\(\int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1498]

Optimal result

Integrand size = 27, antiderivative size = 126 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a-3 b) \log (1+\sin (c+d x))}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d} \]

[Out]

-1/8*a*(4*a+3*b)*ln(1-sin(d*x+c))/d+a^2*ln(sin(d*x+c))/d-1/8*a*(4*a-3*b)*ln(1+sin(d*x+c))/d+1/4*a*sec(d*x+c)^2
*(2*a+3*b*sin(d*x+c))/d+1/4*sec(d*x+c)^4*(a^2+b^2+2*a*b*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2916, 12, 1819, 837, 815} \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}-\frac {a (4 a-3 b) \log (\sin (c+d x)+1)}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d} \]

[In]

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

[Out]

-1/8*(a*(4*a + 3*b)*Log[1 - Sin[c + d*x]])/d + (a^2*Log[Sin[c + d*x]])/d - (a*(4*a - 3*b)*Log[1 + Sin[c + d*x]
])/(8*d) + (a*Sec[c + d*x]^2*(2*a + 3*b*Sin[c + d*x]))/(4*d) + (Sec[c + d*x]^4*(a^2 + b^2 + 2*a*b*Sin[c + d*x]
))/(4*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 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && 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 {b (a+x)^2}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^6 \text {Subst}\left (\int \frac {(a+x)^2}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {b^4 \text {Subst}\left (\int \frac {-4 a^2-6 a x}{x \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {\text {Subst}\left (\int \frac {-8 a^2 b^2-6 a b^2 x}{x \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {\text {Subst}\left (\int \left (-\frac {a (4 a+3 b)}{b-x}-\frac {8 a^2}{x}+\frac {a (4 a-3 b)}{b+x}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a-3 b) \log (1+\sin (c+d x))}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.09 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-2 a (4 a+3 b) \log (1-\sin (c+d x))+16 a^2 \log (\sin (c+d x))-2 a (4 a-3 b) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}-\frac {(a+b) (5 a+b)}{-1+\sin (c+d x)}+\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(a-b) (5 a-b)}{1+\sin (c+d x)}}{16 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(99\)
default	\(\frac {a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(99\)
parallelrisch	\(\frac {-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{4}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (a -\frac {3 b}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-4 a^{2}-4 b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-3 a^{2}-b^{2}\right ) \cos \left (4 d x +4 c \right )+22 a b \sin \left (d x +c \right )+6 a b \sin \left (3 d x +3 c \right )+7 a^{2}+5 b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(227\)
risch	\(-\frac {i \left (4 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+16 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+11 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-11 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(235\)
norman	\(\frac {\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 b^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 a^{2}+b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 a^{2}+b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (4 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}-\frac {a \left (4 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\)	\(328\)

[In]

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

[Out]

1/d*(a^2*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+2*a*b*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x
+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+1/4*b^2/cos(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} + 2 \, {\left (3 \, a b \cos \left (d x + c\right )^{2} + 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

integrate(csc(d*x+c)*sec(d*x+c)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.03 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a b \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 5 \, a b \sin \left (d x + c\right ) - 3 \, a^{2} - b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right )^{4} - 3 \, a b \sin \left (d x + c\right )^{3} - 8 \, a^{2} \sin \left (d x + c\right )^{2} + 5 \, a b \sin \left (d x + c\right ) + 6 \, a^{2} + b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.04 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}+\frac {-\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {3\,a^2}{4}-\frac {3\,a\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {5\,a\,b\,\sin \left (c+d\,x\right )}{4}+\frac {b^2}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (4\,a+3\,b\right )}{8\,d}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (4\,a-3\,b\right )}{8\,d} \]

[In]

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

[Out]

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