3.3.73 \(\int \frac {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{5/2}} \, dx\) [273]

Optimal. Leaf size=329 \[ -\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}+\frac {d^2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}+\frac {d^2 \sqrt {d \csc (a+b x)} \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \sqrt {d \csc (a+b x)} \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}} \]

[Out]

-2/3*d*(d*csc(b*x+a))^(3/2)/b/c/(c*sec(b*x+a))^(3/2)-1/2*d^2*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*2^(1/2)/(c*sec(b*x+a))^(1/2)-1/2*d^2*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*2^(1/2)/(c*sec(b*x+a))^(1/2)+1/4*d^2*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*2^(1/2)/(c*sec(b*x+a))^(1/2)-1/4*d^2*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*2^(1/2)/(c*sec(b*x+a))^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 329, 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 = {2703, 2709, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {d^2 \sqrt {\tan (a+b x)} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)}}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \sqrt {\tan (a+b x)} \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right ) \sqrt {d \csc (a+b x)}}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}+\frac {d^2 \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 )}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \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 )}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(d*Csc[a + b*x])^(3/2))/(3*b*c*(c*Sec[a + b*x])^(3/2)) + (d^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqr
t[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]])/(Sqrt[2]*b*c^2*Sqrt[c*Sec[a + b*x]]) - (d^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[
a + b*x]]]*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]])/(Sqrt[2]*b*c^2*Sqrt[c*Sec[a + b*x]]) + (d^2*Sqrt[d*Csc[a +
b*x]]*Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[Tan[a + b*x]])/(2*Sqrt[2]*b*c^2*Sqrt[c*Sec[a +
b*x]]) - (d^2*Sqrt[d*Csc[a + b*x]]*Log[1 + Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[Tan[a + b*x]])/(2*S
qrt[2]*b*c^2*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 2703

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)/(f*b*(m - 1))), x] + Dist[a^2*((n + 1)/(b^2*(m - 1))), Int[(a*Csc[e
+ f*x])^(m - 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && Inte
gersQ[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 {(d \csc (a+b x))^{5/2}}{(c \sec (a+b x))^{5/2}} \, dx &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}-\frac {d^2 \int \frac {\sqrt {d \csc (a+b x)}}{\sqrt {c \sec (a+b x)}} \, dx}{c^2}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}-\frac {\left (d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}\right ) \int \frac {1}{\sqrt {\tan (a+b x)}} \, dx}{c^2 \sqrt {c \sec (a+b x)}}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}-\frac {\left (d^2 \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 )}{b c^2 \sqrt {c \sec (a+b x)}}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}-\frac {\left (2 d^2 \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 )}{b c^2 \sqrt {c \sec (a+b x)}}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}-\frac {\left (d^2 \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 )}{b c^2 \sqrt {c \sec (a+b x)}}-\frac {\left (d^2 \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 )}{b c^2 \sqrt {c \sec (a+b x)}}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}-\frac {\left (d^2 \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 )}{2 b c^2 \sqrt {c \sec (a+b x)}}-\frac {\left (d^2 \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 )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {\left (d^2 \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 )}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}+\frac {\left (d^2 \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 )}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}+\frac {d^2 \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)}}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \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)}}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {\left (d^2 \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 )}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}+\frac {\left (d^2 \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 )}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}\\ &=-\frac {2 d (d \csc (a+b x))^{3/2}}{3 b c (c \sec (a+b x))^{3/2}}+\frac {d^2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{\sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}+\frac {d^2 \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)}}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}-\frac {d^2 \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)}}{2 \sqrt {2} b c^2 \sqrt {c \sec (a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 1.67, size = 154, normalized size = 0.47 \begin {gather*} -\frac {d^3 \left (4 \cot ^2(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 ) \sqrt [4]{\cot ^2(a+b x)}+3 \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)}}{6 b c^3 \sqrt {d \csc (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(d^3*(4*Cot[a + b*x]^2 - 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)^(1/4) + 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)^(1/4))*Sqrt[c*Sec[a + b*x]])/(b*c^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 = 37.47, size = 1259, normalized size = 3.83

method	result	size
default	\(\text {Expression too large to display}\)	\(1259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/b*(3*I*cos(b*x+a)*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))-3*I*cos(b*x+a)*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))+3*I*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+si
n(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))-3*I*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+s
in(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))+3*cos(b*x+a)*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))+3*cos(b*x+a)*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))-6*cos(b*x+a)*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))+3*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-s
in(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*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-si
n(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*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))+2*cos(b*x+a)^2*2^(1/2))*(d/sin(b*x+a))^(5/2)*sin(b*x+a)/cos(b*x+a)^3/(c
/cos(b*x+a))^(5/2)*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((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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