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

Optimal. Leaf size=155 \[ -\frac{a \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{3/2}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{3/2}}-\frac{2 a E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a}{d e \sqrt{e \sin (c+d x)}}-\frac{2 a \cos (c+d x)}{d e \sqrt{e \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.199393, antiderivative size = 155, 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, 298, 203, 206, 2636, 2640, 2639} \[ -\frac{a \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{3/2}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{3/2}}-\frac{2 a E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a}{d e \sqrt{e \sin (c+d x)}}-\frac{2 a \cos (c+d x)}{d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2))) + (a*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2)
) - (2*a)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*EllipticE[(c - Pi/
2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt[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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/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 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 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 2640

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

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]

Rubi steps

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

Mathematica [A]  time = 0.362867, size = 143, normalized size = 0.92 \[ -\frac{a \sin ^{\frac{3}{2}}(c+d x) (\cos (c+d x)+1) \sec \left (\frac{1}{2} (c+d x)\right ) \left (2 \sqrt{\sin (c+d x)} \csc \left (\frac{1}{2} (c+d x)\right )-2 \sec \left (\frac{1}{2} (c+d x)\right ) E\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+\sec \left (\frac{1}{2} (c+d x)\right ) \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )-\sec \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )\right )}{2 d (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.326, size = 247, normalized size = 1.6 \begin{align*}{\frac{a}{d}{\it Artanh} \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ){e}^{-{\frac{3}{2}}}}-{\frac{a}{d}\arctan \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ){e}^{-{\frac{3}{2}}}}-2\,{\frac{a}{ed\sqrt{e\sin \left ( dx+c \right ) }}}+2\,{\frac{a\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) }{ed\cos \left ( dx+c \right ) \sqrt{e\sin \left ( dx+c \right ) }}}-{\frac{a}{ed\cos \left ( dx+c \right ) }\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},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}-2\,{\frac{a\cos \left ( dx+c \right ) }{ed\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))^(3/2),x)

[Out]

a*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)-a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)-2*a/d/e/(e*
sin(d*x+c))^(1/2)+2/d*a/e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x
+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-1/d*a/e/cos(d*x+c)/(e*sin(d*x+c))^(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))-2*a*cos(d*x+c)/d/e
/(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))^(3/2),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 )}}{e^{2} \cos \left (d x + c\right )^{2} - e^{2}}, 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))^(3/2),x, algorithm="fricas")

[Out]

integral(-(a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c))/(e^2*cos(d*x + c)^2 - e^2), 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))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

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