3.71 \(\int \frac{\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=178 \[ -\frac{2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))}-\frac{b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2}-\frac{\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}-\frac{b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac{b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}+\frac{3 b \cot ^2(c+d x)}{2 a^4 d}-\frac{\cot ^3(c+d x)}{3 a^3 d} \]

[Out]

-(((a^2 + 6*b^2)*Cot[c + d*x])/(a^5*d)) + (3*b*Cot[c + d*x]^2)/(2*a^4*d) - Cot[c + d*x]^3/(3*a^3*d) - (b*(3*a^
2 + 10*b^2)*Log[Tan[c + d*x]])/(a^6*d) + (b*(3*a^2 + 10*b^2)*Log[a + b*Tan[c + d*x]])/(a^6*d) - (b*(a^2 + b^2)
)/(2*a^4*d*(a + b*Tan[c + d*x])^2) - (2*b*(a^2 + 2*b^2))/(a^5*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.152575, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3516, 894} \[ -\frac{2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))}-\frac{b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2}-\frac{\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}-\frac{b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac{b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}+\frac{3 b \cot ^2(c+d x)}{2 a^4 d}-\frac{\cot ^3(c+d x)}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((a^2 + 6*b^2)*Cot[c + d*x])/(a^5*d)) + (3*b*Cot[c + d*x]^2)/(2*a^4*d) - Cot[c + d*x]^3/(3*a^3*d) - (b*(3*a^
2 + 10*b^2)*Log[Tan[c + d*x]])/(a^6*d) + (b*(3*a^2 + 10*b^2)*Log[a + b*Tan[c + d*x]])/(a^6*d) - (b*(a^2 + b^2)
)/(2*a^4*d*(a + b*Tan[c + d*x])^2) - (2*b*(a^2 + 2*b^2))/(a^5*d*(a + b*Tan[c + d*x]))

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rule 894

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]))

Rubi steps

\begin{align*} \int \frac{\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{b^2+x^2}{x^4 (a+x)^3} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \left (\frac{b^2}{a^3 x^4}-\frac{3 b^2}{a^4 x^3}+\frac{a^2+6 b^2}{a^5 x^2}+\frac{-3 a^2-10 b^2}{a^6 x}+\frac{a^2+b^2}{a^4 (a+x)^3}+\frac{2 \left (a^2+2 b^2\right )}{a^5 (a+x)^2}+\frac{3 a^2+10 b^2}{a^6 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}+\frac{3 b \cot ^2(c+d x)}{2 a^4 d}-\frac{\cot ^3(c+d x)}{3 a^3 d}-\frac{b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac{b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}-\frac{b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2}-\frac{2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 3.22664, size = 456, normalized size = 2.56 \[ -\frac{b^3 \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{2 a^4 d (a+b \tan (c+d x))^3}+\frac{\sec ^3(c+d x) \left (3 a^2 b^2 \sin (c+d x)+4 b^4 \sin (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{a^6 d (a+b \tan (c+d x))^3}+\frac{\left (-3 a^2 b-10 b^3\right ) \sec ^3(c+d x) \log (\sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{a^6 d (a+b \tan (c+d x))^3}+\frac{\left (3 a^2 b+10 b^3\right ) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \log (a \cos (c+d x)+b \sin (c+d x))}{a^6 d (a+b \tan (c+d x))^3}-\frac{2 \csc (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+9 b^2 \cos (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^3}{3 a^5 d (a+b \tan (c+d x))^3}+\frac{3 b \csc ^2(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{2 a^4 d (a+b \tan (c+d x))^3}-\frac{\csc ^3(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{3 a^3 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

