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

Optimal. Leaf size=169 \[ -\frac {2 a e \sqrt {e \csc (c+d x)}}{d}-\frac {2 a e \cos (c+d x) \sqrt {e \csc (c+d x)}}{d}-\frac {a e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {a e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}-\frac {2 a e \sqrt {\sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 169, 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, 325, 329, 298, 203, 206, 2636, 2639} \[ -\frac {2 a e \sqrt {e \csc (c+d x)}}{d}-\frac {2 a e \cos (c+d x) \sqrt {e \csc (c+d x)}}{d}-\frac {a e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {a e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}-\frac {2 a e \sqrt {\sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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]

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

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Mathematica [C]  time = 1.28, size = 146, normalized size = 0.86 \[ \frac {a (e \csc (c+d x))^{3/2} \left (\frac {2 \sin (2 (c+d x)) \csc ^{\frac {3}{2}}(c+d x) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\csc ^2(c+d x)\right )}{\sqrt {-\cot ^2(c+d x)}}-4 (\cos (c+d x)+1) \sqrt {\csc (c+d x)}-\log \left (1-\sqrt {\csc (c+d x)}\right )+\log \left (\sqrt {\csc (c+d x)}+1\right )+2 \tan ^{-1}\left (\sqrt {\csc (c+d x)}\right )\right )}{2 d \csc ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(e*Csc[c + d*x])^(3/2)*(2*ArcTan[Sqrt[Csc[c + d*x]]] - 4*(1 + Cos[c + d*x])*Sqrt[Csc[c + d*x]] - Log[1 - Sq
rt[Csc[c + d*x]]] + Log[1 + Sqrt[Csc[c + d*x]]] + (2*Csc[c + d*x]^(3/2)*Hypergeometric2F1[-1/4, 1/2, 3/4, Csc[
c + d*x]^2]*Sin[2*(c + d*x)])/Sqrt[-Cot[c + d*x]^2]))/(2*d*Csc[c + d*x]^(3/2))

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fricas [F]  time = 1.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e \csc \left (d x + c\right ) \sec \left (d x + c\right ) + a e \csc \left (d x + c\right )\right )} \sqrt {e \csc \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.66, size = 1522, normalized size = 9.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a/d*(I*cos(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))*(-I*(-1+cos(d*x+c))/si
n(d*x+c))^(1/2)-I*cos(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))*(-I*(-1+cos(d*x
+c))/sin(d*x+c))^(1/2)+cos(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))*(-I*(-1+co
s(d*x+c))/sin(d*x+c))^(1/2)+cos(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))*(-I*(
-1+cos(d*x+c))/sin(d*x+c))^(1/2)+4*cos(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)*EllipticE(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+c
os(d*x+c))/sin(d*x+c))^(1/2)-2*cos(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))*(-I*(-1+cos(d
*x+c))/sin(d*x+c))^(1/2)+I*((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))*(-I*(-1+cos(d*x+c
))/sin(d*x+c))^(1/2)-I*((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))*(-I*(-1+cos(d*x+c))/s
in(d*x+c))^(1/2)+((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))*(-I*(-1+cos(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)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*Elli
pticPi(((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))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(
1/2)+4*((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)*Elliptic
E(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2*((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))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-4*2^(1/2))*(e/sin(d*x+c))
^(3/2)*sin(d*x+c)*2^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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