3.559 \(\int \frac{c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{(3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[Out]

-((3*c + 5*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) - ((c
- d)*Cos[e + f*x])/(4*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*c + 5*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])
^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0979192, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2750, 2650, 2649, 206} \[ -\frac{(3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*c + 5*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) - ((c
- d)*Cos[e + f*x])/(4*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*c + 5*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])
^(3/2))

Rule 2750

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

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

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

Rubi steps

\begin{align*} \int \frac{c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{(3 c+5 d) \int \frac{1}{(a+a \sin (e+f x))^{3/2}} \, dx}{8 a}\\ &=-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac{(3 c+5 d) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(3 c+5 d) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{16 a^2 f}\\ &=-\frac{(3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.352523, size = 227, normalized size = 1.8 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (8 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )-(3 c+5 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (3 c+5 d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (d-c) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (3 c+5 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{16 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(c - d)*Sin[(e + f*x)/2] + 4*(-c + d)*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2]) + 2*(3*c + 5*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (3*c + 5*d)*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*c + 5*d)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4
])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4))/(16*f*(a*(1 + Sin[e + f*x]))^(5/2))

________________________________________________________________________________________

Maple [B]  time = 1.065, size = 279, normalized size = 2.2 \begin{align*} -{\frac{1}{ \left ( 32+32\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 2\,\sin \left ( fx+e \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \sqrt{2}{a}^{3} \left ( 3\,c+5\,d \right ) -{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) \sqrt{2}{a}^{3} \left ( 3\,c+5\,d \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+20\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{5/2}c+12\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{5/2}d-6\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}{a}^{3/2}c-10\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}{a}^{3/2}d+6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3}c+10\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3}d \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(2*sin(f*x+e)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^3*(3*c+5*d)-arctanh(1/2*(a-a
*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^3*(3*c+5*d)*cos(f*x+e)^2+20*(a-a*sin(f*x+e))^(1/2)*a^(5/2)*c+12*
(a-a*sin(f*x+e))^(1/2)*a^(5/2)*d-6*(a-a*sin(f*x+e))^(3/2)*a^(3/2)*c-10*(a-a*sin(f*x+e))^(3/2)*a^(3/2)*d+6*2^(1
/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*c+10*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^
(1/2)/a^(1/2))*a^3*d)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(11/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d \sin \left (f x + e\right ) + c}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.00946, size = 1015, normalized size = 8.06 \begin{align*} \frac{\sqrt{2}{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) +{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) - 12 \, c - 20 \, d\right )} \sin \left (f x + e\right ) - 12 \, c - 20 \, d\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \,{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} +{\left (7 \, c + d\right )} \cos \left (f x + e\right ) +{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) - 4 \, c + 4 \, d\right )} \sin \left (f x + e\right ) + 4 \, c - 4 \, d\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{64 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(2)*((3*c + 5*d)*cos(f*x + e)^3 + 3*(3*c + 5*d)*cos(f*x + e)^2 - 2*(3*c + 5*d)*cos(f*x + e) + ((3*c
+ 5*d)*cos(f*x + e)^2 - 2*(3*c + 5*d)*cos(f*x + e) - 12*c - 20*d)*sin(f*x + e) - 12*c - 20*d)*sqrt(a)*log(-(a*
cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x +
e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x +
e) - 2)) + 4*((3*c + 5*d)*cos(f*x + e)^2 + (7*c + d)*cos(f*x + e) + ((3*c + 5*d)*cos(f*x + e) - 4*c + 4*d)*sin
(f*x + e) + 4*c - 4*d)*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(
f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2