3.49 \(\int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=194 \[ -\frac {a^4 \cot ^5(c+d x)}{5 d}-\frac {a^3 b \cot ^4(c+d x)}{d}+\frac {2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}-\frac {2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}+\frac {4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}-\frac {\left (a^4+12 a^2 b^2+b^4\right ) \cot (c+d x)}{d}+\frac {2 a b^3 \tan ^2(c+d x)}{d}+\frac {b^4 \tan ^3(c+d x)}{3 d} \]

[Out]

-(a^4+12*a^2*b^2+b^4)*cot(d*x+c)/d-2*a*b*(2*a^2+b^2)*cot(d*x+c)^2/d-2/3*a^2*(a^2+3*b^2)*cot(d*x+c)^3/d-a^3*b*c
ot(d*x+c)^4/d-1/5*a^4*cot(d*x+c)^5/d+4*a*b*(a^2+2*b^2)*ln(tan(d*x+c))/d+2*b^2*(3*a^2+b^2)*tan(d*x+c)/d+2*a*b^3
*tan(d*x+c)^2/d+1/3*b^4*tan(d*x+c)^3/d

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Rubi [A]  time = 0.16, antiderivative size = 194, 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, 948} \[ \frac {2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}-\frac {2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}-\frac {\left (12 a^2 b^2+a^4+b^4\right ) \cot (c+d x)}{d}+\frac {4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}-\frac {a^3 b \cot ^4(c+d x)}{d}-\frac {a^4 \cot ^5(c+d x)}{5 d}+\frac {2 a b^3 \tan ^2(c+d x)}{d}+\frac {b^4 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^4,x]

[Out]

-(((a^4 + 12*a^2*b^2 + b^4)*Cot[c + d*x])/d) - (2*a*b*(2*a^2 + b^2)*Cot[c + d*x]^2)/d - (2*a^2*(a^2 + 3*b^2)*C
ot[c + d*x]^3)/(3*d) - (a^3*b*Cot[c + d*x]^4)/d - (a^4*Cot[c + d*x]^5)/(5*d) + (4*a*b*(a^2 + 2*b^2)*Log[Tan[c
+ d*x]])/d + (2*b^2*(3*a^2 + b^2)*Tan[c + d*x])/d + (2*a*b^3*Tan[c + d*x]^2)/d + (b^4*Tan[c + d*x]^3)/(3*d)

Rule 948

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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 \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^4 \left (b^2+x^2\right )^2}{x^6} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \left (2 \left (3 a^2+b^2\right )+\frac {a^4 b^4}{x^6}+\frac {4 a^3 b^4}{x^5}+\frac {2 a^2 b^2 \left (a^2+3 b^2\right )}{x^4}+\frac {4 a b^2 \left (2 a^2+b^2\right )}{x^3}+\frac {a^4+12 a^2 b^2+b^4}{x^2}+\frac {4 \left (a^3+2 a b^2\right )}{x}+4 a x+x^2\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac {\left (a^4+12 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-\frac {2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}-\frac {2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {a^3 b \cot ^4(c+d x)}{d}-\frac {a^4 \cot ^5(c+d x)}{5 d}+\frac {4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}+\frac {2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {2 a b^3 \tan ^2(c+d x)}{d}+\frac {b^4 \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 4.01, size = 233, normalized size = 1.20 \[ -\frac {(a+b \tan (c+d x))^4 \left (3 a^4 \cot ^5(c+d x)+15 a^3 b \cot ^4(c+d x)-5 b^2 \left (18 a^2+5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)+2 a \cos ^2(c+d x) \left (a \left (2 a^2+15 b^2\right ) \cot ^3(c+d x)+15 b \left (a^2+b^2\right ) \cot ^2(c+d x)-15 b^3\right )+\cos ^4(c+d x) \left (60 a b \left (a^2+2 b^2\right ) (\log (\cos (c+d x))-\log (\sin (c+d x)))+\left (8 a^4+150 a^2 b^2+15 b^4\right ) \cot (c+d x)\right )-\frac {5}{2} b^4 \sin (2 (c+d x))\right )}{15 d (a \cos (c+d x)+b \sin (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^4,x]

