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

Optimal. Leaf size=137 \[ -\frac {a^4 \cot ^3(c+d x)}{3 d}-\frac {2 a^3 b \cot ^2(c+d x)}{d}+\frac {b^2 \left (6 a^2+b^2\right ) \tan (c+d x)}{d}-\frac {a^2 \left (a^2+6 b^2\right ) \cot (c+d x)}{d}+\frac {4 a b \left (a^2+b^2\right ) \log (\tan (c+d x))}{d}+\frac {2 a b^3 \tan ^2(c+d x)}{d}+\frac {b^4 \tan ^3(c+d x)}{3 d} \]

[Out]

-a^2*(a^2+6*b^2)*cot(d*x+c)/d-2*a^3*b*cot(d*x+c)^2/d-1/3*a^4*cot(d*x+c)^3/d+4*a*b*(a^2+b^2)*ln(tan(d*x+c))/d+b
^2*(6*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 = 0.10, antiderivative size = 137, 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} \[ \frac {b^2 \left (6 a^2+b^2\right ) \tan (c+d x)}{d}-\frac {a^2 \left (a^2+6 b^2\right ) \cot (c+d x)}{d}+\frac {4 a b \left (a^2+b^2\right ) \log (\tan (c+d x))}{d}-\frac {2 a^3 b \cot ^2(c+d x)}{d}-\frac {a^4 \cot ^3(c+d x)}{3 d}+\frac {2 a b^3 \tan ^2(c+d x)}{d}+\frac {b^4 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(a^2 + 6*b^2)*Cot[c + d*x])/d) - (2*a^3*b*Cot[c + d*x]^2)/d - (a^4*Cot[c + d*x]^3)/(3*d) + (4*a*b*(a^2
+ b^2)*Log[Tan[c + d*x]])/d + (b^2*(6*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 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 \csc ^4(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 )}{x^4} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \left (6 a^2 \left (1+\frac {b^2}{6 a^2}\right )+\frac {a^4 b^2}{x^4}+\frac {4 a^3 b^2}{x^3}+\frac {a^4+6 a^2 b^2}{x^2}+\frac {4 a \left (a^2+b^2\right )}{x}+4 a x+x^2\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac {a^2 \left (a^2+6 b^2\right ) \cot (c+d x)}{d}-\frac {2 a^3 b \cot ^2(c+d x)}{d}-\frac {a^4 \cot ^3(c+d x)}{3 d}+\frac {4 a b \left (a^2+b^2\right ) \log (\tan (c+d x))}{d}+\frac {b^2 \left (6 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 = 3.84, size = 188, normalized size = 1.37 \[ -\frac {\sin (c+d x) \tan ^3(c+d x) (a \cot (c+d x)+b)^4 \left (-2 b^2 \left (9 a^2+b^2\right ) \sin (c+d x) \cos ^2(c+d x)+2 a \cos ^3(c+d x) \left (a \left (a^2+9 b^2\right ) \cot (c+d x)+6 b \left (a^2+b^2\right ) (\log (\cos (c+d x))-\log (\sin (c+d x)))\right )+\cos (c+d x) \left (a^4 \cot ^3(c+d x)+6 a^3 b \cot ^2(c+d x)-6 a b^3\right )+b^4 (-\sin (c+d x))\right )}{3 d (a \cos (c+d x)+b \sin (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((b + a*Cot[c + d*x])^4*Sin[c + d*x]*(Cos[c + d*x]*(-6*a*b^3 + 6*a^3*b*Cot[c + d*x]^2 + a^4*Cot[c + d*x]^
3) + 2*a*Cos[c + d*x]^3*(a*(a^2 + 9*b^2)*Cot[c + d*x] + 6*b*(a^2 + b^2)*(Log[Cos[c + d*x]] - Log[Sin[c + d*x]]
)) - b^4*Sin[c + d*x] - 2*b^2*(9*a^2 + b^2)*Cos[c + d*x]^2*Sin[c + d*x])*Tan[c + d*x]^3)/(d*(a*Cos[c + d*x] +
b*Sin[c + d*x])^4)

