\(\int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) [1360]

Optimal result

Integrand size = 29, antiderivative size = 221 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^7 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )^3 d}-\frac {\sin (c+d x)}{b d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \]

[Out]

-1/16*(24*a^2+37*a*b+15*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(24*a^2-37*a*b+15*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d
+a^7*ln(a+b*sin(d*x+c))/b^2/(a^2-b^2)^3/d-sin(d*x+c)/b/d+1/4*sec(d*x+c)^4*(a-b*sin(d*x+c))/(a^2-b^2)/d-1/8*sec
(d*x+c)^2*(4*a*(3*a^2-2*b^2)-b*(13*a^2-9*b^2)*sin(d*x+c))/(a^2-b^2)^2/d

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 12, 1661, 1643} \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a^7 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )^3}-\frac {\sin (c+d x)}{b d} \]

[In]

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

[Out]

-1/16*((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/((a + b)^3*d) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + S
in[c + d*x]])/(16*(a - b)^3*d) + (a^7*Log[a + b*Sin[c + d*x]])/(b^2*(a^2 - b^2)^3*d) - Sin[c + d*x]/(b*d) + (S
ec[c + d*x]^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*d) - (Sec[c + d*x]^2*(4*a*(3*a^2 - 2*b^2) - b*(13*a^2 - 9*b
^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

Rule 12

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

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x^7}{b^7 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x^7}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b^2 d} \\ & = \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \frac {\frac {a b^8}{a^2-b^2}-\frac {b^6 \left (4 a^2-b^2\right ) x}{a^2-b^2}-4 b^4 x^3-4 b^2 x^5}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^4 d} \\ & = \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\text {Subst}\left (\int \frac {-\frac {a b^8 \left (11 a^2-7 b^2\right )}{\left (a^2-b^2\right )^2}+\frac {b^6 \left (16 a^4-19 a^2 b^2+7 b^4\right ) x}{\left (a^2-b^2\right )^2}+8 b^4 x^3}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^6 d} \\ & = \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\text {Subst}\left (\int \left (-8 b^4+\frac {b^6 \left (24 a^2+37 a b+15 b^2\right )}{2 (a+b)^3 (b-x)}+\frac {8 a^7 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac {b^6 \left (24 a^2-37 a b+15 b^2\right )}{2 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^6 d} \\ & = -\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^7 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )^3 d}-\frac {\sin (c+d x)}{b d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{(a-b)^3}+\frac {16 a^7 \log (a+b \sin (c+d x))}{(a-b)^3 b^2 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {11 a+9 b}{(a+b)^2 (-1+\sin (c+d x))}-\frac {16 \sin (c+d x)}{b}+\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {-11 a+9 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \]

[In]

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

[Out]

