3.286 \(\int \frac{a+a \sec (c+d x)}{(e \csc (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=197 \[ -\frac{2 a \sin (c+d x)}{3 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 a \sin (c+d x) \cos (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}-\frac{a \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{a \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{6 a E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]
[Out]
-((a*ArcTan[Sqrt[Sin[c + d*x]]])/(d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])) + (a*ArcTanh[Sqrt[Sin[c + d*
x]]])/(d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (6*a*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*d*e^2*Sqrt[e
*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) - (2*a*Sin[c + d*x])/(3*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*a*Cos[c + d*x]*Sin
[c + d*x])/(5*d*e^2*Sqrt[e*Csc[c + d*x]])
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Rubi [A] time = 0.172174, antiderivative size = 197, normalized size of antiderivative =
1., number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) =
0.478, Rules used = {3878, 3872, 2838, 2564, 321, 329, 298, 203, 206, 2635, 2639}
\[ -\frac{2 a \sin (c+d x)}{3 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 a \sin (c+d x) \cos (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}-\frac{a \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{a \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{6 a E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])/(e*Csc[c + d*x])^(5/2),x]
[Out]
-((a*ArcTan[Sqrt[Sin[c + d*x]]])/(d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])) + (a*ArcTanh[Sqrt[Sin[c + d*
x]]])/(d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (6*a*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*d*e^2*Sqrt[e
*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) - (2*a*Sin[c + d*x])/(3*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*a*Cos[c + d*x]*Sin
[c + d*x])/(5*d*e^2*Sqrt[e*Csc[c + d*x]])
Rule 3878
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[g^Int
Part[p]*(g*Sec[e + f*x])^FracPart[p]*Cos[e + f*x]^FracPart[p], Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x],
x] /; FreeQ[{a, b, e, f, g, m, p}, x] && !IntegerQ[p]
Rule 3872
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Rule 2838
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Rule 2564
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
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 298
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
& !GtQ[a/b, 0]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \frac{a+a \sec (c+d x)}{(e \csc (c+d x))^{5/2}} \, dx &=\frac{\int (a+a \sec (c+d x)) \sin ^{\frac{5}{2}}(c+d x) \, dx}{e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{\int (-a-a \cos (c+d x)) \sec (c+d x) \sin ^{\frac{5}{2}}(c+d x) \, dx}{e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{a \int \sin ^{\frac{5}{2}}(c+d x) \, dx}{e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a \int \sec (c+d x) \sin ^{\frac{5}{2}}(c+d x) \, dx}{e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{2 a \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}+\frac{(3 a) \int \sqrt{\sin (c+d x)} \, dx}{5 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a \operatorname{Subst}\left (\int \frac{x^{5/2}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{6 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 a \sin (c+d x)}{3 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 a \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}+\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{6 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 a \sin (c+d x)}{3 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 a \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{6 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 a \sin (c+d x)}{3 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 a \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}+\frac{a \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{a \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{6 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 a \sin (c+d x)}{3 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 a \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt{e \csc (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.39265, size = 165, normalized size = 0.84 \[ \frac{a \left (-72 \cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},\csc ^2(c+d x)\right )-2 \sqrt{-\cot ^2(c+d x)} \left (20 \sin (c+d x)+6 \sin (2 (c+d x))+15 \sqrt{\csc (c+d x)} \left (\log \left (1-\sqrt{\csc (c+d x)}\right )-\log \left (\sqrt{\csc (c+d x)}+1\right )\right )-30 \sqrt{\csc (c+d x)} \tan ^{-1}\left (\sqrt{\csc (c+d x)}\right )\right )\right )}{60 d e^2 \sqrt{-\cot ^2(c+d x)} \sqrt{e \csc (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sec[c + d*x])/(e*Csc[c + d*x])^(5/2),x]
[Out]
(a*(-72*Cot[c + d*x]*Hypergeometric2F1[-1/4, 1/2, 3/4, Csc[c + d*x]^2] - 2*Sqrt[-Cot[c + d*x]^2]*(-30*ArcTan[S
qrt[Csc[c + d*x]]]*Sqrt[Csc[c + d*x]] + 15*Sqrt[Csc[c + d*x]]*(Log[1 - Sqrt[Csc[c + d*x]]] - Log[1 + Sqrt[Csc[
c + d*x]]]) + 20*Sin[c + d*x] + 6*Sin[2*(c + d*x)])))/(60*d*e^2*Sqrt[-Cot[c + d*x]^2]*Sqrt[e*Csc[c + d*x]])
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Maple [C] time = 0.227, size = 1565, normalized size = 7.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))/(e*csc(d*x+c))^(5/2),x)
[Out]
1/30*a/d*2^(1/2)*(-15*I*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x
+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((I*cos(d*x+c)+sin
(d*x+c)-I)/sin(d*x+c))^(1/2)-15*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+
c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((I*cos(d*x+c)+sin(
d*x+c)-I)/sin(d*x+c))^(1/2)+15*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c)
)^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1
/2),1/2+1/2*I,1/2*2^(1/2))+15*I*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/
sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(
d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+18*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d
*x+c)-I)/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-
I)/sin(d*x+c))^(1/2),1/2*2^(1/2))+15*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+
c)-I)/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+15*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-
sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2
))*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)-36*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos
(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticE(((I*cos(d*x+c
)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))+6*cos(d*x+c)^3*2^(1/2)+18*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*
2^(1/2))*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)+15*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+
c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(
1/2))*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)+15*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-
sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2
))*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)-36*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin
(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticE(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))*((I*cos(d*x
+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)+10*cos(d*x+c)^2*2^(1/2)-24*cos(d*x+c)*2^(1/2)+8*2^(1/2))/(e/sin(d*x+c))^(5
/2)/sin(d*x+c)^3
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Maxima [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((a+a*sec(d*x+c))/(e*csc(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e \csc \left (d x + c\right )}{\left (a \sec \left (d x + c\right ) + a\right )}}{e^{3} \csc \left (d x + c\right )^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))/(e*csc(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(e*csc(d*x + c))*(a*sec(d*x + c) + a)/(e^3*csc(d*x + c)^3), x)
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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((a+a*sec(d*x+c))/(e*csc(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a \sec \left (d x + c\right ) + a}{\left (e \csc \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))/(e*csc(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)/(e*csc(d*x + c))^(5/2), x)