3.3.79 \(\int \frac {1}{(d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2}} \, dx\) [279]

Optimal. Leaf size=406 \[ -\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}-\frac {5 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {5 \sqrt {d \csc (a+b x)} \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \sqrt {d \csc (a+b x)} \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}} \]

[Out]

-1/6*c/b/d/(d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(7/2)-5/48*c/b/d^3/(c*sec(b*x+a))^(7/2)/(d*csc(b*x+a))^(1/2)+5/
192/b/c/d^3/(c*sec(b*x+a))^(3/2)/(d*csc(b*x+a))^(1/2)+5/256*arctan(-1+2^(1/2)*tan(b*x+a)^(1/2))*(d*csc(b*x+a))
^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^4*2^(1/2)/(c*sec(b*x+a))^(1/2)+5/256*arctan(1+2^(1/2)*tan(b*x+a)^(1/2))*(d*csc
(b*x+a))^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^4*2^(1/2)/(c*sec(b*x+a))^(1/2)-5/512*ln(1-2^(1/2)*tan(b*x+a)^(1/2)+tan
(b*x+a))*(d*csc(b*x+a))^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^4*2^(1/2)/(c*sec(b*x+a))^(1/2)+5/512*ln(1+2^(1/2)*tan(b
*x+a)^(1/2)+tan(b*x+a))*(d*csc(b*x+a))^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^4*2^(1/2)/(c*sec(b*x+a))^(1/2)

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Rubi [A]
time = 0.27, antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2707, 2708, 2709, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {5 \sqrt {\tan (a+b x)} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)}}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \sqrt {\tan (a+b x)} \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right ) \sqrt {d \csc (a+b x)}}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {5 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {5 c}{48 b d^3 (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}+\frac {5}{192 b c d^3 (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}-\frac {c}{6 b d (c \sec (a+b x))^{7/2} (d \csc (a+b x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Csc[a + b*x])^(7/2)*(c*Sec[a + b*x])^(5/2)),x]

[Out]

-1/6*c/(b*d*(d*Csc[a + b*x])^(5/2)*(c*Sec[a + b*x])^(7/2)) - (5*c)/(48*b*d^3*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b
*x])^(7/2)) + 5/(192*b*c*d^3*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(3/2)) - (5*ArcTan[1 - Sqrt[2]*Sqrt[Tan[a +
b*x]]]*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]])/(128*Sqrt[2]*b*c^2*d^4*Sqrt[c*Sec[a + b*x]]) + (5*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]])/(128*Sqrt[2]*b*c^2*d^4*Sqrt[c*Sec[a + b*x
]]) - (5*Sqrt[d*Csc[a + b*x]]*Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[Tan[a + b*x]])/(256*Sqrt
[2]*b*c^2*d^4*Sqrt[c*Sec[a + b*x]]) + (5*Sqrt[d*Csc[a + b*x]]*Log[1 + Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x
]]*Sqrt[Tan[a + b*x]])/(256*Sqrt[2]*b*c^2*d^4*Sqrt[c*Sec[a + b*x]])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2707

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[b*(a*Csc[e +
f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/(a*f*(m + n))), x] + Dist[(m + 1)/(a^2*(m + n)), Int[(a*Csc[e + f*x])
^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n, 0] && IntegersQ[2
*m, 2*n]

Rule 2708

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a)*(a*Csc[
e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n + 1)/(b*f*(m + n))), x] + Dist[(n + 1)/(b^2*(m + n)), Int[(a*Csc[e + f*
x])^m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n, 0] && Integers
Q[2*m, 2*n]

