3.3.48 \(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\) [248]

Optimal. Leaf size=98 \[ -\frac {a (B c-(A+B) d) x}{d^2}+\frac {2 a (c-d) (B c-A d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^2 \sqrt {c^2-d^2} f}-\frac {a B \cos (e+f x)}{d f} \]

[Out]

-a*(B*c-(A+B)*d)*x/d^2-a*B*cos(f*x+e)/d/f+2*a*(c-d)*(-A*d+B*c)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2)
)/d^2/f/(c^2-d^2)^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3047, 3102, 2814, 2739, 632, 210} \begin {gather*} \frac {2 a (c-d) (B c-A d) \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d^2 f \sqrt {c^2-d^2}}-\frac {a x (B c-d (A+B))}{d^2}-\frac {a B \cos (e+f x)}{d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(B*c - (A + B)*d)*x)/d^2) + (2*a*(c - d)*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d
^2*Sqrt[c^2 - d^2]*f) - (a*B*Cos[e + f*x])/(d*f)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2814

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

Rule 3047

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 3102

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx &=\int \frac {a A+(a A+a B) \sin (e+f x)+a B \sin ^2(e+f x)}{c+d \sin (e+f x)} \, dx\\ &=-\frac {a B \cos (e+f x)}{d f}+\frac {\int \frac {a A d-a (B c-(A+B) d) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{d}\\ &=-\frac {a (B c-(A+B) d) x}{d^2}-\frac {a B \cos (e+f x)}{d f}+\frac {(a (c-d) (B c-A d)) \int \frac {1}{c+d \sin (e+f x)} \, dx}{d^2}\\ &=-\frac {a (B c-(A+B) d) x}{d^2}-\frac {a B \cos (e+f x)}{d f}+\frac {(2 a (c-d) (B c-A d)) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{d^2 f}\\ &=-\frac {a (B c-(A+B) d) x}{d^2}-\frac {a B \cos (e+f x)}{d f}-\frac {(4 a (c-d) (B c-A d)) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d^2 f}\\ &=-\frac {a (B c-(A+B) d) x}{d^2}+\frac {2 a (c-d) (B c-A d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^2 \sqrt {c^2-d^2} f}-\frac {a B \cos (e+f x)}{d f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.45, size = 196, normalized size = 2.00 \begin {gather*} \frac {a \left (A d x+B (-c+d) x-\frac {B d \cos (e) \cos (f x)}{f}+\frac {2 (c-d) (B c-A d) \tan ^{-1}\left (\frac {\sec \left (\frac {f x}{2}\right ) (\cos (e)-i \sin (e)) \left (d \cos \left (e+\frac {f x}{2}\right )+c \sin \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) (\cos (e)-i \sin (e))}{\sqrt {c^2-d^2} f \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {B d \sin (e) \sin (f x)}{f}\right ) (1+\sin (e+f x))}{d^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*d*x + B*(-c + d)*x - (B*d*Cos[e]*Cos[f*x])/f + (2*(c - d)*(B*c - A*d)*ArcTan[(Sec[(f*x)/2]*(Cos[e] - I*S
in[e])*(d*Cos[e + (f*x)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(Cos[e] - I*Sin[e
]))/(Sqrt[c^2 - d^2]*f*Sqrt[(Cos[e] - I*Sin[e])^2]) + (B*d*Sin[e]*Sin[f*x])/f)*(1 + Sin[e + f*x]))/(d^2*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])^2)

