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

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

[Out]

-c/(4*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*Sec
[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]])

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Rubi [A]  time = 0.297432, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {2627, 2628, 2629, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{3 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (a+b x)}\right ) \sqrt{c \sec (a+b x)}}{32 \sqrt{2} b c^2 d^2 \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{3 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (a+b x)}+1\right ) \sqrt{c \sec (a+b x)}}{32 \sqrt{2} b c^2 d^2 \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{3 \sqrt{c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt{2} \sqrt{\tan (a+b x)}+1\right )}{64 \sqrt{2} b c^2 d^2 \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{3 \sqrt{c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt{2} \sqrt{\tan (a+b x)}+1\right )}{64 \sqrt{2} b c^2 d^2 \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{c}{4 b d (c \sec (a+b x))^{5/2} (d \csc (a+b x))^{3/2}}+\frac{3}{16 b c d \sqrt{c \sec (a+b x)} (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(4*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*Sec
[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 2627

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 2628

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] && IntegersQ[
2*m, 2*n]

Rule 2629

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 3476

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]

Rule 329

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 297

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 2.30413, size = 246, normalized size = 0.66 \[ \frac{\sqrt{d \csc (a+b x)} \left (8 \sqrt [4]{\cot ^2(a+b x)}-12 \cos (2 (a+b x)) \sqrt [4]{\cot ^2(a+b x)}+4 \cos (4 (a+b x)) \sqrt [4]{\cot ^2(a+b x)}+3 \sqrt{2} \log \left (\sqrt{\cot ^2(a+b x)}-\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}+1\right )-3 \sqrt{2} \log \left (\sqrt{\cot ^2(a+b x)}+\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}+1\right )+6 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}\right )-6 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}+1\right )\right )}{128 b c d^3 \sqrt [4]{\cot ^2(a+b x)} \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*Csc[a + b*x]]*(6*Sqrt[2]*ArcTan[1 - Sqrt[2]*(Cot[a + b*x]^2)^(1/4)] - 6*Sqrt[2]*ArcTan[1 + Sqrt[2]*(Co
t[a + b*x]^2)^(1/4)] + 8*(Cot[a + b*x]^2)^(1/4) - 12*Cos[2*(a + b*x)]*(Cot[a + b*x]^2)^(1/4) + 4*Cos[4*(a + b*
x)]*(Cot[a + b*x]^2)^(1/4) + 3*Sqrt[2]*Log[1 - Sqrt[2]*(Cot[a + b*x]^2)^(1/4) + Sqrt[Cot[a + b*x]^2]] - 3*Sqrt
[2]*Log[1 + Sqrt[2]*(Cot[a + b*x]^2)^(1/4) + Sqrt[Cot[a + b*x]^2]]))/(128*b*c*d^3*(Cot[a + b*x]^2)^(1/4)*Sqrt[
c*Sec[a + b*x]])

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Maple [C]  time = 0.174, size = 556, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

-1/64/b*2^(1/2)*(8*2^(1/2)*cos(b*x+a)^4-3*I*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+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))+3*I*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)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(
b*x+a))/sin(b*x+a))^(1/2)-8*cos(b*x+a)^3*2^(1/2)-3*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x
+a)+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))/s
in(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-3*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)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1
+cos(b*x+a))/sin(b*x+a))^(1/2)-6*cos(b*x+a)^2*2^(1/2)+6*cos(b*x+a)*2^(1/2))/(-1+cos(b*x+a))/cos(b*x+a)^2/sin(b
*x+a)/(d/sin(b*x+a))^(5/2)/(c/cos(b*x+a))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \csc \left (b x + a\right )\right )^{\frac{5}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \csc \left (b x + a\right )\right )^{\frac{5}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)