\(\int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1364]

Optimal result

Integrand size = 29, antiderivative size = 182 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {(3 a-b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^3 b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \]

[Out]

1/16*b*(3*a+b)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(3*a-b)*b*ln(1+sin(d*x+c))/(a-b)^3/d+a^3*b^2*ln(a+b*sin(d*x+c))
/(a^2-b^2)^3/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-b*(5*a^2-b^2)*sin(d*x+c))
/(a^2-b^2)^2/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2916, 12, 1661, 837, 815} \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {a^3 b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {b (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {b (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

[In]

Int[(Sec[c + d*x]^2*Tan[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

(b*(3*a + b)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - ((3*a - b)*b*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) +
(a^3*b^2*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^3*d) + (Sec[c + d*x]^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*d)
- (Sec[c + d*x]^2*(4*a^3 - b*(5*a^2 - 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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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^3}{b^3 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^2 \text {Subst}\left (\int \frac {x^3}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{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^4}{a^2-b^2}-\frac {b^2 \left (4 a^2-b^2\right ) x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{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 (\frac {4 a^3}{a^2-b^2}-\frac {b \left (5 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{8 \left (a^2-b^2\right ) d}-\frac {\text {Subst}\left (\int \frac {\frac {a b^4 \left (3 a^2+b^2\right )}{a^2-b^2}-\frac {b^4 \left (5 a^2-b^2\right ) x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^2 \left (a^2-b^2\right ) 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 (\frac {4 a^3}{a^2-b^2}-\frac {b \left (5 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{8 \left (a^2-b^2\right ) d}-\frac {\text {Subst}\left (\int \left (-\frac {b^3 (-a+b) (3 a+b)}{2 (a+b)^2 (b-x)}-\frac {8 a^3 b^4}{(a-b)^2 (a+b)^2 (a+x)}+\frac {(3 a-b) b^3 (a+b)}{2 (a-b)^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^2 \left (a^2-b^2\right ) d} \\ & = \frac {b (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {(3 a-b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^3 b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 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 (\frac {4 a^3}{a^2-b^2}-\frac {b \left (5 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{8 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {(3 a-b) b \log (1+\sin (c+d x))}{(a-b)^3}+\frac {16 a^3 b^2 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {3 a+b}{(a+b)^2 (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {-3 a+b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \]

[In]

Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

((b*(3*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^3 - ((3*a - b)*b*Log[1 + Sin[c + d*x]])/(a - b)^3 + (16*a^3*b^2*L
og[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (3*a + b)/((a + b)^2*(-1 +
Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])^2) + (-3*a + b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\left (3 a -b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{3} b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-3 a -b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (3 a +b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\)	\(176\)
default	\(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\left (3 a -b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{3} b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-3 a -b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (3 a +b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\)	\(176\)
parallelrisch	\(\frac {8 b^{2} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+3 b \left (a +\frac {b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \left (\frac {3 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {b}{3}\right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (a -b \right ) \left (\left (a^{3}-a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\frac {\left (-a^{3}-a \,b^{2}\right ) \cos \left (4 d x +4 c \right )}{4}+\frac {\left (-5 a^{2} b +b^{3}\right ) \sin \left (3 d x +3 c \right )}{4}+\frac {\left (3 a^{2} b -7 b^{3}\right ) \sin \left (d x +c \right )}{4}-\frac {3 a^{3}}{4}+\frac {5 a \,b^{2}}{4}\right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(304\)
norman	\(\frac {-\frac {2 a \,b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 a \,b^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {4 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (11 a^{2}-7 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (11 a^{2}-7 b^{2}\right ) b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{3} b^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {\left (3 a -b \right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {\left (3 a +b \right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}\)	\(460\)
risch	\(\frac {3 i a b c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {2 i a^{3} b^{2} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {i b^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {3 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i a^{3} b^{2} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {i \left (8 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-5 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+16 i b^{2} a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-7 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+8 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+5 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} d}+\frac {3 \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 {\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 b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {\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 {a^{3} b^{2} \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 )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(752\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(3*a-b)/(a-b)^2/(1+sin(d*x+c))-1/16*(3*a-b)/(a-b)^3*b*ln(1+sin(d*x+c)
)+a^3/(a+b)^3*b^2/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(-3*a-b)/(a+b)^2/(sin(d*x+c)-
1)+1/16*(3*a+b)/(a+b)^3*b*ln(sin(d*x+c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a^{3} b^{2} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{4} b + 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{4} b - 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/16*(16*a^3*b^2*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^4*b + 8*a^3*b^2 + 6*a^2*b^3 - b^5)*cos(d*x + c)
^4*log(sin(d*x + c) + 1) + (3*a^4*b - 8*a^3*b^2 + 6*a^2*b^3 - b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a
^5 - 8*a^3*b^2 + 4*a*b^4 - 8*(a^5 - a^3*b^2)*cos(d*x + c)^2 - 2*(2*a^4*b - 4*a^2*b^3 + 2*b^5 - (5*a^4*b - 6*a^
2*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{3} b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a b - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a^{3} \sin \left (d x + c\right )^{2} - {\left (5 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 2 \, a b^{2} + {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*(16*a^3*b^2*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (3*a*b - b^2)*log(sin(d*x + c)
+ 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a*b + b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2
*(4*a^3*sin(d*x + c)^2 - (5*a^2*b - b^3)*sin(d*x + c)^3 - 2*a^3 - 2*a*b^2 + (3*a^2*b + 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))/d

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.79 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{3} b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (3 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} - 5 \, a^{4} b \sin \left (d x + c\right )^{3} + 6 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - b^{5} \sin \left (d x + c\right )^{3} + 4 \, a^{5} \sin \left (d x + c\right )^{2} - 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 2 \, a^{2} b^{3} \sin \left (d x + c\right ) - b^{5} \sin \left (d x + c\right ) - 2 \, a^{5} + 6 \, a^{3} b^{2} + 2 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*(16*a^3*b^3*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (3*a*b - b^2)*log(abs(si
n(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a*b + b^2)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3
*a*b^2 + b^3) + 2*(6*a^3*b^2*sin(d*x + c)^4 - 5*a^4*b*sin(d*x + c)^3 + 6*a^2*b^3*sin(d*x + c)^3 - b^5*sin(d*x
+ c)^3 + 4*a^5*sin(d*x + c)^2 - 16*a^3*b^2*sin(d*x + c)^2 + 3*a^4*b*sin(d*x + c) - 2*a^2*b^3*sin(d*x + c) - b^
5*sin(d*x + c) - 2*a^5 + 6*a^3*b^2 + 2*a*b^4)/((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 = 12.48 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.59 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (3\,a+b\right )}{8\,d\,{\left (a+b\right )}^3}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (11\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^2\,b+b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (11\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {3\,b}{8\,{\left (a-b\right )}^2}\right )}{d}+\frac {a^3\,b^2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

[In]

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

[Out]

(b*log(tan(c/2 + (d*x)/2) - 1)*(3*a + b))/(8*d*(a + b)^3) - ((tan(c/2 + (d*x)/2)^3*(11*a^2*b - 7*b^3))/(4*(a^4
+ b^4 - 2*a^2*b^2)) - (4*a^3*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^7*(3*a^2*b +
b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(11*a^2*b - 7*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (
2*a*b^2*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) + (2*a*b^2*tan(c/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2)
- (b*tan(c/2 + (d*x)/2)*(3*a^2 + b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 +
(d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b
)^3) + (3*b)/(8*(a - b)^2)))/d + (a^3*b^2*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^6 -
b^6 + 3*a^2*b^4 - 3*a^4*b^2))