[Out]

-1/15*((15*a^3*b*Cot[c + d*x]^4 + 3*a^4*Cot[c + d*x]^5 + 2*a*Cos[c + d*x]^2*(-15*b^3 + 15*b*(a^2 + b^2)*Cot[c
+ d*x]^2 + a*(2*a^2 + 15*b^2)*Cot[c + d*x]^3) + Cos[c + d*x]^4*((8*a^4 + 150*a^2*b^2 + 15*b^4)*Cot[c + d*x] +
60*a*b*(a^2 + 2*b^2)*(Log[Cos[c + d*x]] - Log[Sin[c + d*x]])) - 5*b^2*(18*a^2 + 5*b^2)*Cos[c + d*x]^3*Sin[c +
d*x] - (5*b^4*Sin[2*(c + d*x)])/2)*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)

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fricas [B]  time = 0.48, size = 386, normalized size = 1.99 \[ -\frac {8 \, {\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{8} - 20 \, {\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} + 15 \, {\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - 5 \, b^{4} - 10 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 2 \, a b^{3} \cos \left (d x + c\right ) - 3 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{7} - 2 \, d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/15*(8*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^8 - 20*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^6 + 15*(a^4 +
30*a^2*b^2 + 5*b^4)*cos(d*x + c)^4 - 5*b^4 - 10*(9*a^2*b^2 + b^4)*cos(d*x + c)^2 + 30*((a^3*b + 2*a*b^3)*cos(d
*x + c)^7 - 2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + (a^3*b + 2*a*b^3)*cos(d*x + c)^3)*log(cos(d*x + c)^2)*sin(d*x
+ c) - 30*((a^3*b + 2*a*b^3)*cos(d*x + c)^7 - 2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + (a^3*b + 2*a*b^3)*cos(d*x
+ c)^3)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c) - 15*(2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + 2*a*b^3*cos(d*x
+ c) - 3*(a^3*b + 2*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/((d*cos(d*x + c)^7 - 2*d*cos(d*x + c)^5 + d*cos(d*x
+ c)^3)*sin(d*x + c))

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giac [A]  time = 3.64, size = 235, normalized size = 1.21 \[ \frac {5 \, b^{4} \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{2} + 90 \, a^{2} b^{2} \tan \left (d x + c\right ) + 30 \, b^{4} \tan \left (d x + c\right ) + 60 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {137 \, a^{3} b \tan \left (d x + c\right )^{5} + 274 \, a b^{3} \tan \left (d x + c\right )^{5} + 15 \, a^{4} \tan \left (d x + c\right )^{4} + 180 \, a^{2} b^{2} \tan \left (d x + c\right )^{4} + 15 \, b^{4} \tan \left (d x + c\right )^{4} + 60 \, a^{3} b \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 10 \, a^{4} \tan \left (d x + c\right )^{2} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/15*(5*b^4*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^2 + 90*a^2*b^2*tan(d*x + c) + 30*b^4*tan(d*x + c) + 60*(a^3
*b + 2*a*b^3)*log(abs(tan(d*x + c))) - (137*a^3*b*tan(d*x + c)^5 + 274*a*b^3*tan(d*x + c)^5 + 15*a^4*tan(d*x +
c)^4 + 180*a^2*b^2*tan(d*x + c)^4 + 15*b^4*tan(d*x + c)^4 + 60*a^3*b*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^3
+ 10*a^4*tan(d*x + c)^2 + 30*a^2*b^2*tan(d*x + c)^2 + 15*a^3*b*tan(d*x + c) + 3*a^4)/tan(d*x + c)^5)/d