-(b^3*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(2*a^4*d*(a + b*Tan[c + d*x])^3) - (Csc[c + d*x]^3*Sec
[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(3*a^3*d*(a + b*Tan[c + d*x])^3) - (2*(a^2*Cos[c + d*x] + 9*b
^2*Cos[c + d*x])*Csc[c + d*x]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(3*a^5*d*(a + b*Tan[c + d*x]
)^3) + (3*b*Csc[c + d*x]^2*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(2*a^4*d*(a + b*Tan[c + d*x])^3
) + ((-3*a^2*b - 10*b^3)*Log[Sin[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(a^6*d*(a + b*T
an[c + d*x])^3) + ((3*a^2*b + 10*b^3)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*
Sin[c + d*x])^3)/(a^6*d*(a + b*Tan[c + d*x])^3) + (Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2*(3*a^2*b
^2*Sin[c + d*x] + 4*b^4*Sin[c + d*x]))/(a^6*d*(a + b*Tan[c + d*x])^3)

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Maple [A]  time = 0.13, size = 234, normalized size = 1.3 \begin{align*} -{\frac{1}{3\,d{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{d{a}^{3}\tan \left ( dx+c \right ) }}-6\,{\frac{{b}^{2}}{d{a}^{5}\tan \left ( dx+c \right ) }}+{\frac{3\,b}{2\,d{a}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{b\ln \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{4}}}-10\,{\frac{{b}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{6}}}+3\,{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{a}^{4}}}+10\,{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{a}^{6}}}-{\frac{b}{2\,{a}^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{3}}{2\,d{a}^{4} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-2\,{\frac{b}{d{a}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-4\,{\frac{{b}^{3}}{d{a}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x)

[Out]

-1/3/d/a^3/tan(d*x+c)^3-1/d/a^3/tan(d*x+c)-6/d/a^5/tan(d*x+c)*b^2+3/2/d/a^4*b/tan(d*x+c)^2-3*b*ln(tan(d*x+c))/
a^4/d-10/d*b^3/a^6*ln(tan(d*x+c))+3*b*ln(a+b*tan(d*x+c))/a^4/d+10/d*b^3/a^6*ln(a+b*tan(d*x+c))-1/2*b/a^2/d/(a+
b*tan(d*x+c))^2-1/2/d*b^3/a^4/(a+b*tan(d*x+c))^2-2*b/a^3/d/(a+b*tan(d*x+c))-4/d*b^3/a^5/(a+b*tan(d*x+c))

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Maxima [A]  time = 1.20175, size = 259, normalized size = 1.46 \begin{align*} \frac{\frac{5 \, a^{3} b \tan \left (d x + c\right ) - 6 \,{\left (3 \, a^{2} b^{2} + 10 \, b^{4}\right )} \tan \left (d x + c\right )^{4} - 2 \, a^{4} - 9 \,{\left (3 \, a^{3} b + 10 \, a b^{3}\right )} \tan \left (d x + c\right )^{3} - 2 \,{\left (3 \, a^{4} + 10 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{5} b^{2} \tan \left (d x + c\right )^{5} + 2 \, a^{6} b \tan \left (d x + c\right )^{4} + a^{7} \tan \left (d x + c\right )^{3}} + \frac{6 \,{\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6}} - \frac{6 \,{\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/6*((5*a^3*b*tan(d*x + c) - 6*(3*a^2*b^2 + 10*b^4)*tan(d*x + c)^4 - 2*a^4 - 9*(3*a^3*b + 10*a*b^3)*tan(d*x +
c)^3 - 2*(3*a^4 + 10*a^2*b^2)*tan(d*x + c)^2)/(a^5*b^2*tan(d*x + c)^5 + 2*a^6*b*tan(d*x + c)^4 + a^7*tan(d*x +
c)^3) + 6*(3*a^2*b + 10*b^3)*log(b*tan(d*x + c) + a)/a^6 - 6*(3*a^2*b + 10*b^3)*log(tan(d*x + c))/a^6)/d

