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

Optimal. Leaf size=154 \[ \frac{3 a^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right ) \sqrt{e \csc (c+d x)}}{d}+\frac{a^2 \tan (c+d x) \sqrt{e \csc (c+d x)}}{d}+\frac{2 a^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d}+\frac{2 a^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.262707, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3878, 3872, 2873, 2641, 2564, 329, 212, 206, 203, 2571} \[ \frac{a^2 \tan (c+d x) \sqrt{e \csc (c+d x)}}{d}+\frac{2 a^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d}+\frac{2 a^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d}+\frac{3 a^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \csc (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2571

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

Rubi steps

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

Mathematica [C]  time = 2.43572, size = 168, normalized size = 1.09 \[ -\frac{2 a^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{e \csc (c+d x)} \sec ^4\left (\frac{1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (3 \sqrt{-\cot ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},\csc ^2(c+d x)\right )+2 \sqrt{\cos ^2(c+d x)} \sqrt{\csc (c+d x)} \tan ^{-1}\left (\sqrt{\csc (c+d x)}\right )-2 \sqrt{\cos ^2(c+d x)} \sqrt{\csc (c+d x)} \tanh ^{-1}\left (\sqrt{\csc (c+d x)}\right )-1\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*a^2*Cos[(c + d*x)/2]^5*Sqrt[e*Csc[c + d*x]]*(-1 + 2*ArcTan[Sqrt[Csc[c + d*x]]]*Sqrt[Cos[c + d*x]^2]*Sqrt[C
sc[c + d*x]] - 2*ArcTanh[Sqrt[Csc[c + d*x]]]*Sqrt[Cos[c + d*x]^2]*Sqrt[Csc[c + d*x]] + 3*Sqrt[-Cot[c + d*x]^2]
*Hypergeometric2F1[1/4, 1/2, 5/4, Csc[c + d*x]^2])*Sec[c + d*x]*Sec[ArcCsc[Csc[c + d*x]]/2]^4*Sin[(c + d*x)/2]
)/d

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Maple [C]  time = 0.242, size = 729, normalized size = 4.7 \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))^2*(e*csc(d*x+c))^(1/2),x)

[Out]

1/2*a^2/d*2^(1/2)*(-1+cos(d*x+c))*(-2*I*((-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)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*cos(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))-2*I*((-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)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*cos(d*x+c)*Ellip
ticPi(((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*((-I*cos(d*x+c)+sin(d*x+c)+I)/si
n(d*x+c))^(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)*sin(d*x+c
)*cos(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))+2*sin(d*x+c)*cos(d*x+c)*Ell
ipticPi(((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)/si
n(d*x+c))^(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)-2*sin(d*x
+c)*cos(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*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)+cos(d*x+c)*2^(1/2)-2^(1/2))*(cos(d*x+c)+1)^2*(e/sin(d*x+c))^(1/2)/sin(d*x+c)^3/cos(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((a+a*sec(d*x+c))^2*(e*csc(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{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((a+a*sec(d*x+c))^2*(e*csc(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((a+a*sec(d*x+c))^2*(e*csc(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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