3.1507 \(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=171 \[ -\frac {a^3 \csc (c+d x)}{d}+\frac {b \sec ^4(c+d x) \left (a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{8 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a (5 a-3 b) (a-b) \log (\sin (c+d x)+1)}{16 d} \]

[Out]

-a^3*csc(d*x+c)/d-3/16*a*(a+b)*(5*a+3*b)*ln(1-sin(d*x+c))/d+3*a^2*b*ln(sin(d*x+c))/d+3/16*a*(5*a-3*b)*(a-b)*ln
(1+sin(d*x+c))/d+1/4*b*sec(d*x+c)^4*(3*a^2+b^2+a*(3+1/b^2*a^2)*b*sin(d*x+c))/d+1/8*a*b*sec(d*x+c)^2*(12*a+(9+7
/b^2*a^2)*b*sin(d*x+c))/d

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Rubi [A]  time = 0.37, antiderivative size = 171, normalized size of antiderivative = 1.00, 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} \[ \frac {b \sec ^4(c+d x) \left (a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{8 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a (5 a-3 b) (a-b) \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Csc[c + d*x])/d) - (3*a*(a + b)*(5*a + 3*b)*Log[1 - Sin[c + d*x]])/(16*d) + (3*a^2*b*Log[Sin[c + d*x]])
/d + (3*a*(5*a - 3*b)*(a - b)*Log[1 + Sin[c + d*x]])/(16*d) + (b*Sec[c + d*x]^4*(3*a^2 + b^2 + a*(3 + a^2/b^2)
*b*Sin[c + d*x]))/(4*d) + (a*b*Sec[c + d*x]^2*(12*a + (9 + (7*a^2)/b^2)*b*Sin[c + d*x]))/(8*d)

Rule 12

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

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]

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 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]

Rubi steps

\begin {align*} \int \csc ^2(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^2 (a+x)^3}{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)^3}{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 (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}-\frac {b^5 \operatorname {Subst}\left (\int \frac {-4 a^3-12 a^2 x-3 a \left (3+\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 ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {8 a^3+24 a^2 x+a \left (9+\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 ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b^3 \operatorname {Subst}\left (\int \left (\frac {3 a (a+b) (5 a+3 b)}{2 b^3 (b-x)}+\frac {8 a^3}{b^2 x^2}+\frac {24 a^2}{b^2 x}+\frac {3 a (5 a-3 b) (a-b)}{2 b^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}+\frac {3 a (5 a-3 b) (a-b) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}\\ \end {align*}

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Mathematica [A]  time = 1.45, size = 161, normalized size = 0.94 \[ -\frac {16 a^3 \csc (c+d x)-48 a^2 b \log (\sin (c+d x))+\frac {(a+b)^2 (7 a+b)}{\sin (c+d x)-1}+\frac {(a-b)^2 (7 a-b)}{\sin (c+d x)+1}-\frac {(a+b)^3}{(\sin (c+d x)-1)^2}+\frac {(a-b)^3}{(\sin (c+d x)+1)^2}+3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))-3 a (5 a-3 b) (a-b) \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(16*a^3*Csc[c + d*x] + 3*a*(a + b)*(5*a + 3*b)*Log[1 - Sin[c + d*x]] - 48*a^2*b*Log[Sin[c + d*x]] - 3*a*
(5*a - 3*b)*(a - b)*Log[1 + Sin[c + d*x]] - (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(7*a + b))/(-1 + Sin[
c + d*x]) + (a - b)^3/(1 + Sin[c + d*x])^2 + ((a - b)^2*(7*a - b))/(1 + Sin[c + d*x]))/d

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fricas [A]  time = 0.47, size = 226, normalized size = 1.32 \[ \frac {48 \, a^{2} b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, a^{3} + 12 \, a b^{2} + 2 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (6 \, a^{2} b \cos \left (d x + c\right )^{2} + 3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]

