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

Optimal. Leaf size=322 \[ -\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{4 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right ) \sqrt {c \sec (a+b x)}}{4 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{8 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {\sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{8 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {d}{2 b c \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{3/2}} \]

[Out]

1/2*d/b/c/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(1/2)+1/8*arctan(-1+2^(1/2)*tan(b*x+a)^(1/2))*(c*sec(b*x+a))^(1/
2)/b/c^2*2^(1/2)/(d*csc(b*x+a))^(1/2)/tan(b*x+a)^(1/2)+1/8*arctan(1+2^(1/2)*tan(b*x+a)^(1/2))*(c*sec(b*x+a))^(
1/2)/b/c^2*2^(1/2)/(d*csc(b*x+a))^(1/2)/tan(b*x+a)^(1/2)+1/16*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*2^(1/2)/(d*csc(b*x+a))^(1/2)/tan(b*x+a)^(1/2)-1/16*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*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 = 322, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2628, 2629, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{4 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right ) \sqrt {c \sec (a+b x)}}{4 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{8 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {\sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{8 \sqrt {2} b c^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {d}{2 b c \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

d/(2*b*c*(d*Csc[a + b*x])^(3/2)*Sqrt[c*Sec[a + b*x]]) - (ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a +
b*x]])/(4*Sqrt[2]*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]]) + (ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sq
rt[c*Sec[a + b*x]])/(4*Sqrt[2]*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]]) + (Log[1 - Sqrt[2]*Sqrt[Tan[a +
b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(8*Sqrt[2]*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]]) - (Log[1
+ Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(8*Sqrt[2]*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt
[Tan[a + b*x]])

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 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 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 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 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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}} \, dx &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\int \frac {\sqrt {c \sec (a+b x)}}{\sqrt {d \csc (a+b x)}} \, dx}{4 c^2}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \int \sqrt {\tan (a+b x)} \, dx}{4 c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (a+b x)\right )}{4 b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{2 b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}-\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{4 b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{4 b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{8 b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{8 b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{8 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{8 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{8 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{8 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{4 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\sqrt {c \sec (a+b x)} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (a+b x)}\right )}{4 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {d}{2 b c (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{4 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{4 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{8 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{8 \sqrt {2} b c^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 2.02, size = 223, normalized size = 0.69 \[ \frac {\sqrt {d \csc (a+b x)} \left (4 \sqrt [4]{\cot ^2(a+b x)}-4 \cos (2 (a+b x)) \sqrt [4]{\cot ^2(a+b x)}+\sqrt {2} \log \left (\sqrt {\cot ^2(a+b x)}-\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}+1\right )-\sqrt {2} \log \left (\sqrt {\cot ^2(a+b x)}+\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}+1\right )+2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}+1\right )\right )}{16 b c d \sqrt [4]{\cot ^2(a+b x)} \sqrt {c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*Csc[a + b*x]]*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*(Cot[a + b*x]^2)^(1/4)] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*(Co
t[a + b*x]^2)^(1/4)] + 4*(Cot[a + b*x]^2)^(1/4) - 4*Cos[2*(a + b*x)]*(Cot[a + b*x]^2)^(1/4) + Sqrt[2]*Log[1 -
Sqrt[2]*(Cot[a + b*x]^2)^(1/4) + Sqrt[Cot[a + b*x]^2]] - Sqrt[2]*Log[1 + Sqrt[2]*(Cot[a + b*x]^2)^(1/4) + Sqrt
[Cot[a + b*x]^2]]))/(16*b*c*d*(Cot[a + b*x]^2)^(1/4)*Sqrt[c*Sec[a + b*x]])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(d*csc(b*x + a))*(c*sec(b*x + a))^(3/2)), x)

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maple [C]  time = 0.99, size = 520, normalized size = 1.61 \[ -\frac {\left (i \sqrt {\frac {1-\cos \left (b x +a \right )+\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 )}{\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 )-i \sqrt {\frac {1-\cos \left (b x +a \right )+\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 )}{\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 )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\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 )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\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 )-2 \sqrt {2}\, \left (\cos ^{2}\left (b x +a \right )\right )+2 \cos \left (b x +a \right ) \sqrt {2}\right ) \sin \left (b x +a \right ) \sqrt {2}}{8 b \left (-1+\cos \left (b x +a \right )\right ) \cos \left (b x +a \right )^{2} \sqrt {\frac {d}{\sin \left (b x +a \right )}}\, \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/b*(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)-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))-((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)*El
lipticPi(((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)*EllipticPi(((1-co
s(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*2^(1/2)*cos(b*x+a)^2+2*cos(b*x+a)*2^(1/2))*sin
(b*x+a)/(-1+cos(b*x+a))/cos(b*x+a)^2/(d/sin(b*x+a))^(1/2)/(c/cos(b*x+a))^(3/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(d*csc(b*x + a))*(c*sec(b*x + a))^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{3/2}\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}}} \,d x \]

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))^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \sec {\left (a + b x \right )}\right )^{\frac {3}{2}} \sqrt {d \csc {\left (a + b x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*csc(b*x+a))**(1/2)/(c*sec(b*x+a))**(3/2),x)

[Out]

Integral(1/((c*sec(a + b*x))**(3/2)*sqrt(d*csc(a + b*x))), x)

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