3.1500 \(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=185 \[ -\frac{\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac{\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac{\left (12 a^2-15 a b+4 b^2\right ) \log (\sin (c+d x)+1)}{8 d}+\frac{\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac{\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a b \csc (c+d x)}{d} \]

[Out]

(-2*a*b*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - ((12*a^2 + 15*a*b + 4*b^2)*Log[1 - Sin[c + d*x]])/(8*d)
+ ((3*a^2 + b^2)*Log[Sin[c + d*x]])/d - ((12*a^2 - 15*a*b + 4*b^2)*Log[1 + Sin[c + d*x]])/(8*d) + (Sec[c + d*
x]^4*(a^2 + b^2 + 2*a*b*Sin[c + d*x]))/(4*d) + (Sec[c + d*x]^2*(2*(2*a^2 + b^2) + 7*a*b*Sin[c + d*x]))/(4*d)

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Rubi [A]  time = 0.38378, antiderivative size = 185, 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} \[ -\frac{\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac{\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac{\left (12 a^2-15 a b+4 b^2\right ) \log (\sin (c+d x)+1)}{8 d}+\frac{\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac{\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a b \csc (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(-2*a*b*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - ((12*a^2 + 15*a*b + 4*b^2)*Log[1 - Sin[c + d*x]])/(8*d)
+ ((3*a^2 + b^2)*Log[Sin[c + d*x]])/d - ((12*a^2 - 15*a*b + 4*b^2)*Log[1 + Sin[c + d*x]])/(8*d) + (Sec[c + d*
x]^4*(a^2 + b^2 + 2*a*b*Sin[c + d*x]))/(4*d) + (Sec[c + d*x]^2*(2*(2*a^2 + b^2) + 7*a*b*Sin[c + d*x]))/(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 ^3(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^3 (a+x)^2}{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)^2}{x^3 \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^6 \operatorname{Subst}\left (\int \frac{-4 a^2-8 a x-4 \left (1+\frac{a^2}{b^2}\right ) x^2-\frac{6 a x^3}{b^2}}{x^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac{\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{8 a^2+16 a x+8 \left (1+\frac{2 a^2}{b^2}\right ) x^2+\frac{14 a x^3}{b^2}}{x^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac{\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}+\frac{b^4 \operatorname{Subst}\left (\int \left (\frac{12 a^2+15 a b+4 b^2}{b^4 (b-x)}+\frac{8 a^2}{b^2 x^3}+\frac{16 a}{b^2 x^2}+\frac{8 \left (3 a^2+b^2\right )}{b^4 x}+\frac{-12 a^2+15 a b-4 b^2}{b^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{2 a b \csc (c+d x)}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac{\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac{\left (12 a^2-15 a b+4 b^2\right ) \log (1+\sin (c+d x))}{8 d}+\frac{\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac{\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 3.75378, size = 182, normalized size = 0.98 \[ \frac{-2 \left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))+16 \left (3 a^2+b^2\right ) \log (\sin (c+d x))-2 \left (12 a^2-15 a b+4 b^2\right ) \log (\sin (c+d x)+1)-8 a^2 \csc ^2(c+d x)+\frac{(a-b)^2}{(\sin (c+d x)+1)^2}+\frac{(9 a-5 b) (a-b)}{\sin (c+d x)+1}-\frac{(a+b) (9 a+5 b)}{\sin (c+d x)-1}+\frac{(a+b)^2}{(\sin (c+d x)-1)^2}-32 a b \csc (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(-32*a*b*Csc[c + d*x] - 8*a^2*Csc[c + d*x]^2 - 2*(12*a^2 + 15*a*b + 4*b^2)*Log[1 - Sin[c + d*x]] + 16*(3*a^2 +
b^2)*Log[Sin[c + d*x]] - 2*(12*a^2 - 15*a*b + 4*b^2)*Log[1 + Sin[c + d*x]] + (a + b)^2/(-1 + Sin[c + d*x])^2
- ((a + b)*(9*a + 5*b))/(-1 + Sin[c + d*x]) + (a - b)^2/(1 + Sin[c + d*x])^2 + ((9*a - 5*b)*(a - b))/(1 + Sin[
c + d*x]))/(16*d)