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Maple [A]
time = 0.30, size = 120, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {2 a \left (\frac {\left (-A c d +A \,d^{2}+B \,c^{2}-B c d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{2} \sqrt {c^{2}-d^{2}}}+\frac {-\frac {B d}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+\left (A d -B c +B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{2}}\right )}{f}\)	\(120\)
default	\(\frac {2 a \left (\frac {\left (-A c d +A \,d^{2}+B \,c^{2}-B c d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{2} \sqrt {c^{2}-d^{2}}}+\frac {-\frac {B d}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+\left (A d -B c +B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{2}}\right )}{f}\)	\(120\)
risch	\(\frac {a x A}{d}-\frac {a x B c}{d^{2}}+\frac {a x B}{d}-\frac {B a \,{\mathrm e}^{i \left (f x +e \right )}}{2 d f}-\frac {B a \,{\mathrm e}^{-i \left (f x +e \right )}}{2 d f}+\frac {\sqrt {-\left (c -d \right ) \left (c +d \right )}\, a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {-i c +\sqrt {-\left (c -d \right ) \left (c +d \right )}}{d}\right ) A}{\left (c +d \right ) f d}-\frac {\sqrt {-\left (c -d \right ) \left (c +d \right )}\, a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {-i c +\sqrt {-\left (c -d \right ) \left (c +d \right )}}{d}\right ) B c}{\left (c +d \right ) f \,d^{2}}-\frac {\sqrt {-\left (c -d \right ) \left (c +d \right )}\, a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c -d \right ) \left (c +d \right )}}{d}\right ) A}{\left (c +d \right ) f d}+\frac {\sqrt {-\left (c -d \right ) \left (c +d \right )}\, a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c -d \right ) \left (c +d \right )}}{d}\right ) B c}{\left (c +d \right ) f \,d^{2}}\)	\(301\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2/f*a*((-A*c*d+A*d^2+B*c^2-B*c*d)/d^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))
+1/d^2*(-B*d/(1+tan(1/2*f*x+1/2*e)^2)+(A*d-B*c+B*d)*arctan(tan(1/2*f*x+1/2*e))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.41, size = 303, normalized size = 3.09 \begin {gather*} \left [-\frac {2 \, B a d \cos \left (f x + e\right ) + 2 \, {\left (B a c - {\left (A + B\right )} a d\right )} f x - {\left (B a c - A a d\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (-\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} - 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right )}{2 \, d^{2} f}, -\frac {B a d \cos \left (f x + e\right ) + {\left (B a c - {\left (A + B\right )} a d\right )} f x + {\left (B a c - A a d\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right )}{d^{2} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*B*a*d*cos(f*x + e) + 2*(B*a*c - (A + B)*a*d)*f*x - (B*a*c - A*a*d)*sqrt(-(c - d)/(c + d))*log(-((2*c^
2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 - 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d
^2)*cos(f*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)))/(d^2*f), -(B
*a*d*cos(f*x + e) + (B*a*c - (A + B)*a*d)*f*x + (B*a*c - A*a*d)*sqrt((c - d)/(c + d))*arctan(-(c*sin(f*x + e)
+ d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))))/(d^2*f)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 5508 vs. \(2 (82) = 164\).
time = 172.70, size = 5508, normalized size = 56.20 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(A + B*sin(e))*(a*sin(e) + a)/sin(e), Eq(c, 0) & Eq(d, 0) & Eq(f, 0)), (A*a*d**2*f*x*tan(e/2
+ f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d
**2)**(3/2)) + A*a*d**2*f*x*tan(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)*
*(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 2*A*a*d**2*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d
**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 2*A*a*d**2/(d**3*f*tan(e/2 +
f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) - A*a*d*f*x*sqrt
(d**2)*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f
*x/2)**2 - f*(d**2)**(3/2)) - A*a*d*f*x*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(
d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 2*A*a*d*sqrt(d**2)*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 +
f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 2*A*a*d*sqrt(d
**2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(
3/2)) + B*a*d**2*f*x*tan(e/2 + f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/
2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) - B*a*d**2*f*x*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**
3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + B*a*d**2*f*x*tan(e/2 + f*x/2)/
(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2))
- B*a*d**2*f*x/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 -
f*(d**2)**(3/2)) + 2*B*a*d**2*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d
**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) - 2*B*a*d**2*tan(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2)**3 +
d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 2*B*a*d**2/(d**3*f*tan(e/2
+ f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + B*a*d*f*x*sq
rt(d**2)*tan(e/2 + f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 +
f*x/2)**2 - f*(d**2)**(3/2)) - B*a*d*f*x*sqrt(d**2)*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*
tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + B*a*d*f*x*sqrt(d**2)*tan(e/2 + f*x
/2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3
/2)) - B*a*d*f*x*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 +
f*x/2)**2 - f*(d**2)**(3/2)) + 2*B*a*d*sqrt(d**2)*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan
(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 4*B*a*d*sqrt(d**2)/(d**3*f*tan(e/2 +
f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)), Eq(c, -sqrt(d**2
))), (A*a*d**2*f*x*tan(e/2 + f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)
*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) + A*a*d**2*f*x*tan(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*t
an(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) + 2*A*a*d**2*tan(e/2 + f*x/2)**2/(d**
3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) + 2
*A*a*d**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**
2)**(3/2)) + A*a*d*f*x*sqrt(d**2)*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) +
f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) + A*a*d*f*x*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2)**3 + d*
*3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) - 2*A*a*d*sqrt(d**2)*tan(e/2 +
f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**
2)**(3/2)) - 2*A*a*d*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/
2 + f*x/2)**2 + f*(d**2)**(3/2)) + B*a*d**2*f*x*tan(e/2 + f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e
/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) - B*a*d**2*f*x*tan(e/2 + f*x/2)**2/(d**3*
f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) + B*a
*d**2*f*x*tan(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f
*x/2)**2 + f*(d**2)**(3/2)) - B*a*d**2*f*x/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(
3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) + 2*B*a*d**2*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**
3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 + f*(d**2)**(3/2)) - 2*B*a*d**2*tan(e/2 + f*x/2)/(d
**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*...

