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

Optimal. Leaf size=151 \[ -\frac{(3 A c+5 A d+5 B c+19 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(A-B) (c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[Out]

-((3*A*c + 5*B*c + 5*A*d + 19*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqr
t[2]*a^(5/2)*f) - ((A - B)*(c - d)*Cos[e + f*x])/(4*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*A*c + 5*B*c + 5*A*d -
13*B*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.28746, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2968, 3019, 2750, 2649, 206} \[ -\frac{(3 A c+5 A d+5 B c+19 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(A-B) (c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A*c + 5*B*c + 5*A*d + 19*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqr
t[2]*a^(5/2)*f) - ((A - B)*(c - d)*Cos[e + f*x])/(4*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*A*c + 5*B*c + 5*A*d -
13*B*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])^(3/2))

Rule 2968

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

Rule 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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 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{(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^{5/2}} \, dx &=\int \frac{A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\\ &=-\frac{(A-B) (c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{\int \frac{-\frac{1}{2} a (3 A c+5 B c+5 A d-5 B d)-4 a B d \sin (e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) (c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac{(3 A c+5 B c+5 A d+19 B d) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B) (c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(3 A c+5 B c+5 A d+19 B 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 A c+5 B c+5 A d+19 B 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{(A-B) (c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.767319, size = 267, normalized size = 1.77 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (8 (A-B) (c-d) \sin \left (\frac{1}{2} (e+f x)\right )-(3 A c+5 A d+5 B c-13 B d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (3 A c+5 A d+5 B c-13 B d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-4 (A-B) (c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (3 A c+5 A d+5 B c+19 B d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{16 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(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*(A - B)*(c - d)*Sin[(e + f*x)/2] - 4*(A - B)*(c - d)*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2]) + 2*(3*A*c + 5*B*c + 5*A*d - 13*B*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)
/2])^2 - (3*A*c + 5*B*c + 5*A*d - 13*B*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*A*c
+ 5*B*c + 5*A*d + 19*B*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))

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Maple [B]  time = 1.544, size = 449, normalized size = 3. \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 ) \sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( 3\,Ac+5\,Ad+5\,Bc+19\,Bd \right ) -\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2} \left ( 3\,Ac+5\,Ad+5\,Bc+19\,Bd \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+6\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}c+10\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}d+20\,A\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}c+12\,A\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}d-6\,A \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}c-10\,A \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}d+10\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}c+38\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}d+12\,B\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}c-44\,B\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}d-10\,B \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}c+26\,B \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}d \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{9}{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((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x)

[Out]

-1/32*(2*sin(f*x+e)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*(3*A*c+5*A*d+5*B*c+19*B*d)
-2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*(3*A*c+5*A*d+5*B*c+19*B*d)*cos(f*x+e)^2+6*A*2
^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c+10*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/
2)*2^(1/2)/a^(1/2))*a^2*d+20*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c+12*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d-6*A*(a-a
*sin(f*x+e))^(3/2)*a^(1/2)*c-10*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d+10*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(
1/2)*2^(1/2)/a^(1/2))*a^2*c+38*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d+12*B*(a-a*s
in(f*x+e))^(1/2)*a^(3/2)*c-44*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d-10*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c+26*B*(a
-a*sin(f*x+e))^(3/2)*a^(1/2)*d)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(9/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^
(1/2)/f

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

[Out]

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

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Fricas [B]  time = 2.08343, size = 1343, normalized size = 8.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(2)*(((3*A + 5*B)*c + (5*A + 19*B)*d)*cos(f*x + e)^3 + 3*((3*A + 5*B)*c + (5*A + 19*B)*d)*cos(f*x +
e)^2 - 4*(3*A + 5*B)*c - 4*(5*A + 19*B)*d - 2*((3*A + 5*B)*c + (5*A + 19*B)*d)*cos(f*x + e) + (((3*A + 5*B)*c
+ (5*A + 19*B)*d)*cos(f*x + e)^2 - 4*(3*A + 5*B)*c - 4*(5*A + 19*B)*d - 2*((3*A + 5*B)*c + (5*A + 19*B)*d)*cos
(f*x + e))*sin(f*x + e))*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 - (co
s(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(((3*A + 5*B)*c + (5*A - 13*B)*d)*cos(f*x + e)^2 + 4*(A
- B)*c - 4*(A - B)*d + ((7*A + B)*c + (A - 9*B)*d)*cos(f*x + e) - (4*(A - B)*c - 4*(A - B)*d - ((3*A + 5*B)*c
+ (5*A - 13*B)*d)*cos(f*x + e))*sin(f*x + e))*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))

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

[Out]

Timed out

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

[Out]

sage2