3.2.94 \(\int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [194]

Optimal. Leaf size=149 \[ -\frac {\log (1-\sin (c+d x))}{2 (a+b)^3 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b)^3 d}+\frac {a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {a}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a^2+b^2}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2800, 815} \begin {gather*} -\frac {a}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {a^2+b^2}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)^3}-\frac {\log (\sin (c+d x)+1)}{2 d (a-b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 815

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]

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]
time = 1.46, size = 213, normalized size = 1.43 \begin {gather*} \frac {-\frac {\log (1-\sin (c+d x))}{(a+b)^2}+\frac {\log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 a b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {2 b}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}+a \left (\frac {\log (1-\sin (c+d x))}{(a+b)^3}-\frac {\log (1+\sin (c+d x))}{(a-b)^3}+\frac {b \left (2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))+\frac {\left (a^2-b^2\right ) \left (-5 a^2+b^2-4 a b \sin (c+d x)\right )}{(a+b \sin (c+d x))^2}\right )}{\left (a^2-b^2\right )^3}\right )}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.52, size = 134, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {a}{2 \left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a^{2}+b^{2}}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a \left (a^{2}+3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}}{d}\)	\(134\)
default	\(\frac {-\frac {a}{2 \left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a^{2}+b^{2}}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a \left (a^{2}+3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}}{d}\)	\(134\)
risch	\(\frac {i x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {i c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i c}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {2 i a^{3} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {2 i a^{3} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {6 i a \,b^{2} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {6 i a \,b^{2} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {2 \left (-i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+i b^{3} {\mathrm e}^{i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} \left (a^{2}-b^{2}\right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(601\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 228, normalized size = 1.53 \begin {gather*} \frac {\frac {2 \, {\left (a^{3} + 3 \, a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {3 \, a^{3} + a b^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (143) = 286\).
time = 0.42, size = 462, normalized size = 3.10 \begin {gather*} \frac {3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4} - 2 \, {\left (a^{5} + 4 \, a^{3} b^{2} + 3 \, a b^{4} - {\left (a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{5} + 3 \, a^{4} b + 4 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{5} - 3 \, a^{4} b + 4 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5} - {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} b - b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a^5 - 2*a^3*b^2 - a*b^4 - 2*(a^5 + 4*a^3*b^2 + 3*a*b^4 - (a^3*b^2 + 3*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b
+ 3*a^2*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (a^5 + 3*a^4*b + 4*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4 + b^5 -
(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*sin(d*x + c))
*log(sin(d*x + c) + 1) + (a^5 - 3*a^4*b + 4*a^3*b^2 - 4*a^2*b^3 + 3*a*b^4 - b^5 - (a^3*b^2 - 3*a^2*b^3 + 3*a*b
^4 - b^5)*cos(d*x + c)^2 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 2*
(a^4*b - b^5)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^7*b - 3*a^5*b^3 +
3*a^3*b^5 - a*b^7)*d*sin(d*x + c) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)/(a + b*sin(c + d*x))**3, x)

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Giac [A]
time = 13.36, size = 257, normalized size = 1.72 \begin {gather*} \frac {\frac {2 \, {\left (a^{3} b + 3 \, a b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {3 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 9 \, a b^{4} \sin \left (d x + c\right )^{2} + 8 \, a^{4} b \sin \left (d x + c\right ) + 18 \, a^{2} b^{3} \sin \left (d x + c\right ) - 2 \, b^{5} \sin \left (d x + c\right ) + 6 \, a^{5} + 7 \, a^{3} b^{2} - a b^{4}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^3*b + 3*a*b^3)*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - log(abs(sin(d*x
+ c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (3*a
^3*b^2*sin(d*x + c)^2 + 9*a*b^4*sin(d*x + c)^2 + 8*a^4*b*sin(d*x + c) + 18*a^2*b^3*sin(d*x + c) - 2*b^5*sin(d*
x + c) + 6*a^5 + 7*a^3*b^2 - a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(b*sin(d*x + c) + a)^2))/d

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Mupad [B]
time = 6.85, size = 304, normalized size = 2.04 \begin {gather*} \frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^2\,b^2+b^4\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,{\left (a+b\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,{\left (a-b\right )}^3}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^3+3\,a\,b^2\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)/(a + b*sin(c + d*x))^3,x)

[Out]

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

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