3.14.50 \(\int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1350]

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

[Out]

-csc(d*x+c)/a/d-1/4*(3*a+4*b)*ln(1-sin(d*x+c))/(a+b)^2/d-b*ln(sin(d*x+c))/a^2/d+1/4*(3*a-4*b)*ln(1+sin(d*x+c))
/(a-b)^2/d+b^5*ln(a+b*sin(d*x+c))/a^2/(a^2-b^2)^2/d+1/4/(a+b)/d/(1-sin(d*x+c))-1/4/(a-b)/d/(1+sin(d*x+c))

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Rubi [A]
time = 0.18, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 908} \begin {gather*} \frac {b^5 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^2}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {1}{4 d (a+b) (1-\sin (c+d x))}-\frac {1}{4 d (a-b) (\sin (c+d x)+1)}-\frac {(3 a+4 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(3 a-4 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2}-\frac {\csc (c+d x)}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) - ((3*a + 4*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) - (b*Log[Sin[c + d*x]])/(a^2*d) +
((3*a - 4*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) + (b^5*Log[a + b*Sin[c + d*x]])/(a^2*(a^2 - b^2)^2*d) + 1/
(4*(a + b)*d*(1 - Sin[c + d*x])) - 1/(4*(a - b)*d*(1 + Sin[c + d*x]))

Rule 12

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

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

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/b)*x)^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 \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^3 \text {Subst}\left (\int \frac {b^2}{x^2 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^5 \text {Subst}\left (\int \frac {1}{x^2 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^5 \text {Subst}\left (\int \left (\frac {1}{4 b^4 (a+b) (b-x)^2}+\frac {3 a+4 b}{4 b^5 (a+b)^2 (b-x)}+\frac {1}{a b^4 x^2}-\frac {1}{a^2 b^4 x}+\frac {1}{a^2 (a-b)^2 (a+b)^2 (a+x)}+\frac {1}{4 (a-b) b^4 (b+x)^2}+\frac {3 a-4 b}{4 (a-b)^2 b^5 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\csc (c+d x)}{a d}-\frac {(3 a+4 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {(3 a-4 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^5 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}-\frac {1}{4 (a-b) d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 174, normalized size = 1.02 \begin {gather*} -\frac {\csc (c+d x) (a+b \sin (c+d x)) \left (\frac {4 \csc (c+d x)}{a}+\frac {(3 a+4 b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {4 b \log (\sin (c+d x))}{a^2}-\frac {(3 a-4 b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 b^5 \log (a+b \sin (c+d x))}{a^2 (a-b)^2 (a+b)^2}+\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}\right )}{4 d (b+a \csc (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Csc[c + d*x]*(a + b*Sin[c + d*x])*((4*Csc[c + d*x])/a + ((3*a + 4*b)*Log[1 - Sin[c + d*x]])/(a + b)^2 +
(4*b*Log[Sin[c + d*x]])/a^2 - ((3*a - 4*b)*Log[1 + Sin[c + d*x]])/(a - b)^2 - (4*b^5*Log[a + b*Sin[c + d*x]])/
(a^2*(a - b)^2*(a + b)^2) + 1/((a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x]))))/(d*(b + a*Csc[c
+ d*x]))

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Maple [A]
time = 0.49, size = 152, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a -4 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a -4 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\)	\(152\)
default	\(\frac {-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a -4 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a -4 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\)	\(152\)
norman	\(\frac {-\frac {1}{2 a d}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}+\frac {\left (3 a^{2}-b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a \left (a^{2}-b^{2}\right )}+\frac {\left (3 a^{2}-b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a \left (a^{2}-b^{2}\right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b^{5} \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}+\frac {\left (3 a -4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (3 a +4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\)	\(315\)
risch	\(-\frac {2 i b^{5} x}{a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {3 i a c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {2 i b c}{a^{2} d}+\frac {2 i b x}{a^{2}}+\frac {2 i b x}{a^{2}-2 a b +b^{2}}+\frac {2 i b c}{\left (a^{2}-2 a b +b^{2}\right ) d}-\frac {3 i a x}{2 \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 i a c}{2 \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {2 i b x}{a^{2}+2 a b +b^{2}}+\frac {2 i b c}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {3 i a x}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 i b^{5} c}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {i \left (3 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{2}-4 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-2 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(597\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/a/sin(d*x+c)-1/a^2*b*ln(sin(d*x+c))+b^5/a^2/(a+b)^2/(a-b)^2*ln(a+b*sin(d*x+c))-1/(4*a-4*b)/(1+sin(d*x+
c))+1/4*(3*a-4*b)/(a-b)^2*ln(1+sin(d*x+c))-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a+b)^2*(-3*a-4*b)*ln(sin(d*x+c)-1))

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Maxima [A]
time = 0.29, size = 200, normalized size = 1.17 \begin {gather*} \frac {\frac {4 \, b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}} + \frac {{\left (3 \, a - 4 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (3 \, a + 4 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a b \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2} - 2 \, b^{2}\right )}}{{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )} - \frac {4 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.82, size = 287, normalized size = 1.68 \begin {gather*} \frac {4 \, b^{5} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} - 4 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (3 \, a^{5} + 2 \, a^{4} b - 5 \, a^{3} b^{2} - 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (3 \, a^{5} - 2 \, a^{4} b - 5 \, a^{3} b^{2} + 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \, {\left (3 \, a^{5} - 5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Integral(csc(c + d*x)**2*sec(c + d*x)**3/(a + b*sin(c + d*x)), x)

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Giac [A]
time = 0.48, size = 279, normalized size = 1.63 \begin {gather*} \frac {\frac {12 \, b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}} + \frac {3 \, {\left (3 \, a - 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {3 \, {\left (3 \, a + 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (2 \, b^{5} \sin \left (d x + c\right )^{3} - 9 \, a^{5} \sin \left (d x + c\right )^{2} + 15 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 6 \, a b^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 3 \, a^{2} b^{3} \sin \left (d x + c\right ) - 2 \, b^{5} \sin \left (d x + c\right ) + 6 \, a^{5} - 12 \, a^{3} b^{2} + 6 \, a b^{4}\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 12.40, size = 195, normalized size = 1.14 \begin {gather*} \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (3\,a-4\,b\right )}{4\,d\,{\left (a-b\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {3}{4\,\left (a+b\right )}\right )}{d}-\frac {\frac {1}{a}+\frac {b\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (3\,a^2-2\,b^2\right )}{2\,a\,\left (a^2-b^2\right )}}{d\,\left (\sin \left (c+d\,x\right )-{\sin \left (c+d\,x\right )}^3\right )}-\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2\,d}+\frac {b^5\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a^2\,d\,{\left (a^2-b^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)

[Out]

(log(sin(c + d*x) + 1)*(3*a - 4*b))/(4*d*(a - b)^2) - (log(sin(c + d*x) - 1)*(b/(4*(a + b)^2) + 3/(4*(a + b)))
)/d - (1/a + (b*sin(c + d*x))/(2*(a^2 - b^2)) - (sin(c + d*x)^2*(3*a^2 - 2*b^2))/(2*a*(a^2 - b^2)))/(d*(sin(c
+ d*x) - sin(c + d*x)^3)) - (b*log(sin(c + d*x)))/(a^2*d) + (b^5*log(a + b*sin(c + d*x)))/(a^2*d*(a^2 - b^2)^2
)

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