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maple [A]  time = 0.59, size = 301, normalized size = 1.55 \[ -\frac {8 a^{4} \cot \left (d x +c \right )}{15 d}-\frac {a^{4} \cot \left (d x +c \right ) \left (\csc ^{4}\left (d x +c \right )\right )}{5 d}-\frac {4 a^{4} \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{15 d}-\frac {a^{3} b}{d \sin \left (d x +c \right )^{4}}-\frac {2 a^{3} b}{d \sin \left (d x +c \right )^{2}}+\frac {4 a^{3} b \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 a^{2} b^{2}}{d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {8 a^{2} b^{2}}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {16 a^{2} b^{2} \cot \left (d x +c \right )}{d}+\frac {2 a \,b^{3}}{d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {4 a \,b^{3}}{d \sin \left (d x +c \right )^{2}}+\frac {8 a \,b^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b^{4}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4 b^{4}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 b^{4} \cot \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6*(a+b*tan(d*x+c))^4,x)

[Out]

-8/15*a^4*cot(d*x+c)/d-1/5/d*a^4*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^4*cot(d*x+c)*csc(d*x+c)^2-1/d*a^3*b/sin(d*x+
c)^4-2/d*a^3*b/sin(d*x+c)^2+4*a^3*b*ln(tan(d*x+c))/d-2/d*a^2*b^2/sin(d*x+c)^3/cos(d*x+c)+8/d*a^2*b^2/sin(d*x+c
)/cos(d*x+c)-16/d*a^2*b^2*cot(d*x+c)+2/d*a*b^3/sin(d*x+c)^2/cos(d*x+c)^2-4/d*a*b^3/sin(d*x+c)^2+8/d*a*b^3*ln(t
an(d*x+c))+1/3/d*b^4/sin(d*x+c)/cos(d*x+c)^3+4/3/d*b^4/sin(d*x+c)/cos(d*x+c)-8/3/d*b^4*cot(d*x+c)

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maxima [A]  time = 0.63, size = 171, normalized size = 0.88 \[ \frac {5 \, b^{4} \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{2} + 60 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + 30 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right ) - \frac {15 \, a^{3} b \tan \left (d x + c\right ) + 15 \, {\left (a^{4} + 12 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} + 3 \, a^{4} + 30 \, {\left (2 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )^{3} + 10 \, {\left (a^{4} + 3 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/15*(5*b^4*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^2 + 60*(a^3*b + 2*a*b^3)*log(tan(d*x + c)) + 30*(3*a^2*b^2
+ b^4)*tan(d*x + c) - (15*a^3*b*tan(d*x + c) + 15*(a^4 + 12*a^2*b^2 + b^4)*tan(d*x + c)^4 + 3*a^4 + 30*(2*a^3*
b + a*b^3)*tan(d*x + c)^3 + 10*(a^4 + 3*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c)^5)/d

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mupad [B]  time = 3.83, size = 181, normalized size = 0.93 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (4\,a^3\,b+8\,a\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {2\,a^4}{3}+2\,a^2\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,a^3\,b+2\,a\,b^3\right )+\frac {a^4}{5}+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4+12\,a^2\,b^2+b^4\right )+a^3\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (6\,a^2\,b^2+2\,b^4\right )}{d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^4/sin(c + d*x)^6,x)

[Out]

(log(tan(c + d*x))*(8*a*b^3 + 4*a^3*b))/d - (cot(c + d*x)^5*(tan(c + d*x)^2*((2*a^4)/3 + 2*a^2*b^2) + tan(c +
d*x)^3*(2*a*b^3 + 4*a^3*b) + a^4/5 + tan(c + d*x)^4*(a^4 + b^4 + 12*a^2*b^2) + a^3*b*tan(c + d*x)))/d + (b^4*t
an(c + d*x)^3)/(3*d) + (tan(c + d*x)*(2*b^4 + 6*a^2*b^2))/d + (2*a*b^3*tan(c + d*x)^2)/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**6*(a+b*tan(d*x+c))**4,x)

[Out]

Timed out

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