3.870 \(\int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=114 \[ \frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {5 a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \sin ^3(c+d x)}{3 d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {6 a^3 \sin (c+d x)}{d}-\frac {10 a^3 \log (1-\sin (c+d x))}{d} \]

[Out]

-10*a^3*ln(1-sin(d*x+c))/d-6*a^3*sin(d*x+c)/d-3/2*a^3*sin(d*x+c)^2/d-1/3*a^3*sin(d*x+c)^3/d+1/2*a^5/d/(a-a*sin
(d*x+c))^2-5*a^4/d/(a-a*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2707, 43} \[ -\frac {a^3 \sin ^3(c+d x)}{3 d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {5 a^4}{d (a-a \sin (c+d x))}-\frac {6 a^3 \sin (c+d x)}{d}-\frac {10 a^3 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

(-10*a^3*Log[1 - Sin[c + d*x]])/d - (6*a^3*Sin[c + d*x])/d - (3*a^3*Sin[c + d*x]^2)/(2*d) - (a^3*Sin[c + d*x]^
3)/(3*d) + a^5/(2*d*(a - a*Sin[c + d*x])^2) - (5*a^4)/(d*(a - a*Sin[c + d*x]))

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 2707

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

Rubi steps

\begin {align*} \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-6 a^2+\frac {a^5}{(a-x)^3}-\frac {5 a^4}{(a-x)^2}+\frac {10 a^3}{a-x}-3 a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {10 a^3 \log (1-\sin (c+d x))}{d}-\frac {6 a^3 \sin (c+d x)}{d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {5 a^4}{d (a-a \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.31, size = 73, normalized size = 0.64 \[ -\frac {a^3 \left (2 \sin ^3(c+d x)+9 \sin ^2(c+d x)+36 \sin (c+d x)+\frac {27-30 \sin (c+d x)}{(\sin (c+d x)-1)^2}+60 \log (1-\sin (c+d x))\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

-1/6*(a^3*(60*Log[1 - Sin[c + d*x]] + (27 - 30*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 36*Sin[c + d*x] + 9*Sin[c
+ d*x]^2 + 2*Sin[c + d*x]^3))/d

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 141, normalized size = 1.24 \[ \frac {10 \, a^{3} \cos \left (d x + c\right )^{4} + 115 \, a^{3} \cos \left (d x + c\right )^{2} - 80 \, a^{3} - 120 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 24 \, a^{3} \cos \left (d x + c\right )^{2} + 37 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(10*a^3*cos(d*x + c)^4 + 115*a^3*cos(d*x + c)^2 - 80*a^3 - 120*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) -
2*a^3)*log(-sin(d*x + c) + 1) + 2*(2*a^3*cos(d*x + c)^4 - 24*a^3*cos(d*x + c)^2 + 37*a^3)*sin(d*x + c))/(d*co
s(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)

________________________________________________________________________________________

giac [B]  time = 0.29, size = 242, normalized size = 2.12 \[ \frac {30 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 60 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 183 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 183 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 55 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac {125 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 524 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 804 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 524 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 125 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(30*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 60*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (55*a^3*tan(1/2*d*x
+ 1/2*c)^6 + 36*a^3*tan(1/2*d*x + 1/2*c)^5 + 183*a^3*tan(1/2*d*x + 1/2*c)^4 + 80*a^3*tan(1/2*d*x + 1/2*c)^3 +
183*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a^3*tan(1/2*d*x + 1/2*c) + 55*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3 + (125*a
^3*tan(1/2*d*x + 1/2*c)^4 - 524*a^3*tan(1/2*d*x + 1/2*c)^3 + 804*a^3*tan(1/2*d*x + 1/2*c)^2 - 524*a^3*tan(1/2*
d*x + 1/2*c) + 125*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d

________________________________________________________________________________________

maple [B]  time = 0.31, size = 325, normalized size = 2.85 \[ \frac {a^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {5 a^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {5 a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d}-\frac {2 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{d}-\frac {10 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {10 a^{3} \sin \left (d x +c \right )}{d}+\frac {10 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {3 a^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {9 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {9 a^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {10 a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {9 a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^9/cos(d*x+c)^4-5/8/d*a^3*sin(d*x+c)^9/cos(d*x+c)^2-5/8*a^3*sin(d*x+c)^7/d-2*a^3*sin(d*x+c
)^5/d-10/3*a^3*sin(d*x+c)^3/d-10*a^3*sin(d*x+c)/d+10/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*a^3*sin(d*x+c)^8/co
s(d*x+c)^4-3/2/d*a^3*sin(d*x+c)^8/cos(d*x+c)^2-3/2*a^3*sin(d*x+c)^6/d-9/4*a^3*sin(d*x+c)^4/d-9/2*a^3*sin(d*x+c
)^2/d-10/d*a^3*ln(cos(d*x+c))+3/4/d*a^3*sin(d*x+c)^7/cos(d*x+c)^4-9/8/d*a^3*sin(d*x+c)^7/cos(d*x+c)^2+1/4/d*a^
3*tan(d*x+c)^4-1/2/d*a^3*tan(d*x+c)^2

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 96, normalized size = 0.84 \[ -\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 36 \, a^{3} \sin \left (d x + c\right ) - \frac {3 \, {\left (10 \, a^{3} \sin \left (d x + c\right ) - 9 \, a^{3}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c) - 1) + 36*a^3*sin(d*x + c) - 3*(10
*a^3*sin(d*x + c) - 9*a^3)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

________________________________________________________________________________________

mupad [B]  time = 11.56, size = 321, normalized size = 2.82 \[ \frac {10\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-60\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {320\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {500\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+184\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {500\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {320\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-60\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+20\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {20\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^5*(a + a*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(10*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ((320*a^3*tan(c/2 + (d*x)/2)^3)/3 - 60*a^3*tan(c/2 + (d*x)/2)^2 - (
500*a^3*tan(c/2 + (d*x)/2)^4)/3 + 184*a^3*tan(c/2 + (d*x)/2)^5 - (500*a^3*tan(c/2 + (d*x)/2)^6)/3 + (320*a^3*t
an(c/2 + (d*x)/2)^7)/3 - 60*a^3*tan(c/2 + (d*x)/2)^8 + 20*a^3*tan(c/2 + (d*x)/2)^9 + 20*a^3*tan(c/2 + (d*x)/2)
)/(d*(9*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 16*tan(c/2 + (d*x)/2)^3 + 22*tan(c/2 + (d*x)/2)^4 - 24*t
an(c/2 + (d*x)/2)^5 + 22*tan(c/2 + (d*x)/2)^6 - 16*tan(c/2 + (d*x)/2)^7 + 9*tan(c/2 + (d*x)/2)^8 - 4*tan(c/2 +
(d*x)/2)^9 + tan(c/2 + (d*x)/2)^10 + 1)) - (20*a^3*log(tan(c/2 + (d*x)/2) - 1))/d

________________________________________________________________________________________

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(sec(d*x+c)**5*sin(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________