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))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 210, normalized size = 1.23 \[ \frac {48 \, a^{2} b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {16 \, {\left (3 \, a^{2} b \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} - 7 \, a^{3} \sin \left (d x + c\right )^{3} - 9 \, a b^{2} \sin \left (d x + c\right )^{3} - 48 \, a^{2} b \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 15 \, a b^{2} \sin \left (d x + c\right ) + 36 \, a^{2} b + 2 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/16*(48*a^2*b*log(abs(sin(d*x + c))) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(5*a^3 +
8*a^2*b + 3*a*b^2)*log(abs(sin(d*x + c) - 1)) - 16*(3*a^2*b*sin(d*x + c) + a^3)/sin(d*x + c) + 2*(18*a^2*b*sin
(d*x + c)^4 - 7*a^3*sin(d*x + c)^3 - 9*a*b^2*sin(d*x + c)^3 - 48*a^2*b*sin(d*x + c)^2 + 9*a^3*sin(d*x + c) + 1
5*a*b^2*sin(d*x + c) + 36*a^2*b + 2*b^3)/(sin(d*x + c)^2 - 1)^2)/d

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maple [A]  time = 0.64, size = 221, normalized size = 1.29 \[ \frac {a^{3}}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 a^{3}}{8 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 a^{3}}{8 d \sin \left (d x +c \right )}+\frac {15 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{2} b}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{2} b}{2 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{2} b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {9 a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3}}{4 d \cos \left (d x +c \right )^{4}} \]

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))^3,x)

[Out]

1/4/d*a^3/sin(d*x+c)/cos(d*x+c)^4+5/8/d*a^3/sin(d*x+c)/cos(d*x+c)^2-15/8/d*a^3/sin(d*x+c)+15/8/d*a^3*ln(sec(d*
x+c)+tan(d*x+c))+3/4/d*a^2*b/cos(d*x+c)^4+3/2/d*a^2*b/cos(d*x+c)^2+3/d*a^2*b*ln(tan(d*x+c))+3/4/d*a*b^2*tan(d*
x+c)*sec(d*x+c)^3+9/8/d*a*b^2*tan(d*x+c)*sec(d*x+c)+9/8/d*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*b^3/cos(d*x+c)
^4

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maxima [A]  time = 0.34, size = 188, normalized size = 1.10 \[ \frac {48 \, a^{2} b \log \left (\sin \left (d x + c\right )\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (12 \, a^{2} b \sin \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{4} + 8 \, a^{3} - 5 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - 2 \, {\left (9 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

1/16*(48*a^2*b*log(sin(d*x + c)) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*log(sin(d*x + c) + 1) - 3*(5*a^3 + 8*a^2*b +
3*a*b^2)*log(sin(d*x + c) - 1) - 2*(12*a^2*b*sin(d*x + c)^3 + 3*(5*a^3 + 3*a*b^2)*sin(d*x + c)^4 + 8*a^3 - 5*(
5*a^3 + 3*a*b^2)*sin(d*x + c)^2 - 2*(9*a^2*b + b^3)*sin(d*x + c))/(sin(d*x + c)^5 - 2*sin(d*x + c)^3 + sin(d*x
+ c)))/d

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mupad [B]  time = 0.13, size = 182, normalized size = 1.06 \[ \frac {3\,a^2\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {15\,a^3}{8}+\frac {9\,a\,b^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {25\,a^3}{8}+\frac {15\,a\,b^2}{8}\right )+a^3-\sin \left (c+d\,x\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{4}\right )+\frac {3\,a^2\,b\,{\sin \left (c+d\,x\right )}^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}+\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (5\,a-3\,b\right )}{16\,d}-\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (5\,a+3\,b\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^3/(cos(c + d*x)^5*sin(c + d*x)^2),x)

[Out]

(3*a^2*b*log(sin(c + d*x)))/d - (sin(c + d*x)^4*((9*a*b^2)/8 + (15*a^3)/8) - sin(c + d*x)^2*((15*a*b^2)/8 + (2
5*a^3)/8) + a^3 - sin(c + d*x)*((9*a^2*b)/4 + b^3/4) + (3*a^2*b*sin(c + d*x)^3)/2)/(d*(sin(c + d*x) - 2*sin(c
+ d*x)^3 + sin(c + d*x)^5)) + (3*a*log(sin(c + d*x) + 1)*(a - b)*(5*a - 3*b))/(16*d) - (3*a*log(sin(c + d*x) -
1)*(a + b)*(5*a + 3*b))/(16*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)**2*sec(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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