3.2.7 \(\int \frac {\csc ^6(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) [107]

Optimal. Leaf size=65 \[ -\frac {2 d^5}{15 b (d \tan (a+b x))^{15/2}}-\frac {4 d^3}{11 b (d \tan (a+b x))^{11/2}}-\frac {2 d}{7 b (d \tan (a+b x))^{7/2}} \]

[Out]

-2/15*d^5/b/(d*tan(b*x+a))^(15/2)-4/11*d^3/b/(d*tan(b*x+a))^(11/2)-2/7*d/b/(d*tan(b*x+a))^(7/2)

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Rubi [A]
time = 0.04, antiderivative size = 65, 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 = {2671, 276} \begin {gather*} -\frac {2 d^5}{15 b (d \tan (a+b x))^{15/2}}-\frac {4 d^3}{11 b (d \tan (a+b x))^{11/2}}-\frac {2 d}{7 b (d \tan (a+b x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^6/(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*d^5)/(15*b*(d*Tan[a + b*x])^(15/2)) - (4*d^3)/(11*b*(d*Tan[a + b*x])^(11/2)) - (2*d)/(7*b*(d*Tan[a + b*x])
^(7/2))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2671

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

Rubi steps

\begin {align*} \int \frac {\csc ^6(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=\frac {d \text {Subst}\left (\int \frac {\left (d^2+x^2\right )^2}{x^{17/2}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac {d \text {Subst}\left (\int \left (\frac {d^4}{x^{17/2}}+\frac {2 d^2}{x^{13/2}}+\frac {1}{x^{9/2}}\right ) \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {2 d^5}{15 b (d \tan (a+b x))^{15/2}}-\frac {4 d^3}{11 b (d \tan (a+b x))^{11/2}}-\frac {2 d}{7 b (d \tan (a+b x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 60, normalized size = 0.92 \begin {gather*} \frac {2 (-117+44 \cos (2 (a+b x))-4 \cos (4 (a+b x))) \cot ^4(a+b x) \csc ^4(a+b x) \sqrt {d \tan (a+b x)}}{1155 b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^6/(d*Tan[a + b*x])^(5/2),x]

[Out]

(2*(-117 + 44*Cos[2*(a + b*x)] - 4*Cos[4*(a + b*x)])*Cot[a + b*x]^4*Csc[a + b*x]^4*Sqrt[d*Tan[a + b*x]])/(1155
*b*d^3)

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Maple [A]
time = 0.56, size = 60, normalized size = 0.92

method	result	size
default	\(-\frac {2 \left (32 \left (\cos ^{4}\left (b x +a \right )\right )-120 \left (\cos ^{2}\left (b x +a \right )\right )+165\right ) \cos \left (b x +a \right )}{1155 b \sin \left (b x +a \right )^{5} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/1155/b*(32*cos(b*x+a)^4-120*cos(b*x+a)^2+165)*cos(b*x+a)/sin(b*x+a)^5/(d*sin(b*x+a)/cos(b*x+a))^(5/2)

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Maxima [A]
time = 0.28, size = 48, normalized size = 0.74 \begin {gather*} -\frac {2 \, {\left (165 \, d^{4} \tan \left (b x + a\right )^{4} + 210 \, d^{4} \tan \left (b x + a\right )^{2} + 77 \, d^{4}\right )} d}{1155 \, \left (d \tan \left (b x + a\right )\right )^{\frac {15}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/1155*(165*d^4*tan(b*x + a)^4 + 210*d^4*tan(b*x + a)^2 + 77*d^4)*d/((d*tan(b*x + a))^(15/2)*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (53) = 106\).
time = 0.43, size = 114, normalized size = 1.75 \begin {gather*} -\frac {2 \, {\left (32 \, \cos \left (b x + a\right )^{8} - 120 \, \cos \left (b x + a\right )^{6} + 165 \, \cos \left (b x + a\right )^{4}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{1155 \, {\left (b d^{3} \cos \left (b x + a\right )^{8} - 4 \, b d^{3} \cos \left (b x + a\right )^{6} + 6 \, b d^{3} \cos \left (b x + a\right )^{4} - 4 \, b d^{3} \cos \left (b x + a\right )^{2} + b d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

-2/1155*(32*cos(b*x + a)^8 - 120*cos(b*x + a)^6 + 165*cos(b*x + a)^4)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*d^3
*cos(b*x + a)^8 - 4*b*d^3*cos(b*x + a)^6 + 6*b*d^3*cos(b*x + a)^4 - 4*b*d^3*cos(b*x + a)^2 + b*d^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**6/(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.81, size = 58, normalized size = 0.89 \begin {gather*} -\frac {2 \, {\left (165 \, d^{5} \tan \left (b x + a\right )^{4} + 210 \, d^{5} \tan \left (b x + a\right )^{2} + 77 \, d^{5}\right )}}{1155 \, \sqrt {d \tan \left (b x + a\right )} b d^{7} \tan \left (b x + a\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/1155*(165*d^5*tan(b*x + a)^4 + 210*d^5*tan(b*x + a)^2 + 77*d^5)/(sqrt(d*tan(b*x + a))*b*d^7*tan(b*x + a)^7)

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Mupad [B]
time = 14.01, size = 1132, normalized size = 17.42 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^6*(d*tan(a + b*x))^(5/2)),x)

[Out]

(199232*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(12285*b*
d^3*(exp(a*2i + b*x*2i) - 1)) + (1581376*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i
+ b*x*2i) + 1))^(1/2))/(135135*b*d^3*(exp(a*2i + b*x*2i) - 1)^2) + (4539104*(exp(a*2i + b*x*2i) + 1)*(-(d*(ex
p(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(225225*b*d^3*(exp(a*2i + b*x*2i) - 1)^3) + (1152*
(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(35*b*d^3*(exp(a*
2i + b*x*2i) - 1)^4) + (74528*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i)
+ 1))^(1/2))/(2145*b*d^3*(exp(a*2i + b*x*2i) - 1)^5) + (1088*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)
*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(55*b*d^3*(exp(a*2i + b*x*2i) - 1)^6) + (896*(exp(a*2i + b*x*2i) +
1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(195*b*d^3*(exp(a*2i + b*x*2i) - 1)^7)
- ((exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2)*439808i)/(2702
7*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)) + (1573888*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(
exp(a*2i + b*x*2i) + 1))^(1/2))/(135135*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^2) + ((exp(a*2i + b*x*2i) + 1)*(-(d
*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2)*4557824i)/(225225*b*d^3*(exp(a*2i + b*x*2i)*1i
- 1i)^3) - (7168*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/
(165*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^4) - ((exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp
(a*2i + b*x*2i) + 1))^(1/2)*172288i)/(2145*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^5) + (5376*(exp(a*2i + b*x*2i) +
1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(55*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)
^6) + ((exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2)*12544i)/(19
5*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^7) - (256*(exp(a*2i + b*x*2i) + 1)*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(ex
p(a*2i + b*x*2i) + 1))^(1/2))/(15*b*d^3*(exp(a*2i + b*x*2i)*1i - 1i)^8)

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