\(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\) [96]

Optimal result

Integrand size = 38, antiderivative size = 175 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a^2 (A-11 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} c^{7/2} f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac {a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/6*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(11/2)+1/24*a^2*(A-11*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(7/
2)-1/16*a^2*(A-11*B)*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(3/2)+1/32*a^2*(A-11*B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*
2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(7/2)/f*2^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3046, 2938, 2759, 2728, 212} \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a^2 (A-11 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} c^{7/2} f}+\frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(a^2*(A - 11*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(16*Sqrt[2]*c^(7/2)*f) + (
a^2*(A + B)*c^2*Cos[e + f*x]^5)/(6*f*(c - c*Sin[e + f*x])^(11/2)) + (a^2*(A - 11*B)*Cos[e + f*x]^3)/(24*f*(c -
c*Sin[e + f*x])^(7/2)) - (a^2*(A - 11*B)*Cos[e + f*x])/(16*c^2*f*(c - c*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 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2938

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

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac {1}{12} \left (a^2 (A-11 B) c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac {a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac {\left (a^2 (A-11 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{16 c} \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac {a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (a^2 (A-11 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{32 c^3} \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac {a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^2 (A-11 B)\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{16 c^3 f} \\ & = \frac {a^2 (A-11 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} c^{7/2} f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{6 f (c-c \sin (e+f x))^{11/2}}+\frac {a^2 (A-11 B) \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 (A-11 B) \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.88 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.95 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (32 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-4 (7 A+19 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+3 (A+21 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-(3+3 i) \sqrt [4]{-1} (A-11 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6+64 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-8 (7 A+19 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+6 (A+21 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{48 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^{7/2}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(32*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 4*(7*A + 19*B)*
(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 3*(A + 21*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - (3 + 3*I)*(-1
)^(1/4)*(A - 11*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])
^6 + 64*(A + B)*Sin[(e + f*x)/2] - 8*(7*A + 19*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 6
*(A + 21*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(48*f*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(353\) vs. \(2(152)=304\).

Time = 3.26 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.02

method	result	size
default	\(\frac {a^{2} \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} \left (A -11 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} \left (A -11 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )-12 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} \left (A -11 B \right ) \sin \left (f x +e \right )+6 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c}+32 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-24 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {5}{2}}+126 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c}-352 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}+264 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {5}{2}}+12 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-132 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{96 c^{\frac {13}{2}} \left (\sin \left (f x +e \right )-1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(354\)
parts	\(\text {Expression too large to display}\)	\(1004\)

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/96/c^(13/2)*a^2*(3*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^3*(A-11*B)*cos(f*x+e)^2*sin
(f*x+e)-9*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^3*(A-11*B)*cos(f*x+e)^2-12*2^(1/2)*arc
tanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^3*(A-11*B)*sin(f*x+e)+6*A*(c+c*sin(f*x+e))^(5/2)*c^(1/2)+32
*A*(c+c*sin(f*x+e))^(3/2)*c^(3/2)-24*A*(c+c*sin(f*x+e))^(1/2)*c^(5/2)+126*B*(c+c*sin(f*x+e))^(5/2)*c^(1/2)-352
*B*(c+c*sin(f*x+e))^(3/2)*c^(3/2)+264*B*(c+c*sin(f*x+e))^(1/2)*c^(5/2)+12*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e
))^(1/2)*2^(1/2)/c^(1/2))*c^3-132*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^3)*(c*(1+sin
(f*x+e)))^(1/2)/(sin(f*x+e)-1)^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (152) = 304\).

Time = 0.27 (sec) , antiderivative size = 521, normalized size of antiderivative = 2.98 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {3 \, \sqrt {2} {\left ({\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} - 3 \, {\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - 8 \, {\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 4 \, {\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right ) + 8 \, {\left (A - 11 \, B\right )} a^{2} + {\left ({\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + 4 \, {\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 4 \, {\left (A - 11 \, B\right )} a^{2} \cos \left (f x + e\right ) - 8 \, {\left (A - 11 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\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 (3 \, {\left (A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + {\left (25 \, A + 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (5 \, A + 41 \, B\right )} a^{2} \cos \left (f x + e\right ) - 32 \, {\left (A + B\right )} a^{2} + {\left (3 \, {\left (A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (11 \, A - 25 \, B\right )} a^{2} \cos \left (f x + e\right ) - 32 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{192 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/192*(3*sqrt(2)*((A - 11*B)*a^2*cos(f*x + e)^4 - 3*(A - 11*B)*a^2*cos(f*x + e)^3 - 8*(A - 11*B)*a^2*cos(f*x
+ e)^2 + 4*(A - 11*B)*a^2*cos(f*x + e) + 8*(A - 11*B)*a^2 + ((A - 11*B)*a^2*cos(f*x + e)^3 + 4*(A - 11*B)*a^2*
cos(f*x + e)^2 - 4*(A - 11*B)*a^2*cos(f*x + e) - 8*(A - 11*B)*a^2)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^
2 - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(
f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4
*(3*(A + 21*B)*a^2*cos(f*x + e)^3 + (25*A + 13*B)*a^2*cos(f*x + e)^2 - 2*(5*A + 41*B)*a^2*cos(f*x + e) - 32*(A
+ B)*a^2 + (3*(A + 21*B)*a^2*cos(f*x + e)^2 - 2*(11*A - 25*B)*a^2*cos(f*x + e) - 32*(A + B)*a^2)*sin(f*x + e)
)*sqrt(-c*sin(f*x + e) + c))/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f
*cos(f*x + e) + 8*c^4*f + (c^4*f*cos(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e) - 8*c^4*f)*sin
(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(7/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (152) = 304\).

Time = 0.74 (sec) , antiderivative size = 658, normalized size of antiderivative = 3.76 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

1/768*(12*sqrt(2)*(A*a^2*sqrt(c) - 11*B*a^2*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi +
1/2*f*x + 1/2*e) + 1))/(c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*(A*a^2*sqrt(c) + B*a^2*sqrt(c) + 3*
A*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 15*B*a^2*sqrt(c)*(co
s(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 3*A*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*
x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 93*B*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) -
1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 22*A*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/
4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 242*B*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f
*x + 1/2*e) + 1)^3)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3/(c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin
(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*(3*A*a^2*c^(17/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi +
1/2*f*x + 1/2*e) + 1) - 93*B*a^2*c^(17/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e)
+ 1) - 3*A*a^2*c^(17/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 15*B*
a^2*c^(17/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - A*a^2*c^(17/2)*(c
os(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 - B*a^2*c^(17/2)*(cos(-1/4*pi + 1/
2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3)/(c^12*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(7/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(7/2), x)