(-(((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + Sin[c +
d*x]])/(a - b)^3 + (16*a^7*Log[a + b*Sin[c + d*x]])/((a - b)^3*b^2*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x]
)^2) + (11*a + 9*b)/((a + b)^2*(-1 + Sin[c + d*x])) - (16*Sin[c + d*x])/b + 1/((a - b)*(1 + Sin[c + d*x])^2) +
(-11*a + 9*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {11 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-24 a^{2}+37 a b -15 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-11 a -9 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-24 a^{2}-37 a b -15 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {a^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(203\)
default	\(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {11 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-24 a^{2}+37 a b -15 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-11 a -9 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-24 a^{2}-37 a b -15 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {a^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(203\)
parallelrisch	\(\frac {8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{7} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{3} \left (a -b \right )^{3} a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (24 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {37}{24} a b +\frac {5}{8} b^{2}\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (24 b \left (a^{2}-\frac {37}{24} a b +\frac {5}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (2 \left (a^{3} b -a \,b^{3}\right ) \cos \left (2 d x +2 c \right )+\frac {\left (-5 a^{3} b +3 a \,b^{3}\right ) \cos \left (4 d x +4 c \right )}{2}+\left (3 a^{4}-\frac {25}{2} a^{2} b^{2}+\frac {15}{2} b^{4}\right ) \sin \left (3 d x +3 c \right )+\left (a -b \right )^{2} \left (a +b \right )^{2} \sin \left (5 d x +5 c \right )+\left (2 a^{4}-\frac {13}{2} a^{2} b^{2}+\frac {5}{2} b^{4}\right ) \sin \left (d x +c \right )+\frac {b a \left (a^{2}+b^{2}\right )}{2}\right ) \left (a -b \right )\right ) \left (a +b \right )\right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} b^{2} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(412\)
norman	\(\frac {\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}+\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}-\frac {2 \left (2 a^{3}-a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {2 \left (2 a^{3}-a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {4 \left (4 a^{3}-3 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (8 a^{4}-27 a^{2} b^{2}+15 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (8 a^{4}-27 a^{2} b^{2}+15 b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (8 a^{4}+7 a^{2} b^{2}-11 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (8 a^{4}+7 a^{2} b^{2}-11 b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {\left (24 a^{4}-45 a^{2} b^{2}+25 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {\left (24 a^{4}-45 a^{2} b^{2}+25 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{7} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {\left (24 a^{2}-37 a b +15 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {\left (24 a^{2}+37 a b +15 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}\)	\(749\)
risch	\(-\frac {37 i a b c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {37 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i a^{7} c}{b^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {i a x}{b^{2}}+\frac {15 i b^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a^{2} x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {15 i b^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i a^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {37 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {37 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {a^{7} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {i \left (24 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-13 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+32 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-16 i b^{2} a \,{\mathrm e}^{4 i \left (d x +c \right )}-5 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+24 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+13 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left (-a^{2}+b^{2}\right )^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 i a^{2} c}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {37 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {15 i b^{2} c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {3 i a^{2} c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {37 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 i b^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i a^{7} x}{b^{2} \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}\)	\(1045\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-sin(d*x+c)/b+1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(11*a-9*b)/(a-b)^2/(1+sin(d*x+c))+1/16/(a-b)^3*(-24*a^2
+37*a*b-15*b^2)*ln(1+sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(-11*a-9*b)/(a+b)^2/(sin(d*x+c)-1)+1/16/(
a+b)^3*(-24*a^2-37*a*b-15*b^2)*ln(sin(d*x+c)-1)+1/b^2*a^7/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.59 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{5} b^{2} - 8 \, a^{3} b^{4} + 4 \, a b^{6} - {\left (24 \, a^{5} b^{2} + 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} - 42 \, a^{2} b^{5} + 8 \, a b^{6} + 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (24 \, a^{5} b^{2} - 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} + 42 \, a^{2} b^{5} + 8 \, a b^{6} - 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (3 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + 2 \, b^{7} + 8 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (13 \, a^{4} b^{3} - 22 \, a^{2} b^{5} + 9 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(16*a^7*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^5*b^2 - 8*a^3*b^4 + 4*a*b^6 - (24*a^5*b^2 + 35*a^4*b
^3 - 24*a^3*b^4 - 42*a^2*b^5 + 8*a*b^6 + 15*b^7)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (24*a^5*b^2 - 35*a^4*b
^3 - 24*a^3*b^4 + 42*a^2*b^5 + 8*a*b^6 - 15*b^7)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*(3*a^5*b^2 - 5*a^3*
b^4 + 2*a*b^6)*cos(d*x + c)^2 - 2*(2*a^4*b^3 - 4*a^2*b^5 + 2*b^7 + 8*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos
(d*x + c)^4 - (13*a^4*b^3 - 22*a^2*b^5 + 9*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^
6 - b^8)*d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**7/(a+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.37 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac {{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (13 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 10 \, a^{3} - 6 \, a b^{2} - 4 \, {\left (3 \, a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} - \frac {16 \, \sin \left (d x + c\right )}{b}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(16*a^7*log(b*sin(d*x + c) + a)/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8) - (24*a^2 - 37*a*b + 15*b^2)*log(
sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*
a^2*b + 3*a*b^2 + b^3) - 2*((13*a^2*b - 9*b^3)*sin(d*x + c)^3 + 10*a^3 - 6*a*b^2 - 4*(3*a^3 - 2*a*b^2)*sin(d*x
+ c)^2 - (11*a^2*b - 7*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2
*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2) - 16*sin(d*x + c)/b)/d

