3.1499 \(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=168 \[ -\frac{\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac{b \sec ^4(c+d x) \left (\frac{\left (a^2+b^2\right ) \sin (c+d x)}{b}+2 a\right )}{4 d}+\frac{b \sec ^2(c+d x) \left (b \left (\frac{7 a^2}{b^2}+3\right ) \sin (c+d x)+8 a\right )}{8 d}-\frac{a^2 \csc (c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.353948, antiderivative size = 168, 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} \[ -\frac{\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac{b \sec ^4(c+d x) \left (\frac{\left (a^2+b^2\right ) \sin (c+d x)}{b}+2 a\right )}{4 d}+\frac{b \sec ^2(c+d x) \left (b \left (\frac{7 a^2}{b^2}+3\right ) \sin (c+d x)+8 a\right )}{8 d}-\frac{a^2 \csc (c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Csc[c + d*x])/d) - ((15*a^2 + 16*a*b + 3*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + (2*a*b*Log[Sin[c + d*x]])
/d + ((15*a^2 - 16*a*b + 3*b^2)*Log[1 + Sin[c + d*x]])/(16*d) + (b*Sec[c + d*x]^2*(8*a + (3 + (7*a^2)/b^2)*b*S
in[c + d*x]))/(8*d) + (b*Sec[c + d*x]^4*(2*a + ((a^2 + b^2)*Sin[c + d*x])/b))/(4*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 ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{b^2 (a+x)^2}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^7 \operatorname{Subst}\left (\int \frac{(a+x)^2}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b \sec ^4(c+d x) \left (2 a+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}-\frac{b^5 \operatorname{Subst}\left (\int \frac{-4 a^2-8 a x-3 \left (1+\frac{a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{b \sec ^2(c+d x) \left (8 a+\left (3+\frac{7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac{b \sec ^4(c+d x) \left (2 a+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{8 a^2+16 a x+\left (3+\frac{7 a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{b \sec ^2(c+d x) \left (8 a+\left (3+\frac{7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac{b \sec ^4(c+d x) \left (2 a+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}+\frac{b^3 \operatorname{Subst}\left (\int \left (\frac{15 a^2+16 a b+3 b^2}{2 b^3 (b-x)}+\frac{8 a^2}{b^2 x^2}+\frac{16 a}{b^2 x}+\frac{15 a^2-16 a b+3 b^2}{2 b^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{a^2 \csc (c+d x)}{d}-\frac{\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{\left (15 a^2-16 a b+3 b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac{b \sec ^2(c+d x) \left (8 a+\left (3+\frac{7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac{b \sec ^4(c+d x) \left (2 a+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.87704, size = 162, normalized size = 0.96 \[ -\frac{\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))-\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)+16 a^2 \csc (c+d x)+\frac{(a+b) (7 a+3 b)}{\sin (c+d x)-1}+\frac{(7 a-3 b) (a-b)}{\sin (c+d x)+1}-\frac{(a+b)^2}{(\sin (c+d x)-1)^2}+\frac{(a-b)^2}{(\sin (c+d x)+1)^2}-32 a b \log (\sin (c+d x))}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(16*a^2*Csc[c + d*x] + (15*a^2 + 16*a*b + 3*b^2)*Log[1 - Sin[c + d*x]] - 32*a*b*Log[Sin[c + d*x]] - (15*a^2 -
16*a*b + 3*b^2)*Log[1 + Sin[c + d*x]] - (a + b)^2/(-1 + Sin[c + d*x])^2 + ((a + b)*(7*a + 3*b))/(-1 + Sin[c +
d*x]) + (a - b)^2/(1 + Sin[c + d*x])^2 + ((7*a - 3*b)*(a - b))/(1 + Sin[c + d*x]))/(16*d)

