3.291 \(\int \frac{(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=222 \[ -\frac{a^2 \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}-\frac{4 a^2}{d e \sqrt{e \csc (c+d x)}}-\frac{2 a^2 \cos (c+d x)}{3 d e \sqrt{e \csc (c+d x)}}+\frac{a^2 \sec (c+d x)}{d e \sqrt{e \csc (c+d x)}}+\frac{2 a^2 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{2 a^2 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

[Out]

(-4*a^2)/(d*e*Sqrt[e*Csc[c + d*x]]) - (2*a^2*Cos[c + d*x])/(3*d*e*Sqrt[e*Csc[c + d*x]]) + (a^2*Sec[c + d*x])/(
d*e*Sqrt[e*Csc[c + d*x]]) + (2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/(d*e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) +
(2*a^2*ArcTanh[Sqrt[Sin[c + d*x]]])/(d*e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) - (a^2*EllipticF[(c - Pi/2
+ d*x)/2, 2])/(3*d*e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.312573, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {3878, 3872, 2873, 2635, 2641, 2564, 321, 329, 212, 206, 203, 2566} \[ -\frac{4 a^2}{d e \sqrt{e \csc (c+d x)}}-\frac{2 a^2 \cos (c+d x)}{3 d e \sqrt{e \csc (c+d x)}}+\frac{a^2 \sec (c+d x)}{d e \sqrt{e \csc (c+d x)}}+\frac{2 a^2 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{2 a^2 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}-\frac{a^2 F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2/(e*Csc[c + d*x])^(3/2),x]

[Out]

(-4*a^2)/(d*e*Sqrt[e*Csc[c + d*x]]) - (2*a^2*Cos[c + d*x])/(3*d*e*Sqrt[e*Csc[c + d*x]]) + (a^2*Sec[c + d*x])/(
d*e*Sqrt[e*Csc[c + d*x]]) + (2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/(d*e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) +
(2*a^2*ArcTanh[Sqrt[Sin[c + d*x]]])/(d*e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) - (a^2*EllipticF[(c - Pi/2
+ d*x)/2, 2])/(3*d*e*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 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 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 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 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 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 2566

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(a*Sin[e
+ f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Sin[e +
f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rubi steps

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

Mathematica [C]  time = 7.61748, size = 164, normalized size = 0.74 \[ \frac{2 a^2 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tan (c+d x) \sqrt{e \csc (c+d x)} \sec ^4\left (\frac{1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (-6 \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},\csc ^2(c+d x)\right )+3 \sqrt{-\cot ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},\csc ^2(c+d x)\right )+\sin ^2(c+d x) \sqrt{-\cot ^2(c+d x)} \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{3}{2},\frac{1}{4},\csc ^2(c+d x)\right )+3\right )}{3 d e^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^2/(e*Csc[c + d*x])^(3/2),x]

[Out]

(2*a^2*Cos[(c + d*x)/2]^4*Sqrt[e*Csc[c + d*x]]*Sec[ArcCsc[Csc[c + d*x]]/2]^4*(3 - 6*Sqrt[Cos[c + d*x]^2]*Hyper
geometric2F1[-1/4, 1, 3/4, Csc[c + d*x]^2] + 3*Sqrt[-Cot[c + d*x]^2]*Hypergeometric2F1[1/4, 1/2, 5/4, Csc[c +
d*x]^2] + Sqrt[-Cot[c + d*x]^2]*Hypergeometric2F1[-3/4, 3/2, 1/4, Csc[c + d*x]^2]*Sin[c + d*x]^2)*Tan[c + d*x]
)/(3*d*e^2)

________________________________________________________________________________________

Maple [C]  time = 0.218, size = 763, normalized size = 3.4 \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))^(3/2),x)

[Out]

-1/6*a^2/d*2^(1/2)*(6*I*sin(d*x+c)*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)/s
in(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-13*I*sin(d*x+c)*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*c
os(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))+6*I*sin(d*x+c)*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))-6*sin(d*x+c)*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))+6*sin(d*x+c)*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))+2*co
s(d*x+c)^3*2^(1/2)+10*cos(d*x+c)^2*2^(1/2)-15*cos(d*x+c)*2^(1/2)+3*2^(1/2))/(-1+cos(d*x+c))/sin(d*x+c)/cos(d*x
+c)/(e/sin(d*x+c))^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^2/(e*csc(d*x + c))^(3/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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 )}}{e^{2} \csc \left (d x + c\right )^{2}}, 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))^(3/2),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))/(e^2*csc(d*x + c)^2), x)

________________________________________________________________________________________

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))**2/(e*csc(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^2/(e*csc(d*x + c))^(3/2), x)