin(d*x + c) - 1) + 16*a*log(sin(d*x + c)) - 2*(15*a*sin(d*x + c)^3
- 11*a*sin(d*x + c)^2 - 14*a*sin(d*x + c) + 8*a)/(sin(d*x + c)^4 - sin(d*x + c)^3 - sin(d*x + c)^2 + sin(d*x
+ c)))/d

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mupad [B]  time = 0.10, size = 118, normalized size = 0.91 \[ \frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {11\,a\,{\sin \left (c+d\,x\right )}^2}{8}-\frac {7\,a\,\sin \left (c+d\,x\right )}{4}+a}{d\,\left ({\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^3-{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\right )}-\frac {23\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}+\frac {7\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*log(sin(c + d*x)))/d - (a - (7*a*sin(c + d*x))/4 - (11*a*sin(c + d*x)^2)/8 + (15*a*sin(c + d*x)^3)/8)/(d*(s
in(c + d*x) - sin(c + d*x)^2 - sin(c + d*x)^3 + sin(c + d*x)^4)) - (23*a*log(sin(c + d*x) - 1))/(16*d) + (7*a*
log(sin(c + d*x) + 1))/(16*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)**2*sec(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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