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fricas [B]  time = 0.46, size = 267, normalized size = 1.95 \[ -\frac {2 \, {\left (a^{4} + 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 18 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{4} + 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + b^{4} + 6 \, {\left ({\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 6 \, {\left ({\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) + 6 \, {\left (a b^{3} \cos \left (d x + c\right ) - {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{5} - d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*(a^4 + 18*a^2*b^2 + b^4)*cos(d*x + c)^6 + 18*a^2*b^2*cos(d*x + c)^2 - 3*(a^4 + 18*a^2*b^2 + b^4)*cos(d
*x + c)^4 + b^4 + 6*((a^3*b + a*b^3)*cos(d*x + c)^5 - (a^3*b + a*b^3)*cos(d*x + c)^3)*log(cos(d*x + c)^2)*sin(
d*x + c) - 6*((a^3*b + a*b^3)*cos(d*x + c)^5 - (a^3*b + a*b^3)*cos(d*x + c)^3)*log(-1/4*cos(d*x + c)^2 + 1/4)*
sin(d*x + c) + 6*(a*b^3*cos(d*x + c) - (a^3*b + a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/((d*cos(d*x + c)^5 - d*co
s(d*x + c)^3)*sin(d*x + c))

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giac [A]  time = 7.80, size = 161, normalized size = 1.18 \[ \frac {b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \tan \left (d x + c\right ) + 3 \, b^{4} \tan \left (d x + c\right ) + 12 \, {\left (a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {22 \, a^{3} b \tan \left (d x + c\right )^{3} + 22 \, a b^{3} \tan \left (d x + c\right )^{3} + 3 \, a^{4} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 6 \, a^{3} b \tan \left (d x + c\right ) + a^{4}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^4*tan(d*x + c)^3 + 6*a*b^3*tan(d*x + c)^2 + 18*a^2*b^2*tan(d*x + c) + 3*b^4*tan(d*x + c) + 12*(a^3*b +
a*b^3)*log(abs(tan(d*x + c))) - (22*a^3*b*tan(d*x + c)^3 + 22*a*b^3*tan(d*x + c)^3 + 3*a^4*tan(d*x + c)^2 + 18
*a^2*b^2*tan(d*x + c)^2 + 6*a^3*b*tan(d*x + c) + a^4)/tan(d*x + c)^3)/d

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maple [A]  time = 0.66, size = 184, normalized size = 1.34 \[ -\frac {2 a^{4} \cot \left (d x +c \right )}{3 d}-\frac {a^{4} \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{3 d}-\frac {2 a^{3} b}{d \sin \left (d x +c \right )^{2}}+\frac {4 a^{3} b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2}}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {12 a^{2} b^{2} \cot \left (d x +c \right )}{d}+\frac {2 a \,b^{3}}{d \cos \left (d x +c \right )^{2}}+\frac {4 a \,b^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 b^{4} \tan \left (d x +c \right )}{3 d}+\frac {b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^4*cot(d*x+c)/d-1/3/d*a^4*cot(d*x+c)*csc(d*x+c)^2-2/d*a^3*b/sin(d*x+c)^2+4*a^3*b*ln(tan(d*x+c))/d+6/d*a^
2*b^2/sin(d*x+c)/cos(d*x+c)-12/d*a^2*b^2*cot(d*x+c)+2/d*a*b^3/cos(d*x+c)^2+4/d*a*b^3*ln(tan(d*x+c))+2/3/d*b^4*
tan(d*x+c)+1/3/d*b^4*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.40, size = 120, normalized size = 0.88 \[ \frac {b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 12 \, {\left (a^{3} b + a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + 3 \, {\left (6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right ) - \frac {6 \, a^{3} b \tan \left (d x + c\right ) + a^{4} + 3 \, {\left (a^{4} + 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^4*tan(d*x + c)^3 + 6*a*b^3*tan(d*x + c)^2 + 12*(a^3*b + a*b^3)*log(tan(d*x + c)) + 3*(6*a^2*b^2 + b^4)*
tan(d*x + c) - (6*a^3*b*tan(d*x + c) + a^4 + 3*(a^4 + 6*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c)^3)/d

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mupad [B]  time = 3.76, size = 132, normalized size = 0.96 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (4\,a^3\,b+4\,a\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^4+6\,a^2\,b^2\right )+\frac {a^4}{3}+2\,a^3\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (6\,a^2\,b^2+b^4\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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