3.65 \(\int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=140 \[ \frac {b \cot ^2(c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{a^5 d}+\frac {2 b \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5 d}-\frac {b \left (a^2+b^2\right )}{a^4 d (a+b \tan (c+d x))}-\frac {\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 140, normalized size of antiderivative = 1.00, 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} \[ -\frac {b \left (a^2+b^2\right )}{a^4 d (a+b \tan (c+d x))}-\frac {\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 d}-\frac {2 b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{a^5 d}+\frac {2 b \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5 d}+\frac {b \cot ^2(c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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]))

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]

Rubi steps

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

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Mathematica [A]  time = 2.66, size = 244, normalized size = 1.74 \[ \frac {3 b^2 \left (-2 \left (a^2+2 b^2\right ) \log (\sin (c+d x))+2 a^2 \log (a \cos (c+d x)+b \sin (c+d x))+a^2 \csc ^2(c+d x)+a^2+4 b^2 \log (a \cos (c+d x)+b \sin (c+d x))+b^2\right )+a b \cot (c+d x) \left (-6 \left (a^2+2 b^2\right ) \log (\sin (c+d x))+6 a^2 \log (a \cos (c+d x)+b \sin (c+d x))+2 a^2 \csc ^2(c+d x)-2 a^2+12 b^2 \log (a \cos (c+d x)+b \sin (c+d x))-9 b^2\right )-\cot ^2(c+d x) \left (a^4 \csc ^2(c+d x)+2 a^4+9 a^2 b^2\right )}{3 a^5 d (a \cot (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]

[Out]

(-(Cot[c + d*x]^2*(2*a^4 + 9*a^2*b^2 + a^4*Csc[c + d*x]^2)) + 3*b^2*(a^2 + b^2 + a^2*Csc[c + d*x]^2 - 2*(a^2 +
2*b^2)*Log[Sin[c + d*x]] + 2*a^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]] + 4*b^2*Log[a*Cos[c + d*x] + b*Sin[c +
d*x]]) + a*b*Cot[c + d*x]*(-2*a^2 - 9*b^2 + 2*a^2*Csc[c + d*x]^2 - 6*(a^2 + 2*b^2)*Log[Sin[c + d*x]] + 6*a^2*L
og[a*Cos[c + d*x] + b*Sin[c + d*x]] + 12*b^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]))/(3*a^5*d*(b + a*Cot[c + d*
x]))

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fricas [B]  time = 0.50, size = 442, normalized size = 3.16 \[ \frac {2 \, {\left (a^{4} + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} - 3 \, {\left (a^{4} + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{2} b^{2} + 2 \, b^{4} - 2 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 3 \, {\left ({\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{2} b^{2} + 2 \, b^{4} - 2 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) - 2 \, {\left (6 \, a b^{3} \cos \left (d x + c\right ) - {\left (a^{3} b + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{5} b d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b d \cos \left (d x + c\right )^{2} + a^{5} b d - {\left (a^{6} d \cos \left (d x + c\right )^{3} - a^{6} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.80, size = 203, normalized size = 1.45 \[ -\frac {\frac {6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac {6 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b} + \frac {3 \, {\left (2 \, a^{2} b^{2} \tan \left (d x + c\right ) + 4 \, b^{4} \tan \left (d x + c\right ) + 3 \, a^{3} b + 5 \, a b^{3}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )} a^{5}} - \frac {11 \, a^{2} b \tan \left (d x + c\right )^{3} + 22 \, b^{3} \tan \left (d x + c\right )^{3} - 3 \, a^{3} \tan \left (d x + c\right )^{2} - 9 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) - a^{3}}{a^{5} \tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(6*(a^2*b + 2*b^3)*log(abs(tan(d*x + c)))/a^5 - 6*(a^2*b^2 + 2*b^4)*log(abs(b*tan(d*x + c) + a))/(a^5*b)
+ 3*(2*a^2*b^2*tan(d*x + c) + 4*b^4*tan(d*x + c) + 3*a^3*b + 5*a*b^3)/((b*tan(d*x + c) + a)*a^5) - (11*a^2*b*t
an(d*x + c)^3 + 22*b^3*tan(d*x + c)^3 - 3*a^3*tan(d*x + c)^2 - 9*a*b^2*tan(d*x + c)^2 + 3*a^2*b*tan(d*x + c) -
a^3)/(a^5*tan(d*x + c)^3))/d

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maple [A]  time = 0.51, size = 189, normalized size = 1.35 \[ -\frac {b}{a^{2} d \left (a +b \tan \left (d x +c \right )\right )}-\frac {b^{3}}{d \,a^{4} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{3} d}+\frac {4 b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,a^{5}}-\frac {1}{3 d \,a^{2} \tan \left (d x +c \right )^{3}}-\frac {1}{d \,a^{2} \tan \left (d x +c \right )}-\frac {3 b^{2}}{d \,a^{4} \tan \left (d x +c \right )}+\frac {b}{d \,a^{3} \tan \left (d x +c \right )^{2}}-\frac {2 b \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}-\frac {4 b^{3} \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.42, size = 144, normalized size = 1.03 \[ \frac {\frac {2 \, a^{2} b \tan \left (d x + c\right ) - 6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{3} - a^{3} - 3 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{4} b \tan \left (d x + c\right )^{4} + a^{5} \tan \left (d x + c\right )^{3}} + \frac {6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5}} - \frac {6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{5}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.91, size = 150, normalized size = 1.07 \[ \frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,\left (a^2+2\,b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (2\,a^2\,b+4\,b^3\right )}\right )\,\left (a^2+2\,b^2\right )}{a^5\,d}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+2\,b^2\right )}{a^3}-\frac {2\,b\,\mathrm {tan}\left (c+d\,x\right )}{3\,a^2}+\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2+2\,b^2\right )}{a^4}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4+a\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^4*(a + b*tan(c + d*x))^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4/(a+b*tan(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**4/(a + b*tan(c + d*x))**2, x)

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