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Maple [A]  time = 0.088, size = 195, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,{a}^{2}}{8\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15\,{a}^{2}}{8\,d\sin \left ( dx+c \right ) }}+{\frac{15\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{ab}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{ab}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{ab\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}\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)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^2,x)

[Out]

1/4/d*a^2/sin(d*x+c)/cos(d*x+c)^4+5/8/d*a^2/sin(d*x+c)/cos(d*x+c)^2-15/8/d*a^2/sin(d*x+c)+15/8/d*a^2*ln(sec(d*
x+c)+tan(d*x+c))+1/2/d*a*b/cos(d*x+c)^4+1/d*a*b/cos(d*x+c)^2+2/d*a*b*ln(tan(d*x+c))+1/4/d*b^2*tan(d*x+c)*sec(d
*x+c)^3+3/8/d*b^2*tan(d*x+c)*sec(d*x+c)+3/8/d*b^2*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.01052, size = 220, normalized size = 1.31 \begin{align*} \frac{32 \, a b \log \left (\sin \left (d x + c\right )\right ) +{\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (8 \, a b \sin \left (d x + c\right )^{3} + 3 \,{\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4} - 12 \, a b \sin \left (d x + c\right ) - 5 \,{\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2} + 8 \, a^{2}\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(32*a*b*log(sin(d*x + c)) + (15*a^2 - 16*a*b + 3*b^2)*log(sin(d*x + c) + 1) - (15*a^2 + 16*a*b + 3*b^2)*l
og(sin(d*x + c) - 1) - 2*(8*a*b*sin(d*x + c)^3 + 3*(5*a^2 + b^2)*sin(d*x + c)^4 - 12*a*b*sin(d*x + c) - 5*(5*a
^2 + b^2)*sin(d*x + c)^2 + 8*a^2)/(sin(d*x + c)^5 - 2*sin(d*x + c)^3 + sin(d*x + c)))/d

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Fricas [A]  time = 2.11247, size = 516, normalized size = 3.07 \begin{align*} \frac{32 \, a b \cos \left (d x + c\right )^{4} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) +{\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) -{\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \,{\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, a^{2} + 4 \, b^{2} + 8 \,{\left (2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(32*a*b*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) + (15*a^2 - 16*a*b + 3*b^2)*cos(d*x + c)^4*log(
sin(d*x + c) + 1)*sin(d*x + c) - (15*a^2 + 16*a*b + 3*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d*x + c)
- 6*(5*a^2 + b^2)*cos(d*x + c)^4 + 2*(5*a^2 + b^2)*cos(d*x + c)^2 + 4*a^2 + 4*b^2 + 8*(2*a*b*cos(d*x + c)^2 +
a*b)*sin(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c))

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

[Out]

Timed out

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Giac [A]  time = 1.27147, size = 251, normalized size = 1.49 \begin{align*} \frac{32 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) +{\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{16 \,{\left (2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )} + \frac{2 \,{\left (12 \, a b \sin \left (d x + c\right )^{4} - 7 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, b^{2} \sin \left (d x + c\right )^{3} - 32 \, a b \sin \left (d x + c\right )^{2} + 9 \, a^{2} \sin \left (d x + c\right ) + 5 \, b^{2} \sin \left (d x + c\right ) + 24 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(32*a*b*log(abs(sin(d*x + c))) + (15*a^2 - 16*a*b + 3*b^2)*log(abs(sin(d*x + c) + 1)) - (15*a^2 + 16*a*b
+ 3*b^2)*log(abs(sin(d*x + c) - 1)) - 16*(2*a*b*sin(d*x + c) + a^2)/sin(d*x + c) + 2*(12*a*b*sin(d*x + c)^4 -
7*a^2*sin(d*x + c)^3 - 3*b^2*sin(d*x + c)^3 - 32*a*b*sin(d*x + c)^2 + 9*a^2*sin(d*x + c) + 5*b^2*sin(d*x + c)
+ 24*a*b)/(sin(d*x + c)^2 - 1)^2)/d