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

Optimal. Leaf size=182 \[ -\frac {2 a}{d e \sqrt {e \csc (c+d x)}}-\frac {2 a \cos (c+d x)}{3 d e \sqrt {e \csc (c+d x)}}+\frac {a \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 182, normalized size of antiderivative = 1.00, 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 = {3878, 3872, 2838, 2564, 321, 329, 212, 206, 203, 2635, 2641} \[ -\frac {2 a}{d e \sqrt {e \csc (c+d x)}}-\frac {2 a \cos (c+d x)}{3 d e \sqrt {e \csc (c+d x)}}+\frac {a \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 11.03, size = 135, normalized size = 0.74 \[ -\frac {a \left (4 \cos (c+d x)+3 \sqrt {\csc (c+d x)} \log \left (1-\sqrt {\csc (c+d x)}\right )-3 \sqrt {\csc (c+d x)} \log \left (\sqrt {\csc (c+d x)}+1\right )+6 \sqrt {\csc (c+d x)} \tan ^{-1}\left (\sqrt {\csc (c+d x)}\right )+\frac {4 F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )}{\sqrt {\sin (c+d x)}}+12\right )}{6 d e \sqrt {e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*(12 + 4*Cos[c + d*x] + 6*ArcTan[Sqrt[Csc[c + d*x]]]*Sqrt[Csc[c + d*x]] + 3*Sqrt[Csc[c + d*x]]*Log[1 -
Sqrt[Csc[c + d*x]]] - 3*Sqrt[Csc[c + d*x]]*Log[1 + Sqrt[Csc[c + d*x]]] + (4*EllipticF[(-2*c + Pi - 2*d*x)/4, 2
])/Sqrt[Sin[c + d*x]]))/(d*e*Sqrt[e*Csc[c + d*x]])

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fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \csc \left (d x + c\right )} {\left (a \sec \left (d x + c\right ) + a\right )}}{e^{2} \csc \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.32, size = 710, normalized size = 3.90 \[ -\frac {a \left (3 i \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 i \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+3 i \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+3 \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+4 \cos \left (d x +c \right ) \sqrt {2}-6 \sqrt {2}\right ) \sqrt {2}}{6 d \left (-1+\cos \left (d x +c \right )\right ) \left (\frac {e}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a/d*(3*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*
(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-
1/2*I,1/2*2^(1/2))-4*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c
))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1
/2),1/2*2^(1/2))+3*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))
^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/
2),1/2+1/2*I,1/2*2^(1/2))-3*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(
d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+
c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c
))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/s
in(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+2*cos(d*x+c)^2*2^(1/2)+4*cos(d*x+c)*2^(1/2)-6*2^(1/2))/(-1+cos(d*x+c))
/(e/sin(d*x+c))^(3/2)/sin(d*x+c)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))/(e/sin(c + d*x))^(3/2),x)

[Out]

int((a + a/cos(c + d*x))/(e/sin(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {1}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec {\left (c + d x \right )}}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral((e*csc(c + d*x))**(-3/2), x) + Integral(sec(c + d*x)/(e*csc(c + d*x))**(3/2), x))

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