3.1508 \(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=221 \[ -\frac{3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{3 a \left (a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac{3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac{b^2 \sec ^4(c+d x) \left (b \left (\frac{3 a^2}{b^2}+1\right ) \sin (c+d x)+a \left (\frac{a^2}{b^2}+3\right )\right )}{4 d}+\frac{b^2 \sec ^2(c+d x) \left (3 b \left (\frac{7 a^2}{b^2}+1\right ) \sin (c+d x)+4 a \left (\frac{2 a^2}{b^2}+3\right )\right )}{8 d}-\frac{3 a^2 b \csc (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x)}{2 d} \]

[Out]

(-3*a^2*b*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) - (3*(a + b)*(8*a^2 + 7*a*b + b^2)*Log[1 - Sin[c + d*x]
])/(16*d) + (3*a*(a^2 + b^2)*Log[Sin[c + d*x]])/d - (3*(a - b)*(8*a^2 - 7*a*b + b^2)*Log[1 + Sin[c + d*x]])/(1
6*d) + (b^2*Sec[c + d*x]^4*(a*(3 + a^2/b^2) + (1 + (3*a^2)/b^2)*b*Sin[c + d*x]))/(4*d) + (b^2*Sec[c + d*x]^2*(
4*a*(3 + (2*a^2)/b^2) + 3*(1 + (7*a^2)/b^2)*b*Sin[c + d*x]))/(8*d)

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Rubi [A]  time = 0.434452, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2837, 12, 1805, 1802} \[ -\frac{3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{3 a \left (a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac{3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac{b^2 \sec ^4(c+d x) \left (b \left (\frac{3 a^2}{b^2}+1\right ) \sin (c+d x)+a \left (\frac{a^2}{b^2}+3\right )\right )}{4 d}+\frac{b^2 \sec ^2(c+d x) \left (3 b \left (\frac{7 a^2}{b^2}+1\right ) \sin (c+d x)+4 a \left (\frac{2 a^2}{b^2}+3\right )\right )}{8 d}-\frac{3 a^2 b \csc (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*b*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) - (3*(a + b)*(8*a^2 + 7*a*b + b^2)*Log[1 - Sin[c + d*x]
])/(16*d) + (3*a*(a^2 + b^2)*Log[Sin[c + d*x]])/d - (3*(a - b)*(8*a^2 - 7*a*b + b^2)*Log[1 + Sin[c + d*x]])/(1
6*d) + (b^2*Sec[c + d*x]^4*(a*(3 + a^2/b^2) + (1 + (3*a^2)/b^2)*b*Sin[c + d*x]))/(4*d) + (b^2*Sec[c + d*x]^2*(
4*a*(3 + (2*a^2)/b^2) + 3*(1 + (7*a^2)/b^2)*b*Sin[c + d*x]))/(8*d)

Rule 2837

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*x)/b)^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]

Rule 12

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

Rule 1805

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 1802

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

Rubi steps

\begin{align*} \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{b^3 (a+x)^3}{x^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^8 \operatorname{Subst}\left (\int \frac{(a+x)^3}{x^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^2 \sec ^4(c+d x) \left (a \left (3+\frac{a^2}{b^2}\right )+\left (1+\frac{3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}-\frac{b^6 \operatorname{Subst}\left (\int \frac{-4 a^3-12 a^2 x-4 a \left (3+\frac{a^2}{b^2}\right ) x^2-3 \left (1+\frac{3 a^2}{b^2}\right ) x^3}{x^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{b^2 \sec ^4(c+d x) \left (a \left (3+\frac{a^2}{b^2}\right )+\left (1+\frac{3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac{b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac{2 a^2}{b^2}\right )+3 \left (1+\frac{7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{8 a^3+24 a^2 x+8 a \left (3+\frac{2 a^2}{b^2}\right ) x^2+3 \left (1+\frac{7 a^2}{b^2}\right ) x^3}{x^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{b^2 \sec ^4(c+d x) \left (a \left (3+\frac{a^2}{b^2}\right )+\left (1+\frac{3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac{b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac{2 a^2}{b^2}\right )+3 \left (1+\frac{7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac{b^4 \operatorname{Subst}\left (\int \left (\frac{3 (a+b) \left (8 a^2+7 a b+b^2\right )}{2 b^4 (b-x)}+\frac{8 a^3}{b^2 x^3}+\frac{24 a^2}{b^2 x^2}+\frac{24 a \left (a^2+b^2\right )}{b^4 x}+\frac{3 (a-b) \left (-8 a^2+7 a b-b^2\right )}{2 b^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{3 a^2 b \csc (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x)}{2 d}-\frac{3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{3 a \left (a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac{3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac{b^2 \sec ^4(c+d x) \left (a \left (3+\frac{a^2}{b^2}\right )+\left (1+\frac{3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac{b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac{2 a^2}{b^2}\right )+3 \left (1+\frac{7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}\\ \end{align*}

Mathematica [A]  time = 3.32417, size = 190, normalized size = 0.86 \[ \frac{48 a \left (a^2+b^2\right ) \log (\sin (c+d x))-3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))-3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (\sin (c+d x)+1)-48 a^2 b \csc (c+d x)-8 a^3 \csc ^2(c+d x)-\frac{3 (a+b)^2 (3 a+b)}{\sin (c+d x)-1}+\frac{3 (a-b)^2 (3 a-b)}{\sin (c+d x)+1}+\frac{(a+b)^3}{(\sin (c+d x)-1)^2}+\frac{(a-b)^3}{(\sin (c+d x)+1)^2}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-48*a^2*b*Csc[c + d*x] - 8*a^3*Csc[c + d*x]^2 - 3*(a + b)*(8*a^2 + 7*a*b + b^2)*Log[1 - Sin[c + d*x]] + 48*a*
(a^2 + b^2)*Log[Sin[c + d*x]] - 3*(a - b)*(8*a^2 - 7*a*b + b^2)*Log[1 + Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c
+ d*x])^2 - (3*(a + b)^2*(3*a + b))/(-1 + Sin[c + d*x]) + (a - b)^3/(1 + Sin[c + d*x])^2 + (3*(a - b)^2*(3*a -
b))/(1 + Sin[c + d*x]))/(16*d)

