3.1455 \(\int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=100 \[ \frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d}-\frac {\left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d}-\frac {2 a b \cot (c+d x)}{d} \]

[Out]

-1/2*(3*a^2+2*b^2)*arctanh(cos(d*x+c))/d-2*a*b*cot(d*x+c)/d+1/2*(3*a^2+2*b^2)*sec(d*x+c)/d-1/2*a^2*csc(d*x+c)^
2*sec(d*x+c)/d+2*a*b*tan(d*x+c)/d

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Rubi [A]  time = 0.25, antiderivative size = 124, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2911, 2620, 14, 3201, 446, 78, 51, 63, 206} \[ \frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d}-\frac {\left (3 a^2+2 b^2\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x) \tanh ^{-1}\left (\sqrt {\cos ^2(c+d x)}\right )}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d}-\frac {2 a b \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*Cot[c + d*x])/d + ((3*a^2 + 2*b^2)*Sec[c + d*x])/(2*d) - ((3*a^2 + 2*b^2)*ArcTanh[Sqrt[Cos[c + d*x]^2]
]*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x])/(2*d) - (a^2*Csc[c + d*x]^2*Sec[c + d*x])/(2*d) + (2*a*b*Tan[c + d*x])/d

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 3201

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

Rubi steps

\begin {align*} \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \csc ^2(c+d x) \sec ^2(c+d x) \, dx+\int \csc ^3(c+d x) \sec ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {(2 a b) \operatorname {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (\sqrt {\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {a^2+b^2 x^2}{x^3 \left (1-x^2\right )^{3/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {(2 a b) \operatorname {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (\sqrt {\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {a^2+b^2 x}{(1-x)^{3/2} x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac {2 a b \cot (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d}-\frac {\left (\left (-3 a^2-2 b^2\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{(1-x)^{3/2} x} \, dx,x,\sin ^2(c+d x)\right )}{4 d}\\ &=-\frac {2 a b \cot (c+d x)}{d}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d}-\frac {\left (\left (-3 a^2-2 b^2\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\sin ^2(c+d x)\right )}{4 d}\\ &=-\frac {2 a b \cot (c+d x)}{d}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {\left (\left (-3 a^2-2 b^2\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\cos ^2(c+d x)}\right )}{2 d}\\ &=-\frac {2 a b \cot (c+d x)}{d}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d}-\frac {\left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\sqrt {\cos ^2(c+d x)}\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 0.48, size = 238, normalized size = 2.38 \[ \frac {\csc ^4(c+d x) \left (-2 \left (3 a^2+2 b^2\right ) \cos (2 (c+d x))-\left (3 a^2+2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 a^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-3 a^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2+8 a b \sin (c+d x)-8 a b \sin (3 (c+d x))+2 b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2\right )}{2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^4*(2*a^2 + 4*b^2 - 2*(3*a^2 + 2*b^2)*Cos[2*(c + d*x)] + 3*a^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)
/2]] + 2*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - (3*a^2 + 2*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2]] - Lo
g[Sin[(c + d*x)/2]]) - 3*a^2*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 2*b^2*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)
/2]] + 8*a*b*Sin[c + d*x] - 8*a*b*Sin[3*(c + d*x)]))/(2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))

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fricas [A]  time = 0.46, size = 186, normalized size = 1.86 \[ \frac {2 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, a^{2} - 4 \, b^{2} - {\left ({\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 8 \, {\left (2 \, a b \cos \left (d x + c\right )^{2} - a b\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 157, normalized size = 1.57 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {16 \, {\left (2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} + b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*tan(1/2*d*x + 1/2*c) + 4*(3*a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))
) - 16*(2*a*b*tan(1/2*d*x + 1/2*c) + a^2 + b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (18*a^2*tan(1/2*d*x + 1/2*c)^2
+ 12*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^2)/d

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maple [A]  time = 0.56, size = 140, normalized size = 1.40 \[ -\frac {a^{2}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3 a^{2}}{2 d \cos \left (d x +c \right )}+\frac {3 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}+\frac {2 a b}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {4 a b \cot \left (d x +c \right )}{d}+\frac {b^{2}}{d \cos \left (d x +c \right )}+\frac {b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x)

[Out]

-1/2/d*a^2/sin(d*x+c)^2/cos(d*x+c)+3/2/d*a^2/cos(d*x+c)+3/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))+2/d*a*b/sin(d*x+c)
/cos(d*x+c)-4*a*b*cot(d*x+c)/d+1/d*b^2/cos(d*x+c)+1/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.45, size = 123, normalized size = 1.23 \[ \frac {a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, b^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, a b {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 11.85, size = 148, normalized size = 1.48 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{2}+b^2\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {17\,a^2}{2}+8\,b^2\right )-\frac {a^2}{2}+20\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^2/(cos(c + d*x)^2*sin(c + d*x)^3),x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (log(tan(c/2 + (d*x)/2))*((3*a^2)/2 + b^2))/d + (tan(c/2 + (d*x)/2)^2*((17*
a^2)/2 + 8*b^2) - a^2/2 + 20*a*b*tan(c/2 + (d*x)/2)^3 - 4*a*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 -
4*tan(c/2 + (d*x)/2)^4)) + (a*b*tan(c/2 + (d*x)/2))/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)**3*sec(d*x+c)**2*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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