3.281 \(\int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=190 \[ \frac{2 a e^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right ) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \csc (c+d x) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{a e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d}+\frac{a e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d} \]

[Out]

(-2*a*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*d) - (2*a*e^2*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*d) + (a*e^
2*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])/d + (a*e^2*ArcTanh[Sqrt[Sin[c + d*x]]]*S
qrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])/d + (2*a*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqr
t[Sin[c + d*x]])/(3*d)

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Rubi [A]  time = 0.165207, antiderivative size = 190, 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, 325, 329, 212, 206, 203, 2636, 2641} \[ -\frac{2 a e^2 \csc (c+d x) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{a e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d}+\frac{a e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d}+\frac{2 a e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \csc (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*d) - (2*a*e^2*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*d) + (a*e^
2*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])/d + (a*e^2*ArcTanh[Sqrt[Sin[c + d*x]]]*S
qrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])/d + (2*a*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqr
t[Sin[c + d*x]])/(3*d)

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/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 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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x)) \, dx &=\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{a+a \sec (c+d x)}{\sin ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\left (\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{(-a-a \cos (c+d x)) \sec (c+d x)}{\sin ^{\frac{5}{2}}(c+d x)} \, dx\right )\\ &=\left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sin ^{\frac{5}{2}}(c+d x)} \, dx+\left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sin ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{1}{3} \left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx+\frac{\left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{x^{5/2} \left (1-x^2\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \csc (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{2 a e^2 \sqrt{e \csc (c+d x)} F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d}+\frac{\left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-x^2\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \csc (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{2 a e^2 \sqrt{e \csc (c+d x)} F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d}+\frac{\left (2 a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^4} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d}\\ &=-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \csc (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{2 a e^2 \sqrt{e \csc (c+d x)} F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d}+\frac{\left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d}+\frac{\left (a e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d}\\ &=-\frac{2 a e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{3 d}-\frac{2 a e^2 \csc (c+d x) \sqrt{e \csc (c+d x)}}{3 d}+\frac{a e^2 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right ) \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}{d}+\frac{a e^2 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right ) \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}{d}+\frac{2 a e^2 \sqrt{e \csc (c+d x)} F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.43489, size = 135, normalized size = 0.71 \[ -\frac{a (e \csc (c+d x))^{5/2} \left (4 \sqrt{\sin (c+d x)} \sqrt{\csc (c+d x)} \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )+4 \cot \left (\frac{1}{2} (c+d x)\right ) \sqrt{\csc (c+d x)}+3 \log \left (1-\sqrt{\csc (c+d x)}\right )-3 \log \left (\sqrt{\csc (c+d x)}+1\right )+6 \tan ^{-1}\left (\sqrt{\csc (c+d x)}\right )\right )}{6 d \csc ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(e*Csc[c + d*x])^(5/2)*(6*ArcTan[Sqrt[Csc[c + d*x]]] + 4*Cot[(c + d*x)/2]*Sqrt[Csc[c + d*x]] + 3*Log[1 - S
qrt[Csc[c + d*x]]] - 3*Log[1 + Sqrt[Csc[c + d*x]]] + 4*Sqrt[Csc[c + d*x]]*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*
Sqrt[Sin[c + d*x]]))/(6*d*Csc[c + d*x]^(5/2))

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Maple [C]  time = 0.335, size = 694, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c)),x)

[Out]

1/6*a/d*2^(1/2)*(4*I*sin(d*x+c)*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)*(-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)-3*I*sin(d*x+c)*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)*(-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)-3*I*sin(d*x+c)*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)*(-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)+3*sin(d*x+c)*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)*(-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)-3*sin(d*x+c)*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)*(-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)+2*2^(1/2))*(-1+cos(d*x+c))*(cos(d*x+c)+1)^2*(e/sin(d*x+c))^(5/2
)/sin(d*x+c)

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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((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a e^{2} \csc \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a e^{2} \csc \left (d x + c\right )^{2}\right )} \sqrt{e \csc \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*csc(d*x + c)^2*sec(d*x + c) + a*e^2*csc(d*x + c)^2)*sqrt(e*csc(d*x + c)), 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((e*csc(d*x+c))**(5/2)*(a+a*sec(d*x+c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*csc(d*x + c))^(5/2)*(a*sec(d*x + c) + a), x)