3.1498 \(\int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=126 \[ \frac{\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac{a^2 \log (\sin (c+d x))}{d}-\frac{a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}-\frac{a (4 a-3 b) \log (\sin (c+d x)+1)}{8 d}+\frac{a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.217093, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2837, 12, 1805, 823, 801} \[ \frac{\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac{a^2 \log (\sin (c+d x))}{d}-\frac{a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}-\frac{a (4 a-3 b) \log (\sin (c+d x)+1)}{8 d}+\frac{a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(4*a + 3*b)*Log[1 - Sin[c + d*x]])/(8*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 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 823

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] + Di
st[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 801

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]

Rubi steps

\begin{align*} \int \csc (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 (a+x)^2}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^6 \operatorname{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 \operatorname{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{\operatorname{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{\operatorname{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]  time = 0.917146, size = 137, normalized size = 1.09 \[ \frac{16 a^2 \log (\sin (c+d x))-\frac{(a+b) (5 a+b)}{\sin (c+d x)-1}+\frac{(a-b) (5 a-b)}{\sin (c+d x)+1}+\frac{(a+b)^2}{(\sin (c+d x)-1)^2}+\frac{(a-b)^2}{(\sin (c+d x)+1)^2}-2 a (4 a+3 b) \log (1-\sin (c+d x))-2 a (4 a-3 b) \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.089, size = 125, normalized size = 1. \begin{align*}{\frac{{a}^{2}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{ab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,ab\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{4\,d}}+{\frac{3\,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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.982631, size = 176, normalized size = 1.4 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.02736, size = 360, normalized size = 2.86 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Giac [A]  time = 1.23934, size = 181, normalized size = 1.44 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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