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Maple [A]  time = 0.12, size = 285, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{a}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{2}b}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{15\,{a}^{2}b}{8\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{45\,{a}^{2}b}{8\,d\sin \left ( dx+c \right ) }}+{\frac{45\,{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,a{b}^{2}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,a{b}^{2}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{a{b}^{2}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{3}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*a^3/sin(d*x+c)^2/cos(d*x+c)^4+3/4/d*a^3/sin(d*x+c)^2/cos(d*x+c)^2-3/2/d*a^3/sin(d*x+c)^2+3/d*a^3*ln(tan(
d*x+c))+3/4/d*a^2*b/sin(d*x+c)/cos(d*x+c)^4+15/8/d*a^2*b/sin(d*x+c)/cos(d*x+c)^2-45/8/d*a^2*b/sin(d*x+c)+45/8/
d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*a*b^2/cos(d*x+c)^4+3/2/d*a*b^2/cos(d*x+c)^2+3/d*a*b^2*ln(tan(d*x+c))+1
/4/d*b^3*tan(d*x+c)*sec(d*x+c)^3+3/8/d*b^3*tan(d*x+c)*sec(d*x+c)+3/8/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.00325, size = 293, normalized size = 1.33 \begin{align*} -\frac{3 \,{\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \,{\left (a^{3} + a b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac{2 \,{\left (3 \,{\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{5} + 12 \,{\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )^{4} + 24 \, a^{2} b \sin \left (d x + c\right ) - 5 \,{\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{3} + 4 \, a^{3} - 18 \,{\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )^{2}\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*(8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*log(sin(d*x + c) + 1) + 3*(8*a^3 + 15*a^2*b + 8*a*b^2 + b^3)*log(s
in(d*x + c) - 1) - 48*(a^3 + a*b^2)*log(sin(d*x + c)) + 2*(3*(15*a^2*b + b^3)*sin(d*x + c)^5 + 12*(a^3 + a*b^2
)*sin(d*x + c)^4 + 24*a^2*b*sin(d*x + c) - 5*(15*a^2*b + b^3)*sin(d*x + c)^3 + 4*a^3 - 18*(a^3 + a*b^2)*sin(d*
x + c)^2)/(sin(d*x + c)^6 - 2*sin(d*x + c)^4 + sin(d*x + c)^2))/d

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Fricas [A]  time = 2.2876, size = 791, normalized size = 3.58 \begin{align*} \frac{24 \,{\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, a^{3} - 12 \, a b^{2} - 12 \,{\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \,{\left ({\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{6} -{\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 3 \,{\left ({\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{6} -{\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{6} -{\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \,{\left (15 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b - 2 \, b^{3} -{\left (15 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(24*(a^3 + a*b^2)*cos(d*x + c)^4 - 4*a^3 - 12*a*b^2 - 12*(a^3 + a*b^2)*cos(d*x + c)^2 + 48*((a^3 + a*b^2)
*cos(d*x + c)^6 - (a^3 + a*b^2)*cos(d*x + c)^4)*log(1/2*sin(d*x + c)) - 3*((8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*
cos(d*x + c)^6 - (8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 3*((8*a^3 + 15*a^2
*b + 8*a*b^2 + b^3)*cos(d*x + c)^6 - (8*a^3 + 15*a^2*b + 8*a*b^2 + b^3)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1)
+ 2*(3*(15*a^2*b + b^3)*cos(d*x + c)^4 - 6*a^2*b - 2*b^3 - (15*a^2*b + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*
cos(d*x + c)^6 - d*cos(d*x + c)^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.38735, size = 324, normalized size = 1.47 \begin{align*} -\frac{3 \,{\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \,{\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 48 \,{\left (a^{3} + a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac{2 \,{\left (45 \, a^{2} b \sin \left (d x + c\right )^{5} + 3 \, b^{3} \sin \left (d x + c\right )^{5} + 12 \, a^{3} \sin \left (d x + c\right )^{4} + 12 \, a b^{2} \sin \left (d x + c\right )^{4} - 75 \, a^{2} b \sin \left (d x + c\right )^{3} - 5 \, b^{3} \sin \left (d x + c\right )^{3} - 18 \, a^{3} \sin \left (d x + c\right )^{2} - 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} b \sin \left (d x + c\right ) + 4 \, a^{3}\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*(8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*log(abs(sin(d*x + c) + 1)) + 3*(8*a^3 + 15*a^2*b + 8*a*b^2 + b^3)*
log(abs(sin(d*x + c) - 1)) - 48*(a^3 + a*b^2)*log(abs(sin(d*x + c))) + 2*(45*a^2*b*sin(d*x + c)^5 + 3*b^3*sin(
d*x + c)^5 + 12*a^3*sin(d*x + c)^4 + 12*a*b^2*sin(d*x + c)^4 - 75*a^2*b*sin(d*x + c)^3 - 5*b^3*sin(d*x + c)^3
- 18*a^3*sin(d*x + c)^2 - 18*a*b^2*sin(d*x + c)^2 + 24*a^2*b*sin(d*x + c) + 4*a^3)/(sin(d*x + c)^3 - sin(d*x +
c))^2)/d