3.113 \(\int \frac{a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=160 \[ \frac{2 a \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}+\frac{a \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}-\frac{2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}} \]

[Out]

(a*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) + (a*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) -
(2*a)/(3*d*e*(e*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*EllipticF[(c
- Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]])

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Rubi [A]  time = 0.200585, antiderivative size = 160, 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 = {3872, 2838, 2564, 325, 329, 212, 206, 203, 2636, 2642, 2641} \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{2 a \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) + (a*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) -
(2*a)/(3*d*e*(e*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*EllipticF[(c
- Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]])

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 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]

Rubi steps

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

Mathematica [A]  time = 0.347795, size = 120, normalized size = 0.75 \[ -\frac{a \sqrt{\sin (c+d x)} (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (2 \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )-3 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )-3 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )+\sqrt{\sin (c+d x)} \csc ^2\left (\frac{1}{2} (c+d x)\right )\right )}{6 d e^2 \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-3*ArcTan[Sqrt[Sin[c + d*x]]] - 3*ArcTanh[Sqrt[Sin[c + d*x]]] + 2*E
llipticF[(-2*c + Pi - 2*d*x)/4, 2] + Csc[(c + d*x)/2]^2*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(6*d*e^2*Sqrt[
e*Sin[c + d*x]])

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Maple [A]  time = 1.561, size = 212, normalized size = 1.3 \begin{align*} -{\frac{2\,a}{3\,ed} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{a}{d}{\it Artanh} \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ){e}^{-{\frac{5}{2}}}}+{\frac{a}{d}\arctan \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ){e}^{-{\frac{5}{2}}}}-{\frac{a}{3\,d{e}^{2}\cos \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}+{\frac{2\,a\sin \left ( dx+c \right ) }{3\,d{e}^{2}\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}-{\frac{2\,a}{3\,d{e}^{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{\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))/(e*sin(d*x+c))^(5/2),x)

[Out]

-2/3*a/d/e/(e*sin(d*x+c))^(3/2)+a*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)+a*arctan((e*sin(d*x+c))^(1/2
)/e^(1/2))/d/e^(5/2)-1/3/d*a/e^2*sin(d*x+c)^(1/2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2+2*s
in(d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2/3/d*a/e^2*sin(d*x+c)/cos(d*x+c)/(e*sin(d*x+c))
^(1/2)-2/3/d*a/e^2/sin(d*x+c)/cos(d*x+c)/(e*sin(d*x+c))^(1/2)

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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))/(e*sin(d*x+c))^(5/2),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 \sec \left (d x + c\right ) + a\right )} \sqrt{e \sin \left (d x + c\right )}}{{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\right )} \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))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-(a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c))/((e^3*cos(d*x + c)^2 - e^3)*sin(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((a+a*sec(d*x+c))/(e*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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