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Maple [A]  time = 0.098, size = 209, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{a}^{2}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{ab}{2\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,ab}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15\,ab}{4\,d\sin \left ( dx+c \right ) }}+{\frac{15\,ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{b}^{2}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}\ln \left ( \tan \left ( dx+c \right ) \right ) }{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))^2,x)

[Out]

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

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Maxima [A]  time = 0.992481, size = 247, normalized size = 1.34 \begin{align*} -\frac{{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \,{\left (3 \, a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac{2 \,{\left (15 \, a b \sin \left (d x + c\right )^{5} - 25 \, a b \sin \left (d x + c\right )^{3} + 2 \,{\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4} + 8 \, a b \sin \left (d x + c\right ) - 3 \,{\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{8 \, 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))^2,x, algorithm="maxima")

[Out]

-1/8*((12*a^2 - 15*a*b + 4*b^2)*log(sin(d*x + c) + 1) + (12*a^2 + 15*a*b + 4*b^2)*log(sin(d*x + c) - 1) - 8*(3
*a^2 + b^2)*log(sin(d*x + c)) + 2*(15*a*b*sin(d*x + c)^5 - 25*a*b*sin(d*x + c)^3 + 2*(3*a^2 + b^2)*sin(d*x + c
)^4 + 8*a*b*sin(d*x + c) - 3*(3*a^2 + b^2)*sin(d*x + c)^2 + 2*a^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 = 2.25345, size = 683, normalized size = 3.69 \begin{align*} \frac{4 \,{\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2} + 8 \,{\left ({\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{6} -{\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) -{\left ({\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{6} -{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{6} -{\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (15 \, a b \cos \left (d x + c\right )^{4} - 5 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \,{\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))^2,x, algorithm="fricas")

[Out]

1/8*(4*(3*a^2 + b^2)*cos(d*x + c)^4 - 2*(3*a^2 + b^2)*cos(d*x + c)^2 - 2*a^2 - 2*b^2 + 8*((3*a^2 + b^2)*cos(d*
x + c)^6 - (3*a^2 + b^2)*cos(d*x + c)^4)*log(1/2*sin(d*x + c)) - ((12*a^2 - 15*a*b + 4*b^2)*cos(d*x + c)^6 - (
12*a^2 - 15*a*b + 4*b^2)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - ((12*a^2 + 15*a*b + 4*b^2)*cos(d*x + c)^6 - (
12*a^2 + 15*a*b + 4*b^2)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(15*a*b*cos(d*x + c)^4 - 5*a*b*cos(d*x + c
)^2 - 2*a*b)*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27644, size = 257, normalized size = 1.39 \begin{align*} -\frac{{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 8 \,{\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac{2 \,{\left (15 \, a b \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{4} + 2 \, b^{2} \sin \left (d x + c\right )^{4} - 25 \, a b \sin \left (d x + c\right )^{3} - 9 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} + 8 \, a b \sin \left (d x + c\right ) + 2 \, a^{2}\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{8 \, 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))^2,x, algorithm="giac")

[Out]

-1/8*((12*a^2 - 15*a*b + 4*b^2)*log(abs(sin(d*x + c) + 1)) + (12*a^2 + 15*a*b + 4*b^2)*log(abs(sin(d*x + c) -
1)) - 8*(3*a^2 + b^2)*log(abs(sin(d*x + c))) + 2*(15*a*b*sin(d*x + c)^5 + 6*a^2*sin(d*x + c)^4 + 2*b^2*sin(d*x
+ c)^4 - 25*a*b*sin(d*x + c)^3 - 9*a^2*sin(d*x + c)^2 - 3*b^2*sin(d*x + c)^2 + 8*a*b*sin(d*x + c) + 2*a^2)/(s
in(d*x + c)^3 - sin(d*x + c))^2)/d