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Giac [A]
time = 0.56, size = 141, normalized size = 1.44 \begin {gather*} -\frac {\frac {{\left (B a c - A a d - B a d\right )} {\left (f x + e\right )}}{d^{2}} + \frac {2 \, B a}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} d} - \frac {2 \, {\left (B a c^{2} - A a c d - B a c d + A a d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{2}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((B*a*c - A*a*d - B*a*d)*(f*x + e)/d^2 + 2*B*a/((tan(1/2*f*x + 1/2*e)^2 + 1)*d) - 2*(B*a*c^2 - A*a*c*d - B*a*
c*d + A*a*d^2)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2))
)/(sqrt(c^2 - d^2)*d^2))/f

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Mupad [B]
time = 16.58, size = 2500, normalized size = 25.51 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*a*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(f*(c + d)) + (2*B*a*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*
x)/2)))/(f*(c + d)) - (B*a*cos(e + f*x))/(f*(c + d)) + (2*A*a*c*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(
d*f*(c + d)) - (A*a*atan((A^2*d^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*3i + A^2*d^6*sin(e/2 + (f*x)/2)*(d^2 -
c^2)^(1/2)*1i - B^2*c^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*2i - B^2*c^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)
*2i + B^2*d^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i + A*B*d^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*4i + A^2*
c*d^3*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*1i + A^2*c*d^5*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*1i + B^2*c*d^5*
cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*1i + B^2*c^3*d*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*1i + B^2*c^5*d*cos(e/
2 + (f*x)/2)*(d^2 - c^2)^(1/2)*1i + A^2*c*d^5*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*4i + A^2*c^2*d^4*cos(e/2 +
(f*x)/2)*(d^2 - c^2)^(1/2)*2i + A^2*c^3*d^3*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*1i - B^2*c^3*d^3*cos(e/2 + (f
*x)/2)*(d^2 - c^2)^(1/2)*2i - A^2*c^2*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*2i + A^2*c^2*d^4*sin(e/2 + (f*x
)/2)*(d^2 - c^2)^(1/2)*3i - A^2*c^3*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i - A^2*c^4*d^2*sin(e/2 + (f*x)/
2)*(d^2 - c^2)^(1/2)*2i + B^2*c^2*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*3i - B^2*c^2*d^4*sin(e/2 + (f*x)/2)
*(d^2 - c^2)^(1/2)*6i + B^2*c^4*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*6i + A*B*c*d^5*cos(e/2 + (f*x)/2)*(d^
2 - c^2)^(1/2)*2i - A*B*c*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*6i + A*B*c*d^5*sin(e/2 + (f*x)/2)*(d^2 - c^
2)^(1/2)*6i + A*B*c^3*d*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*4i + A*B*c^5*d*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/
2)*4i - A*B*c^2*d^2*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*2i + A*B*c^2*d^4*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)
*2i - A*B*c^3*d^3*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i - A*B*c^4*d^2*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2
i - A*B*c^2*d^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*6i - A*B*c^3*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*10i
+ A*B*c^4*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i)/(4*A^2*d^7*sin(e/2 + (f*x)/2) + 2*B^2*d^7*sin(e/2 + (f
