3.117 \(\int \frac{(a+a \sec (c+d x))^2}{\sqrt{e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=139 \[ \frac{3 a^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{d \sqrt{e \sin (c+d x)}}+\frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d \sqrt{e}}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d \sqrt{e}}+\frac{a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{d e} \]

[Out]

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

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Rubi [A]  time = 0.307262, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3872, 2873, 2642, 2641, 2564, 329, 212, 206, 203, 2571} \[ \frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d \sqrt{e}}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d \sqrt{e}}+\frac{a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{d e}+\frac{3 a^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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

Mathematica [C]  time = 64.3412, size = 164, normalized size = 1.18 \[ \frac{a^2 \sqrt{\sin (c+d x)} \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac{1}{2} \sin ^{-1}(\sin (c+d x))\right ) \left (3 \sqrt{\sin (c+d x)} \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},\sin ^2(c+d x)\right )+\sqrt{\sin (c+d x)}+2 \sqrt{\cos ^2(c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )+2 \sqrt{\cos ^2(c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )\right )}{d \sqrt{e \sin (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sec[ArcSin[Sin[c + d*x]]/2]^4*(2*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c +
d*x]^2] + 2*ArcTanh[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] + Sqrt[Sin[c + d*x]] + 3*Sqrt[Cos[c + d*x]^2]*Hyp
ergeometric2F1[1/4, 1/2, 5/4, Sin[c + d*x]^2]*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x]])

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Maple [A]  time = 2.059, size = 163, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2}}{2\,d\cos \left ( dx+c \right ) } \left ( 3\,\sqrt{e}\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) -4\,\cos \left ( dx+c \right ) \sqrt{e\sin \left ( dx+c \right ) }{\it Artanh} \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) -4\,\cos \left ( dx+c \right ) \sqrt{e\sin \left ( dx+c \right ) }\arctan \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) -2\,\sqrt{e}\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e}}}{\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/e^(1/2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(3*e^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d
*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-4*cos(d*x+c)*(e*sin(d*x+c))^(1/2)*arctanh((e*sin(d*x+
c))^(1/2)/e^(1/2))-4*cos(d*x+c)*(e*sin(d*x+c))^(1/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))-2*e^(1/2)*sin(d*x+c)
)/d

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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*sin(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 (\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 \sin \left (d x + c\right )}}{e \sin \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*sin(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*sin(d*x + c))/(e*sin(d*x + c)), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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