3.333 \(\int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=121 \[ -\frac {3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {11 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}+\frac {5 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d} \]

[Out]

-3*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-cot(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d+11/3*a^2*c
os(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+5/3*a*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.31, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2716, 2976, 2981, 2773, 206} \[ \frac {11 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {5 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (11*a^2*Cos[c + d*x])/(3*d*Sqrt[a +
a*Sin[c + d*x]]) + (5*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) - (Cot[c + d*x]*(a + a*Sin[c + d*x])^(3/2
))/d

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 2716

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e +
f*x])^m/(f*Tan[e + f*x]), x] + Dist[1/a, Int[((a + b*Sin[e + f*x])^m*(b*m - a*(m + 1)*Sin[e + f*x]))/Sin[e + f
*x], x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] &&  !LtQ[m, -1]

Rule 2773

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 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

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)*Sqr
t[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 \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}+\frac {\int \csc (c+d x) \left (\frac {3 a}{2}-\frac {5}{2} a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{a}\\ &=\frac {5 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}+\frac {2 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {9 a^2}{4}-\frac {11}{4} a^2 \sin (c+d x)\right ) \, dx}{3 a}\\ &=\frac {11 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}+\frac {1}{2} (3 a) \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {11 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {\left (3 a^2\right ) \operatorname {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 {3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {11 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 233, normalized size = 1.93 \[ -\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sin (c+d x)+1)} \left (-14 \sin \left (\frac {1}{2} (c+d x)\right )-9 \sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {5}{2} (c+d x)\right )+14 \cos \left (\frac {1}{2} (c+d x)\right )-9 \cos \left (\frac {3}{2} (c+d x)\right )+\cos \left (\frac {5}{2} (c+d x)\right )+9 \sin (c+d x) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-9 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{3 d \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(14*Cos[(c + d*x)/2] - 9*Cos[(3*(c + d*x))/2] + Cos[(5*(
c + d*x))/2] - 14*Sin[(c + d*x)/2] + 9*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 9*Log[1 - C
os[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 9*Sin[(3*(c + d*x))/2] - Sin[(5*(c + d*x))/2]))/(d*(1 + Cot
[(c + d*x)/2])*(Csc[(c + d*x)/4] - Sec[(c + d*x)/4])*(Csc[(c + d*x)/4] + Sec[(c + d*x)/4]))

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fricas [B]  time = 0.49, size = 315, normalized size = 2.60 \[ \frac {9 \, {\left (a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - 8 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - {\left (2 \, a \cos \left (d x + c\right )^{2} + 10 \, a \cos \left (d x + c\right ) + 11 \, a\right )} \sin \left (d x + c\right ) + 11 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{12 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(9*(a*cos(d*x + c)^2 - (a*cos(d*x + c) + a)*sin(d*x + c) - a)*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)*s
qrt(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 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*(2*a*cos(d*x + c)^3 - 8*a*cos(d*x + c)
^2 + a*cos(d*x + c) - (2*a*cos(d*x + c)^2 + 10*a*cos(d*x + c) + 11*a)*sin(d*x + c) + 11*a)*sqrt(a*sin(d*x + c)
+ a))/(d*cos(d*x + c)^2 - (d*cos(d*x + c) + d)*sin(d*x + c) - d)

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giac [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(cos(d*x+c)^2*csc(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.18, size = 144, normalized size = 1.19 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) \left (2 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-12 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}+9 \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{2}\right )+3 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right )}{3 \sin \left (d x +c \right ) \sqrt {a}\, \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(sin(d*x+c)*(2*(a-a*sin(d*x+c))^(3/2)*a^(1/2)-12*(a-a*sin(d*x+c)
)^(1/2)*a^(3/2)+9*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*a^2)+3*(a-a*sin(d*x+c))^(1/2)*a^(3/2))/sin(d*x+c)/a^
(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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \csc \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^2*csc(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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