3.3.70 \(\int \frac {1}{(d \csc (a+b x))^{5/2} (c \sec (a+b x))^{3/2}} \, dx\) [270]

Optimal. Leaf size=371 \[ -\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}-\frac {3 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {3 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {3 \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {3 \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}} \]

[Out]

-1/4*c/b/d/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(5/2)+3/16/b/c/d/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(1/2)+3/64
*arctan(-1+2^(1/2)*tan(b*x+a)^(1/2))*(c*sec(b*x+a))^(1/2)/b/c^2/d^2*2^(1/2)/(d*csc(b*x+a))^(1/2)/tan(b*x+a)^(1
/2)+3/64*arctan(1+2^(1/2)*tan(b*x+a)^(1/2))*(c*sec(b*x+a))^(1/2)/b/c^2/d^2*2^(1/2)/(d*csc(b*x+a))^(1/2)/tan(b*
x+a)^(1/2)+3/128*ln(1-2^(1/2)*tan(b*x+a)^(1/2)+tan(b*x+a))*(c*sec(b*x+a))^(1/2)/b/c^2/d^2*2^(1/2)/(d*csc(b*x+a
))^(1/2)/tan(b*x+a)^(1/2)-3/128*ln(1+2^(1/2)*tan(b*x+a)^(1/2)+tan(b*x+a))*(c*sec(b*x+a))^(1/2)/b/c^2/d^2*2^(1/
2)/(d*csc(b*x+a))^(1/2)/tan(b*x+a)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 14, 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, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {3 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {3 \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right ) \sqrt {c \sec (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {3 \sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{64 \sqrt {2} b c^2 d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {3 \sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{64 \sqrt {2} b c^2 d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c}{4 b d (c \sec (a+b x))^{5/2} (d \csc (a+b x))^{3/2}}+\frac {3}{16 b c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Csc[a + b*x])^(5/2)*(c*Sec[a + b*x])^(3/2)),x]

[Out]

-1/4*c/(b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(5/2)) + 3/(16*b*c*d*(d*Csc[a + b*x])^(3/2)*Sqrt[c*Sec[a +
b*x]]) - (3*ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a + b*x]])/(32*Sqrt[2]*b*c^2*d^2*Sqrt[d*Csc[a +
b*x]]*Sqrt[Tan[a + b*x]]) + (3*ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a + b*x]])/(32*Sqrt[2]*b*c^2
*d^2*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]]) + (3*Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[c*S
ec[a + b*x]])/(64*Sqrt[2]*b*c^2*d^2*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]]) - (3*Log[1 + Sqrt[2]*Sqrt[Tan[a +
b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(64*Sqrt[2]*b*c^2*d^2*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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))^{5/2} (c \sec (a+b x))^{3/2}} \, dx &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3 \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}} \, dx}{8 d^2}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {3 \int \frac {\sqrt {c \sec (a+b x)}}{\sqrt {d \csc (a+b x)}} \, dx}{32 c^2 d^2}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \int \sqrt {\tan (a+b x)} \, dx}{32 c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (a+b x)\right )}{32 b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{16 b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}-\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{32 b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{32 b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{64 b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{64 b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\left (3 \sqrt {c \sec (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 )}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\left (3 \sqrt {c \sec (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 )}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {3 \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {3 \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{32 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\left (3 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (a+b x)}\right )}{32 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=-\frac {c}{4 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}}+\frac {3}{16 b c d (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {3 \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {3 \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {3 \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 1.63, size = 168, normalized size = 0.45 \begin {gather*} -\frac {\left (2 (\cos (2 (a+b x))+\cos (4 (a+b x)))+3 \sqrt {2} \text {ArcTan}\left (\frac {-1+\sqrt {\cot ^2(a+b x)}}{\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}\right ) \cot ^2(a+b x)^{3/4}+3 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}{1+\sqrt {\cot ^2(a+b x)}}\right ) \cot ^2(a+b x)^{3/4}\right ) \sec ^3(a+b x)}{64 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Csc[a + b*x])^(5/2)*(c*Sec[a + b*x])^(3/2)),x]

[Out]

-1/64*((2*(Cos[2*(a + b*x)] + Cos[4*(a + b*x)]) + 3*Sqrt[2]*ArcTan[(-1 + Sqrt[Cot[a + b*x]^2])/(Sqrt[2]*(Cot[a
+ b*x]^2)^(1/4))]*(Cot[a + b*x]^2)^(3/4) + 3*Sqrt[2]*ArcTanh[(Sqrt[2]*(Cot[a + b*x]^2)^(1/4))/(1 + Sqrt[Cot[a
+ b*x]^2])]*(Cot[a + b*x]^2)^(3/4))*Sec[a + b*x]^3)/(b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(3/2))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 32.35, size = 548, normalized size = 1.48

method	result	size
default	\(\frac {\left (3 i \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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-3 i \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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-8 \sqrt {2}\, \left (\cos ^{4}\left (b x +a \right )\right )+3 \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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 \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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+8 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+6 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-6 \sqrt {2}\, \cos \left (b x +a \right )\right ) \sqrt {2}}{64 b \left (-1+\cos \left (b x +a \right )\right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}\)	\(548\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/64/b*(3*I*((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(((1-cos
(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)-3*I*
((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(((1-cos(b*x+a)+sin(
b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)-8*cos(b*x+a)^4*2
^(1/2)+3*((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(((1-cos(b*
x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)+3*((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(((1-cos(b*x+a)+sin(b*x+a
))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)+8*cos(b*x+a)^3*2^(1/2
)+6*cos(b*x+a)^2*2^(1/2)-6*2^(1/2)*cos(b*x+a))/(-1+cos(b*x+a))/(d/sin(b*x+a))^(5/2)/(c/cos(b*x+a))^(3/2)/sin(b
*x+a)/cos(b*x+a)^2*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))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((d*csc(b*x + a))^(5/2)*(c*sec(b*x + a))^(3/2)), 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))^(5/2)/(c*sec(b*x+a))^(3/2),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))**(5/2)/(c*sec(b*x+a))**(3/2),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))^(5/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((d*csc(b*x + a))^(5/2)*(c*sec(b*x + a))^(3/2)), 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 )}^{3/2}\,{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c/cos(a + b*x))^(3/2)*(d/sin(a + b*x))^(5/2)),x)

[Out]

int(1/((c/cos(a + b*x))^(3/2)*(d/sin(a + b*x))^(5/2)), x)

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