3.303 \(\int \frac{1}{\sqrt{e \csc (c+d x)} (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=199 \[ \frac{4 \csc ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{4 \csc (c+d x)}{a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc ^2(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{16 \cot (c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{28 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 a^2 d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

[Out]

(16*Cot[c + d*x])/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) - (2*Cot[c + d*x]^3)/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) - (4*Csc[
c + d*x])/(a^2*d*Sqrt[e*Csc[c + d*x]]) - (2*Cot[c + d*x]*Csc[c + d*x]^2)/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) + (4*C
sc[c + d*x]^3)/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) + (28*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a^2*d*Sqrt[e*Csc[c +
d*x]]*Sqrt[Sin[c + d*x]])

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Rubi [A]  time = 0.468637, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3878, 3872, 2875, 2873, 2567, 2636, 2639, 2564, 14} \[ \frac{4 \csc ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{4 \csc (c+d x)}{a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc ^2(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{16 \cot (c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{28 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 a^2 d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*Cot[c + d*x])/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) - (2*Cot[c + d*x]^3)/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) - (4*Csc[
c + d*x])/(a^2*d*Sqrt[e*Csc[c + d*x]]) - (2*Cot[c + d*x]*Csc[c + d*x]^2)/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) + (4*C
sc[c + d*x]^3)/(5*a^2*d*Sqrt[e*Csc[c + d*x]]) + (28*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a^2*d*Sqrt[e*Csc[c +
d*x]]*Sqrt[Sin[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 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 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 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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{e \csc (c+d x)} (a+a \sec (c+d x))^2} \, dx &=\frac{\int \frac{\sqrt{\sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \frac{\cos ^2(c+d x) \sqrt{\sin (c+d x)}}{(-a-a \cos (c+d x))^2} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \frac{\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sin ^{\frac{7}{2}}(c+d x)} \, dx}{a^4 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \left (\frac{a^2 \cos ^2(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)}-\frac{2 a^2 \cos ^3(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)}+\frac{a^2 \cos ^4(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)}\right ) \, dx}{a^4 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \frac{\cos ^2(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)} \, dx}{a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{\int \frac{\cos ^4(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)} \, dx}{a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 \int \frac{\cos ^3(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)} \, dx}{a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{2 \cot ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc ^2(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \int \frac{1}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{5 a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{6 \int \frac{\cos ^2(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{5 a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{1-x^2}{x^{7/2}} \, dx,x,\sin (c+d x)\right )}{a^2 d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{16 \cot (c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc ^2(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{2 \int \sqrt{\sin (c+d x)} \, dx}{5 a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{12 \int \sqrt{\sin (c+d x)} \, dx}{5 a^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \left (\frac{1}{x^{7/2}}-\frac{1}{x^{3/2}}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{16 \cot (c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}-\frac{4 \csc (c+d x)}{a^2 d \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc ^2(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{4 \csc ^3(c+d x)}{5 a^2 d \sqrt{e \csc (c+d x)}}+\frac{28 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{5 a^2 d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.61185, size = 252, normalized size = 1.27 \[ \frac{4 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{\csc (c+d x)} \sec ^2(c+d x) \left (-3 \sqrt{\csc (c+d x)} \left ((5 \cos (2 c)-23) \sec (c) \cos (d x)-2 \left (5 \sin (c) \sin (d x)+\sec ^2\left (\frac{1}{2} (c+d x)\right )-10\right )\right )-\frac{28 \sqrt{2} e^{i (c-d x)} \sqrt{\frac{i e^{i (c+d x)}}{-1+e^{2 i (c+d x)}}} \left (\left (1+e^{2 i c}\right ) e^{2 i d x} \sqrt{1-e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )-3 e^{2 i (c+d x)}+3\right )}{1+e^{2 i c}}\right )}{15 a^2 d (\sec (c+d x)+1)^2 \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]^4*Sqrt[Csc[c + d*x]]*Sec[c + d*x]^2*((-28*Sqrt[2]*E^(I*(c - d*x))*Sqrt[(I*E^(I*(c + d*x)))
/(-1 + E^((2*I)*(c + d*x)))]*(3 - 3*E^((2*I)*(c + d*x)) + E^((2*I)*d*x)*(1 + E^((2*I)*c))*Sqrt[1 - E^((2*I)*(c
+ d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))]))/(1 + E^((2*I)*c)) - 3*Sqrt[Csc[c + d*x]]*((-
23 + 5*Cos[2*c])*Cos[d*x]*Sec[c] - 2*(-10 + Sec[(c + d*x)/2]^2 + 5*Sin[c]*Sin[d*x]))))/(15*a^2*d*Sqrt[e*Csc[c
+ d*x]]*(1 + Sec[c + d*x])^2)

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Maple [C]  time = 0.236, size = 793, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/a^2/d*2^(1/2)*(-1+cos(d*x+c))*(28*cos(d*x+c)^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)*((-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))-14*cos(d*x+c)^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)*((-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))+56*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*((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)/s
in(d*x+c))^(1/2),1/2*2^(1/2))-28*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*((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))+28*((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)*EllipticE(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),
1/2*2^(1/2))-14*((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)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))+5
*cos(d*x+c)^2*2^(1/2)+cos(d*x+c)*2^(1/2)-6*2^(1/2))/(e/sin(d*x+c))^(1/2)/sin(d*x+c)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}{a^{2} e \csc \left (d x + c\right ) \sec \left (d x + c\right )^{2} + 2 \, a^{2} e \csc \left (d x + c\right ) \sec \left (d x + c\right ) + a^{2} e \csc \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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