3.316 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=133 \[ -\frac{(A c+3 A d+3 B c-7 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(A-B) (c-d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}-\frac{2 B d \cos (e+f x)}{a f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-((A*c + 3*B*c + 3*A*d - 7*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]
*a^(3/2)*f) - ((A - B)*(c - d)*Cos[e + f*x])/(2*f*(a + a*Sin[e + f*x])^(3/2)) - (2*B*d*Cos[e + f*x])/(a*f*Sqrt
[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.279038, antiderivative size = 133, 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, 2751, 2649, 206} \[ -\frac{(A c+3 A d+3 B c-7 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(A-B) (c-d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}-\frac{2 B d \cos (e+f x)}{a f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-((A*c + 3*B*c + 3*A*d - 7*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]
*a^(3/2)*f) - ((A - B)*(c - d)*Cos[e + f*x])/(2*f*(a + a*Sin[e + f*x])^(3/2)) - (2*B*d*Cos[e + f*x])/(a*f*Sqrt
[a + a*Sin[e + f*x]])

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 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, 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))^{3/2}} \, 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))^{3/2}} \, dx\\ &=-\frac{(A-B) (c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{\int \frac{-\frac{1}{2} a (3 B (c-d)+A (c+3 d))-2 a B d \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) (c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{2 B d \cos (e+f x)}{a f \sqrt{a+a \sin (e+f x)}}+\frac{(A c+3 B c+3 A d-7 B d) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 a}\\ &=-\frac{(A-B) (c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{2 B d \cos (e+f x)}{a f \sqrt{a+a \sin (e+f x)}}-\frac{(A c+3 B c+3 A d-7 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 )}{2 a f}\\ &=-\frac{(A c+3 B c+3 A d-7 B d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(A-B) (c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{2 B d \cos (e+f x)}{a f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.435797, size = 246, normalized size = 1.85 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 (A-B) (c-d) \sin \left (\frac{1}{2} (e+f x)\right )-(A-B) (c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (A c+3 A d+3 B c-7 B d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )-4 B d \cos \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+4 B d \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2\right )}{2 f (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*(c - d)*Sin[(e + f*x)/2] - (A - B)*(c - d)*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2]) + (1 + I)*(-1)^(3/4)*(A*c + 3*B*c + 3*A*d - 7*B*d)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + T
an[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 4*B*d*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^2 + 4*B*d*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2))/(2*f*(a*(1 + Sin[e + f*x]))^(3
/2))

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Maple [B]  time = 0.986, size = 389, normalized size = 2.9 \begin{align*} -{\frac{1}{4\,f\cos \left ( fx+e \right ) } \left ( \sin \left ( fx+e \right ) \left ( A\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) ac+3\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ad+3\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ac-7\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ad+8\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}d \right ) +A\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) ac+3\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ad+3\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ac-7\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ad+2\,A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}c-2\,A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}d-2\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}c+10\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}d \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{5}{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))^(3/2),x)

[Out]

-1/4/a^(5/2)*(sin(f*x+e)*(A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c+3*A*2^(1/2)*arctan
h(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*d+3*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/
2))*a*c-7*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*d+8*B*(a-a*sin(f*x+e))^(1/2)*a^(1/2)
*d)+A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c+3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))
^(1/2)*2^(1/2)/a^(1/2))*a*d+3*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c-7*B*2^(1/2)*ar
ctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*d+2*A*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c-2*A*(a-a*sin(f*x+e)
)^(1/2)*a^(1/2)*d-2*B*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c+10*B*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d)*(-a*(-1+sin(f*x+
e)))^(1/2)/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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(3/2), x)

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Fricas [B]  time = 1.80485, size = 1023, normalized size = 7.69 \begin{align*} -\frac{\sqrt{2}{\left ({\left ({\left (A + 3 \, B\right )} c +{\left (3 \, A - 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + 3 \, B\right )} c - 2 \,{\left (3 \, A - 7 \, B\right )} d -{\left ({\left (A + 3 \, B\right )} c +{\left (3 \, A - 7 \, B\right )} d\right )} \cos \left (f x + e\right ) -{\left (2 \,{\left (A + 3 \, B\right )} c + 2 \,{\left (3 \, A - 7 \, B\right )} d +{\left ({\left (A + 3 \, B\right )} c +{\left (3 \, A - 7 \, B\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\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 (4 \, B d \cos \left (f x + e\right )^{2} +{\left (A - B\right )} c -{\left (A - B\right )} d +{\left ({\left (A - B\right )} c -{\left (A - 5 \, B\right )} d\right )} \cos \left (f x + e\right ) +{\left (4 \, B d \cos \left (f x + e\right ) -{\left (A - B\right )} c +{\left (A - B\right )} d\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{8 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \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))^(3/2),x, algorithm="fricas")

[Out]

-1/8*(sqrt(2)*(((A + 3*B)*c + (3*A - 7*B)*d)*cos(f*x + e)^2 - 2*(A + 3*B)*c - 2*(3*A - 7*B)*d - ((A + 3*B)*c +
(3*A - 7*B)*d)*cos(f*x + e) - (2*(A + 3*B)*c + 2*(3*A - 7*B)*d + ((A + 3*B)*c + (3*A - 7*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 - (cos(f*x + e)
+ 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(4*B*d*cos(f*x + e)^2 + (A - B)*c - (A - B)*d + ((A - B)*c - (A - 5
*B)*d)*cos(f*x + e) + (4*B*d*cos(f*x + e) - (A - B)*c + (A - B)*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^
2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sin{\left (e + f x \right )}\right ) \left (c + d \sin{\left (e + f x \right )}\right )}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \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))**(3/2),x)

[Out]

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [B]  time = 2.51031, size = 1077, normalized size = 8.1 \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))^(3/2),x, algorithm="giac")

[Out]

1/2*(4*(B*d*tan(1/2*f*x + 1/2*e)/(a*sgn(tan(1/2*f*x + 1/2*e) + 1)) - B*d/(a*sgn(tan(1/2*f*x + 1/2*e) + 1)))/sq
rt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sqrt(2)*(A*c + 3*B*c + 3*A*d - 7*B*d)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*
f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*f*x + 1/2*e) +
1)) + 2*(3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*A*c - 3*(sqrt(a)*tan(1/2*f*x
+ 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*B*c - 3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x +
1/2*e)^2 + a))^3*A*d + 3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*B*d + (sqrt(a)*
tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*A*sqrt(a)*c - (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqr
t(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*B*sqrt(a)*c - (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2
+ a))^2*A*sqrt(a)*d + (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*B*sqrt(a)*d - (sq
rt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*A*a*c + (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(
a*tan(1/2*f*x + 1/2*e)^2 + a))*B*a*c + (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*A*a
*d - (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*B*a*d + A*a^(3/2)*c - B*a^(3/2)*c - A
*a^(3/2)*d + B*a^(3/2)*d)/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(a)
*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)^2*a*sgn(tan(1/2*f*x + 1/2*e) + 1)))/f