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Fricas [B]  time = 2.59892, size = 1818, normalized size = 10.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(2*(2*a^7 + 27*a^5*b^2 + a^3*b^4 - 30*a*b^6)*cos(d*x + c)^5 - 2*(3*a^7 + 43*a^5*b^2 - 8*a^3*b^4 - 60*a*b^6
)*cos(d*x + c)^3 + 6*(5*a^5*b^2 - 3*a^3*b^4 - 10*a*b^6)*cos(d*x + c) + 3*(2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6
)*cos(d*x + c)^5 - 4*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^3 + 2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6
)*cos(d*x + c) + (3*a^4*b^3 + 13*a^2*b^5 + 10*b^7 - (3*a^6*b + 10*a^4*b^3 - 3*a^2*b^5 - 10*b^7)*cos(d*x + c)^4
+ (3*a^6*b + 7*a^4*b^3 - 16*a^2*b^5 - 20*b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x +
c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 3*(2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^5 - 4*(3*a^5*b^
2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^3 + 2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c) + (3*a^4*b^3 +
13*a^2*b^5 + 10*b^7 - (3*a^6*b + 10*a^4*b^3 - 3*a^2*b^5 - 10*b^7)*cos(d*x + c)^4 + (3*a^6*b + 7*a^4*b^3 - 16*a
^2*b^5 - 20*b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4) + (24*a^4*b^3 + 30*a^2*b^5 + 4*(
2*a^6*b + 29*a^4*b^3 + 30*a^2*b^5)*cos(d*x + c)^4 - 3*(a^6*b + 45*a^4*b^3 + 50*a^2*b^5)*cos(d*x + c)^2)*sin(d*
x + c))/(2*(a^9*b + a^7*b^3)*d*cos(d*x + c)^5 - 4*(a^9*b + a^7*b^3)*d*cos(d*x + c)^3 + 2*(a^9*b + a^7*b^3)*d*c
os(d*x + c) - ((a^10 - a^6*b^4)*d*cos(d*x + c)^4 - (a^10 - a^8*b^2 - 2*a^6*b^4)*d*cos(d*x + c)^2 - (a^8*b^2 +
a^6*b^4)*d)*sin(d*x + c))

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

[Out]

Timed out

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Giac [A]  time = 1.34649, size = 320, normalized size = 1.8 \begin{align*} -\frac{\frac{6 \,{\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{6}} - \frac{6 \,{\left (3 \, a^{2} b^{2} + 10 \, b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b} + \frac{3 \,{\left (9 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} + 30 \, b^{5} \tan \left (d x + c\right )^{2} + 22 \, a^{3} b^{2} \tan \left (d x + c\right ) + 68 \, a b^{4} \tan \left (d x + c\right ) + 14 \, a^{4} b + 39 \, a^{2} b^{3}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} a^{6}} - \frac{33 \, a^{2} b \tan \left (d x + c\right )^{3} + 110 \, b^{3} \tan \left (d x + c\right )^{3} - 6 \, a^{3} \tan \left (d x + c\right )^{2} - 36 \, a b^{2} \tan \left (d x + c\right )^{2} + 9 \, a^{2} b \tan \left (d x + c\right ) - 2 \, a^{3}}{a^{6} \tan \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(6*(3*a^2*b + 10*b^3)*log(abs(tan(d*x + c)))/a^6 - 6*(3*a^2*b^2 + 10*b^4)*log(abs(b*tan(d*x + c) + a))/(a
^6*b) + 3*(9*a^2*b^3*tan(d*x + c)^2 + 30*b^5*tan(d*x + c)^2 + 22*a^3*b^2*tan(d*x + c) + 68*a*b^4*tan(d*x + c)
+ 14*a^4*b + 39*a^2*b^3)/((b*tan(d*x + c) + a)^2*a^6) - (33*a^2*b*tan(d*x + c)^3 + 110*b^3*tan(d*x + c)^3 - 6*
a^3*tan(d*x + c)^2 - 36*a*b^2*tan(d*x + c)^2 + 9*a^2*b*tan(d*x + c) - 2*a^3)/(a^6*tan(d*x + c)^3))/d