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

Optimal. Leaf size=123 \[ -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}+\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}+\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2874, 2976, 2981, 2773, 206} \[ -\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}+\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (2*a^2*Cos[c + d*x])/(5*d*Sqrt[a + a
*Sin[c + d*x]]) + (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(5*d) + (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/
2))/(5*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 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 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 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 \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac {\int \csc (c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{5/2} \, dx}{a^2}\\ &=\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {2 \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \left (\frac {5 a^2}{2}-\frac {3}{2} a^2 \sin (c+d x)\right ) \, dx}{5 a^2}\\ &=\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {4 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {15 a^3}{4}+\frac {3}{4} a^3 \sin (c+d x)\right ) \, dx}{15 a^2}\\ &=-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+a \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}-\frac {\left (2 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 {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 145, normalized size = 1.18 \[ \frac {(a (\sin (c+d x)+1))^{3/2} \left (5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )+5 \cos \left (\frac {3}{2} (c+d x)\right )-\cos \left (\frac {5}{2} (c+d x)\right )-10 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+10 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{10 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(3/2)*(5*Cos[(3*(c + d*x))/2] - Cos[(5*(c + d*x))/2] - 10*Log[1 + Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 10*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 5*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x)
)/2]))/(10*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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fricas [B]  time = 0.52, size = 282, normalized size = 2.29 \[ \frac {5 \, {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{10 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(5*(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)*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 + (cos(d*x
+ c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) - 4*(a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) -
(a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*sin(d*x + c) + a)*sqrt(a*sin(d*x + c) + a))/(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)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.08, size = 121, normalized size = 0.98 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-5 a^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}-5 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a +5 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{5 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-5*a^(5/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))+(a-a*sin(d*x+
c))^(5/2)-5*(a-a*sin(d*x+c))^(3/2)*a+5*a^2*(a-a*sin(d*x+c))^(1/2))/a/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \csc \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(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), x)

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

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),x)

[Out]

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

[Out]

Timed out

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