3.182 \(\int \frac{\tan ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=161 \[ \frac{a^3}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{a^2 \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac{\sec ^2(c+d x) \left (a^2-2 a b \sin (c+d x)+b^2\right )}{2 d \left (a^2-b^2\right )^2}+\frac{a \log (1-\sin (c+d x))}{2 d (a+b)^3}+\frac{a \log (\sin (c+d x)+1)}{2 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.311696, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2721, 1647, 1629} \[ \frac{a^3}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{a^2 \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac{\sec ^2(c+d x) \left (a^2-2 a b \sin (c+d x)+b^2\right )}{2 d \left (a^2-b^2\right )^2}+\frac{a \log (1-\sin (c+d x))}{2 d (a+b)^3}+\frac{a \log (\sin (c+d x)+1)}{2 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2721

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]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.719396, size = 145, normalized size = 0.9 \[ \frac{\frac{4 a^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{4 a^2 \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}-\frac{1}{(a+b)^2 (\sin (c+d x)-1)}+\frac{1}{(a-b)^2 (\sin (c+d x)+1)}+\frac{2 a \log (1-\sin (c+d x))}{(a+b)^3}+\frac{2 a \log (\sin (c+d x)+1)}{(a-b)^3}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.087, size = 182, normalized size = 1.1 \begin{align*}{\frac{{a}^{3}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{{a}^{4}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}-3\,{\frac{{a}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}-{\frac{1}{4\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{a\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{3}}}+{\frac{1}{4\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{a\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{2\, \left ( a-b \right ) ^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.75269, size = 370, normalized size = 2.3 \begin{align*} -\frac{\frac{2 \,{\left (a^{4} + 3 \, a^{2} 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{a \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{a \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{3 \, a^{3} + a b^{2} - 2 \,{\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )^{2} -{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.3941, size = 869, normalized size = 5.4 \begin{align*} \frac{a^{5} - 2 \, a^{3} b^{2} + a b^{4} + 2 \,{\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left ({\left (a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{5} + 3 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.84256, size = 335, normalized size = 2.08 \begin{align*} -\frac{\frac{2 \,{\left (a^{4} b + 3 \, a^{2} 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{a \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{a \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{2 \, a^{3} \sin \left (d x + c\right )^{2} + 2 \, a b^{2} \sin \left (d x + c\right )^{2} + a^{2} b \sin \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - 3 \, a^{3} - a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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