*x)/2) + 2*A^2*c^2*d^5*cos(e/2 + (f*x)/2) - 2*A^2*c^3*d^4*cos(e/2 + (f*x)/2) - 2*A^2*c^4*d^3*cos(e/2 + (f*x)/2
) - 2*B^2*c^3*d^4*cos(e/2 + (f*x)/2) + B^2*c^5*d^2*cos(e/2 + (f*x)/2) - 4*A^2*c^2*d^5*sin(e/2 + (f*x)/2) - 4*A
^2*c^3*d^4*sin(e/2 + (f*x)/2) - 4*B^2*c^2*d^5*sin(e/2 + (f*x)/2) + 2*B^2*c^4*d^3*sin(e/2 + (f*x)/2) + 4*A*B*d^
7*sin(e/2 + (f*x)/2) + 2*A^2*c*d^6*cos(e/2 + (f*x)/2) + B^2*c*d^6*cos(e/2 + (f*x)/2) + 4*A^2*c*d^6*sin(e/2 + (
f*x)/2) - 4*A*B*c^3*d^4*cos(e/2 + (f*x)/2) + 2*A*B*c^5*d^2*cos(e/2 + (f*x)/2) - 8*A*B*c^2*d^5*sin(e/2 + (f*x)/
2) + 4*A*B*c^4*d^3*sin(e/2 + (f*x)/2) + 2*A*B*c*d^6*cos(e/2 + (f*x)/2)))*(d^2 - c^2)^(1/2)*2i)/(d*f*(c + d)) -
(2*B*a*c^2*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(d^2*f*(c + d)) - (B*a*c*cos(e + f*x))/(d*f*(c + d))
+ (B*a*c*atan((A^2*d^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*3i + A^2*d^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*
1i - B^2*c^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*2i - B^2*c^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i + B^2*d
^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i + A*B*d^6*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*4i + A^2*c*d^3*cos(e
/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*1i + A^2*c*d^5*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*1i + B^2*c*d^5*cos(e/2 + (
f*x)/2)*(d^2 - c^2)^(1/2)*1i + B^2*c^3*d*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*1i + B^2*c^5*d*cos(e/2 + (f*x)/2
)*(d^2 - c^2)^(1/2)*1i + A^2*c*d^5*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*4i + A^2*c^2*d^4*cos(e/2 + (f*x)/2)*(d
^2 - c^2)^(1/2)*2i + A^2*c^3*d^3*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*1i - B^2*c^3*d^3*cos(e/2 + (f*x)/2)*(d^2
- c^2)^(1/2)*2i - A^2*c^2*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*2i + A^2*c^2*d^4*sin(e/2 + (f*x)/2)*(d^2 -
c^2)^(1/2)*3i - A^2*c^3*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i - A^2*c^4*d^2*sin(e/2 + (f*x)/2)*(d^2 - c
^2)^(1/2)*2i + B^2*c^2*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*3i - B^2*c^2*d^4*sin(e/2 + (f*x)/2)*(d^2 - c^2
)^(1/2)*6i + B^2*c^4*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*6i + A*B*c*d^5*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1
/2)*2i - A*B*c*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*6i + A*B*c*d^5*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*6i
+ A*B*c^3*d*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*4i + A*B*c^5*d*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*4i - A*B
*c^2*d^2*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2)*2i + A*B*c^2*d^4*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i - A*B*c
^3*d^3*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i - A*B*c^4*d^2*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i - A*B*c^2
*d^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*6i - A*B*c^3*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*10i + A*B*c^4*
d^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2)*2i)/(4*A^2*d^7*sin(e/2 + (f*x)/2) + 2*B^2*d^7*sin(e/2 + (f*x)/2) + 2*
A^2*c^2*d^5*cos(e/2 + (f*x)/2) - 2*A^2*c^3*d^4*cos(e/2 + (f*x)/2) - 2*A^2*c^4*d^3*cos(e/2 + (f*x)/2) - 2*B^2*c
^3*d^4*cos(e/2 + (f*x)/2) + B^2*c^5*d^2*cos(e/2 + (f*x)/2) - 4*A^2*c^2*d^5*sin(e/2 + (f*x)/2) - 4*A^2*c^3*d^4*
sin(e/2 + (f*x)/2) - 4*B^2*c^2*d^5*sin(e/2 + (f...

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