Rule 2709

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^m*((b*Sec[e + f*x])^n/Tan[e + f*x]^n), Int[Tan[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !Int
egerQ[n] && EqQ[m + n, 0]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {1}{(d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2}} \, dx &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}+\frac {5 \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx}{12 d^2}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5 \int \frac {\sqrt {d \csc (a+b x)}}{(c \sec (a+b x))^{5/2}} \, dx}{96 d^4}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}+\frac {5 \int \frac {\sqrt {d \csc (a+b x)}}{\sqrt {c \sec (a+b x)}} \, dx}{128 c^2 d^4}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \int \frac {1}{\sqrt {\tan (a+b x)}} \, dx}{128 c^2 d^4 \sqrt {c \sec (a+b x)}}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\tan (a+b x)\right )}{128 b c^2 d^4 \sqrt {c \sec (a+b x)}}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{64 b c^2 d^4 \sqrt {c \sec (a+b x)}}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{128 b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{128 b c^2 d^4 \sqrt {c \sec (a+b x)}}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{256 b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{256 b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}-\frac {5 \sqrt {d \csc (a+b x)} \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \sqrt {d \csc (a+b x)} \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {\left (5 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (a+b x)}\right )}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}\\ &=-\frac {c}{6 b d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{7/2}}-\frac {5 c}{48 b d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {5}{192 b c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{128 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}-\frac {5 \sqrt {d \csc (a+b x)} \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}+\frac {5 \sqrt {d \csc (a+b x)} \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{256 \sqrt {2} b c^2 d^4 \sqrt {c \sec (a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 1.97, size = 175, normalized size = 0.43 \begin {gather*} -\frac {\left (28+34 \cos (2 (a+b x))+2 \cos (4 (a+b x))-4 \cos (6 (a+b x))+15 \sqrt {2} \text {ArcTan}\left (\frac {-1+\sqrt {\cot ^2(a+b x)}}{\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}\right ) \sqrt [4]{\cot ^2(a+b x)}-15 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}{1+\sqrt {\cot ^2(a+b x)}}\right ) \sqrt [4]{\cot ^2(a+b x)}\right ) \sqrt {c \sec (a+b x)}}{768 b c^3 d^3 \sqrt {d \csc (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Csc[a + b*x])^(7/2)*(c*Sec[a + b*x])^(5/2)),x]

[Out]

-1/768*((28 + 34*Cos[2*(a + b*x)] + 2*Cos[4*(a + b*x)] - 4*Cos[6*(a + b*x)] + 15*Sqrt[2]*ArcTan[(-1 + Sqrt[Cot
[a + b*x]^2])/(Sqrt[2]*(Cot[a + b*x]^2)^(1/4))]*(Cot[a + b*x]^2)^(1/4) - 15*Sqrt[2]*ArcTanh[(Sqrt[2]*(Cot[a +
b*x]^2)^(1/4))/(1 + Sqrt[Cot[a + b*x]^2])]*(Cot[a + b*x]^2)^(1/4))*Sqrt[c*Sec[a + b*x]])/(b*c^3*d^3*Sqrt[d*Csc
[a + b*x]])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 35.20, size = 722, normalized size = 1.78

method	result	size
default	\(\frac {\left (64 \sqrt {2}\, \left (\cos ^{7}\left (b x +a \right )\right )-64 \sqrt {2}\, \left (\cos ^{6}\left (b x +a \right )\right )-104 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )-15 i \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+15 i \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+104 \sqrt {2}\, \left (\cos ^{4}\left (b x +a \right )\right )-15 \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-15 \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+30 \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+10 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-10 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}\right ) \sqrt {2}}{768 b \left (-1+\cos \left (b x +a \right )\right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {7}{2}} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{3}}\)	\(722\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/768/b*(64*2^(1/2)*cos(b*x+a)^7-64*2^(1/2)*cos(b*x+a)^6-104*2^(1/2)*cos(b*x+a)^5-15*I*sin(b*x+a)*((-1+cos(b*x
+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1
/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))+15*I*sin(b*x+a)*((-1+cos(b
*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^
(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))+104*cos(b*x+a)^4*2^(1/2)
-15*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1
+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))
-15*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1
+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))
+30*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1
+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+10*cos(b*x
+a)^3*2^(1/2)-10*cos(b*x+a)^2*2^(1/2))/(-1+cos(b*x+a))/(d/sin(b*x+a))^(7/2)/(c/cos(b*x+a))^(5/2)/cos(b*x+a)^3/
sin(b*x+a)^3*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((d*csc(b*x + a))^(7/2)*(c*sec(b*x + a))^(5/2)), x)

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Fricas [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(1/(d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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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(1/(d*csc(b*x+a))**(7/2)/(c*sec(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((d*csc(b*x + a))^(7/2)*(c*sec(b*x + a))^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}\,{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(7/2)),x)

[Out]

int(1/((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(7/2)), x)

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