3.4.23 \(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^{5/2}} \, dx\) [323]

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

[Out]

-1/4*(A-B)*(c-d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3*A*c+5*A*d+5*B*c-13*B*d)*cos(f*x+e)/a/f/(a+a*sin(f
*x+e))^(3/2)-1/32*(3*A*c+5*A*d+5*B*c+19*B*d)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^
(5/2)/f*2^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 = {3047, 3098, 2829, 2728, 212} \begin {gather*} -\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}} \end {gather*}

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]

-1/16*((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]])])/(S
qrt[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 212

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

Rule 2728

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 2829

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 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 3098

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]

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) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.51, size = 267, normalized size = 1.77 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (A-B) (c-d) \sin \left (\frac {1}{2} (e+f x)\right )-4 (A-B) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (3 A c+5 B c+5 A d-13 B d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(3 A c+5 B c+5 A d-13 B d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1+i) (-1)^{3/4} (3 A c+5 B c+5 A d+19 B d) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{16 f (a (1+\sin (e+f x)))^{5/2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(447\) vs. \(2(132)=264\).
time = 9.06, size = 448, normalized size = 2.97

method	result	size
default	\(\frac {\left (-2 \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{2} \left (3 A c +5 A d +5 B c +19 B d \right )+\arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{2} \left (3 A c +5 A d +5 B c +19 B d \right ) \left (\cos ^{2}\left (f x +e \right )\right )-6 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c -10 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d +6 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c +10 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, d -20 A \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c -12 A \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} d -10 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c -38 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d +10 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c -26 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, d -12 B \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c +44 B \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} d \right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(448\)

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,method=_RETURNVERBOSE)

[Out]

1/32/a^(9/2)*(-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^2*(3*A*c+5*A*d+5*B*c
+19*B*d)+arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(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+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-2
0*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-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+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-12*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c
+44*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d)*(-a*(sin(f*x+e)-1))^(1/2)/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (138) = 276\).
time = 0.40, size = 564, normalized size = 3.74 \begin {gather*} \frac {\sqrt {2} {\left ({\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A + 19 \, B\right )} d\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A + 19 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (3 \, A + 5 \, B\right )} c - 4 \, {\left (5 \, A + 19 \, B\right )} d - 2 \, {\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A + 19 \, B\right )} d\right )} \cos \left (f x + e\right ) + {\left ({\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A + 19 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (3 \, A + 5 \, B\right )} c - 4 \, {\left (5 \, A + 19 \, B\right )} d - 2 \, {\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A + 19 \, 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 ({\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A - 13 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (A - B\right )} c - 4 \, {\left (A - B\right )} d + {\left ({\left (7 \, A + B\right )} c + {\left (A - 9 \, B\right )} d\right )} \cos \left (f x + e\right ) - {\left (4 \, {\left (A - B\right )} c - 4 \, {\left (A - B\right )} d - {\left ({\left (3 \, A + 5 \, B\right )} c + {\left (5 \, A - 13 \, B\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {5}{2}}}\, dx \end {gather*}

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]

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (138) = 276\).
time = 0.54, size = 378, normalized size = 2.50 \begin {gather*} \frac {\frac {\sqrt {2} {\left (3 \, A \sqrt {a} c + 5 \, B \sqrt {a} c + 5 \, A \sqrt {a} d + 19 \, B \sqrt {a} d\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (3 \, A \sqrt {a} c + 5 \, B \sqrt {a} c + 5 \, A \sqrt {a} d + 19 \, B \sqrt {a} d\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, {\left (3 \, \sqrt {2} A \sqrt {a} c \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, \sqrt {2} B \sqrt {a} c \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, \sqrt {2} A \sqrt {a} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 13 \, \sqrt {2} B \sqrt {a} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, \sqrt {2} A \sqrt {a} c \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, \sqrt {2} B \sqrt {a} c \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, \sqrt {2} A \sqrt {a} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 11 \, \sqrt {2} B \sqrt {a} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{64 \, f} \end {gather*}

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]

1/64*(sqrt(2)*(3*A*sqrt(a)*c + 5*B*sqrt(a)*c + 5*A*sqrt(a)*d + 19*B*sqrt(a)*d)*log(sin(-1/4*pi + 1/2*f*x + 1/2
*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(3*A*sqrt(a)*c + 5*B*sqrt(a)*c + 5*A*sqrt(a)*d +
19*B*sqrt(a)*d)*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*(3*sqrt
(2)*A*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 5*sqrt(2)*B*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 5*
sqrt(2)*A*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 13*sqrt(2)*B*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3
- 5*sqrt(2)*A*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 3*sqrt(2)*B*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e)
- 3*sqrt(2)*A*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 11*sqrt(2)*B*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e
))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (c+d\,\sin \left (e+f\,x\right )\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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