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.74 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac {{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {16 \, \sin \left (d x + c\right )}{b} + \frac {2 \, {\left (18 \, a^{5} \sin \left (d x + c\right )^{4} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} + 6 \, a b^{4} \sin \left (d x + c\right )^{4} - 13 \, a^{4} b \sin \left (d x + c\right )^{3} + 22 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - 9 \, b^{5} \sin \left (d x + c\right )^{3} - 24 \, a^{5} \sin \left (d x + c\right )^{2} + 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 4 \, a b^{4} \sin \left (d x + c\right )^{2} + 11 \, a^{4} b \sin \left (d x + c\right ) - 18 \, a^{2} b^{3} \sin \left (d x + c\right ) + 7 \, b^{5} \sin \left (d x + c\right ) + 8 \, a^{5} - 2 \, a^{3} b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(16*a^7*log(abs(b*sin(d*x + c) + a))/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8) - (24*a^2 - 37*a*b + 15*b^2)
*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(abs(sin(d*x + c)
- 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 16*sin(d*x + c)/b + 2*(18*a^5*sin(d*x + c)^4 - 18*a^3*b^2*sin(d*x + c)
^4 + 6*a*b^4*sin(d*x + c)^4 - 13*a^4*b*sin(d*x + c)^3 + 22*a^2*b^3*sin(d*x + c)^3 - 9*b^5*sin(d*x + c)^3 - 24*
a^5*sin(d*x + c)^2 + 16*a^3*b^2*sin(d*x + c)^2 - 4*a*b^4*sin(d*x + c)^2 + 11*a^4*b*sin(d*x + c) - 18*a^2*b^3*s
in(d*x + c) + 7*b^5*sin(d*x + c) + 8*a^5 - 2*a^3*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1
)^2))/d

Mupad [B] (verification not implemented)

Time = 14.57 (sec) , antiderivative size = 627, normalized size of antiderivative = 2.84 \[ \int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {11\,b}{8\,{\left (a-b\right )}^2}+\frac {3}{a-b}\right )}{d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a\,b^2-4\,a^3\right )}{{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a\,b^2-2\,a^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a\,b^2-4\,a^3\right )}{{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a\,b^2-2\,a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (24\,a^4-29\,a^2\,b^2+9\,b^4\right )}{2\,b\,{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^2-5\,b^2\right )}{b\,\left (a^2-b^2\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (4\,a^2-5\,b^2\right )}{b\,\left (a^2-b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,a^4-27\,a^2\,b^2+15\,b^4\right )}{4\,b\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (8\,a^4-27\,a^2\,b^2+15\,b^4\right )}{4\,b\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3}{a+b}-\frac {11\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d}-\frac {a^7\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (-a^6\,b^2+3\,a^4\,b^4-3\,a^2\,b^6+b^8\right )} \]

[In]

int(sin(c + d*x)^7/(cos(c + d*x)^5*(a + b*sin(c + d*x))),x)

[Out]

- (log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + (11*b)/(8*(a - b)^2) + 3/(a - b)))/d - ((2*tan(c/2 + (d*x)
/2)^4*(3*a*b^2 - 4*a^3))/(a^2 - b^2)^2 - (2*tan(c/2 + (d*x)/2)^2*(a*b^2 - 2*a^3))/(a^2 - b^2)^2 + (2*tan(c/2 +
(d*x)/2)^6*(3*a*b^2 - 4*a^3))/(a^2 - b^2)^2 - (2*tan(c/2 + (d*x)/2)^8*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2
) + (tan(c/2 + (d*x)/2)^5*(24*a^4 + 9*b^4 - 29*a^2*b^2))/(2*b*(a^2 - b^2)^2) - (2*tan(c/2 + (d*x)/2)^3*(4*a^2
- 5*b^2))/(b*(a^2 - b^2)) - (2*tan(c/2 + (d*x)/2)^7*(4*a^2 - 5*b^2))/(b*(a^2 - b^2)) + (tan(c/2 + (d*x)/2)*(8*
a^4 + 15*b^4 - 27*a^2*b^2))/(4*b*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^9*(8*a^4 + 15*b^4 - 27*a^2*b^2
))/(4*b*(a^4 + b^4 - 2*a^2*b^2)))/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6
- 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (log(tan(c/2 + (d*x)/2) - 1)*(3/(a + b) - (11*b)/(8*
(a + b)^2) + b^2/(4*(a + b)^3)))/d - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/(b^2*d) - (a^7*log(a + 2*b*tan(c/2 + (d
*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))