3.1.56 \(\int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [56]

Optimal. Leaf size=98 \[ -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {14 a^3 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \]

[Out]

-2*a^(5/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-14/3*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2
/3*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.13, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2842, 3060, 2852, 212} \begin {gather*} -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {14 a^3 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (14*a^3*Cos[c + d*x])/(3*d*Sqrt[a +
a*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/(
d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d
*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m,
2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {2}{3} \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {3 a^2}{2}+\frac {7}{2} a^2 \sin (c+d x)\right ) \, dx\\ &=-\frac {14 a^3 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+a^2 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {14 a^3 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}\\ &=-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {14 a^3 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 143, normalized size = 1.46 \begin {gather*} -\frac {(a (1+\sin (c+d x)))^{5/2} \left (15 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )+3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((a*(1 + Sin[c + d*x]))^(5/2)*(15*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] + 3*Log[1 + Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] - 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 15*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2
]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]
time = 2.29, size = 103, normalized size = 1.05

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a \left (3 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )-\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}+9 a \sqrt {a -a \sin \left (d x +c \right )}\right )}{3 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a*(3*a^(3/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))-(a-a*sin(d*
x+c))^(3/2)+9*a*(a-a*sin(d*x+c))^(1/2))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (84) = 168\).
time = 0.34, size = 279, normalized size = 2.85 \begin {gather*} \frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2} \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2} \cos \left (d x + c\right ) + 7 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right ) - 7 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/6*(3*(a^2*cos(d*x + c) + a^2*sin(d*x + c) + a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos
(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos
(d*x + c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (co
s(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) - 4*(a^2*cos(d*x + c)^2 + 8*a^2*cos(d*x + c) + 7*a^2 + (a^
2*cos(d*x + c) - 7*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.53, size = 141, normalized size = 1.44 \begin {gather*} -\frac {\sqrt {2} {\left (8 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 36 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(2)*(8*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 3*sqrt(2)*a^2*log(a
bs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-
1/4*pi + 1/2*d*x + 1/2*c)) - 36*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))*sqrt(a
)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x),x)

[Out]

int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x), x)

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