3.736 \(\int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=460 \[ \frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-\left (b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right )\right ) \cos (e+f x)}{15 d f \left (c^2-d^2\right )^3 \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-\left (b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right )\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{5/2}}-\frac {4 \left (4 a c d+b \left (c^2-5 d^2\right )\right ) (b c-a d) \cos (e+f x)}{15 d f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))^{3/2}}+\frac {4 \left (4 a c d+b \left (c^2-5 d^2\right )\right ) (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}} \]

[Out]

2/5*(-a*d+b*c)^2*cos(f*x+e)/d/(c^2-d^2)/f/(c+d*sin(f*x+e))^(5/2)-4/15*(-a*d+b*c)*(4*a*c*d+b*(c^2-5*d^2))*cos(f
*x+e)/d/(c^2-d^2)^2/f/(c+d*sin(f*x+e))^(3/2)+2/15*(a^2*d^2*(23*c^2+9*d^2)-a*b*(6*c^3*d+58*c*d^3)-b^2*(2*c^4-19
*c^2*d^2-15*d^4))*cos(f*x+e)/d/(c^2-d^2)^3/f/(c+d*sin(f*x+e))^(1/2)-2/15*(a^2*d^2*(23*c^2+9*d^2)-a*b*(6*c^3*d+
58*c*d^3)-b^2*(2*c^4-19*c^2*d^2-15*d^4))*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*Ellipti
cE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/d^2/(c^2-d^2)^3/f/((c+d*sin(f*x+e
))/(c+d))^(1/2)-4/15*(-a*d+b*c)*(4*a*c*d+b*(c^2-5*d^2))*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1
/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/d^2/(c^2-d
^2)^2/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.86, antiderivative size = 460, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2790, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )+b^2 \left (-\left (-19 c^2 d^2+2 c^4-15 d^4\right )\right )\right ) \cos (e+f x)}{15 d f \left (c^2-d^2\right )^3 \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )+b^2 \left (-\left (-19 c^2 d^2+2 c^4-15 d^4\right )\right )\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{5/2}}-\frac {4 \left (4 a c d+b \left (c^2-5 d^2\right )\right ) (b c-a d) \cos (e+f x)}{15 d f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))^{3/2}}+\frac {4 \left (4 a c d+b \left (c^2-5 d^2\right )\right ) (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*(b*c - a*d)^2*Cos[e + f*x])/(5*d*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(5/2)) - (4*(b*c - a*d)*(4*a*c*d + b*(c
^2 - 5*d^2))*Cos[e + f*x])/(15*d*(c^2 - d^2)^2*f*(c + d*Sin[e + f*x])^(3/2)) + (2*(a^2*d^2*(23*c^2 + 9*d^2) -
a*b*(6*c^3*d + 58*c*d^3) - b^2*(2*c^4 - 19*c^2*d^2 - 15*d^4))*Cos[e + f*x])/(15*d*(c^2 - d^2)^3*f*Sqrt[c + d*S
in[e + f*x]]) + (2*(a^2*d^2*(23*c^2 + 9*d^2) - a*b*(6*c^3*d + 58*c*d^3) - b^2*(2*c^4 - 19*c^2*d^2 - 15*d^4))*E
llipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(15*d^2*(c^2 - d^2)^3*f*Sqrt[(c + d*Sin[
e + f*x])/(c + d)]) + (4*(b*c - a*d)*(4*a*c*d + b*(c^2 - 5*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*
Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2754

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 + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] - Di
st[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(2*b*c*d - a*(c^2 + d^2)) + (a^2
*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1) + c^2*(m + 2)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx &=\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}+\frac {2 \int \frac {\frac {5}{2} d \left (\left (a^2+b^2\right ) c-2 a b d\right )+\frac {1}{2} \left (6 a b c d-3 a^2 d^2+b^2 \left (2 c^2-5 d^2\right )\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d \left (c^2-d^2\right )}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {4 (b c-a d) \left (4 a c d+b \left (c^2-5 d^2\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 \int \frac {\frac {3}{4} d \left (16 a b c d-a^2 \left (5 c^2+3 d^2\right )-b^2 \left (3 c^2+5 d^2\right )\right )-\frac {1}{2} (b c-a d) \left (b c^2+4 a c d-5 b d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 d \left (c^2-d^2\right )^2}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {4 (b c-a d) \left (4 a c d+b \left (c^2-5 d^2\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}+\frac {8 \int \frac {-\frac {1}{8} d \left (2 a b d \left (27 c^2+5 d^2\right )-b^2 c \left (7 c^2+25 d^2\right )-a^2 \left (15 c^3+17 c d^2\right )\right )+\frac {1}{8} \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d \left (c^2-d^2\right )^3}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {4 (b c-a d) \left (4 a c d+b \left (c^2-5 d^2\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 d^2 \left (c^2-d^2\right )^3}--\frac {\left (8 \left (-\frac {1}{8} d^2 \left (2 a b d \left (27 c^2+5 d^2\right )-b^2 c \left (7 c^2+25 d^2\right )-a^2 \left (15 c^3+17 c d^2\right )\right )-\frac {1}{8} c \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right )\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^2 \left (c^2-d^2\right )^3}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {4 (b c-a d) \left (4 a c d+b \left (c^2-5 d^2\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (\left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{15 d^2 \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}--\frac {\left (8 \left (-\frac {1}{8} d^2 \left (2 a b d \left (27 c^2+5 d^2\right )-b^2 c \left (7 c^2+25 d^2\right )-a^2 \left (15 c^3+17 c d^2\right )\right )-\frac {1}{8} c \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{15 d^2 \left (c^2-d^2\right )^3 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{5 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {4 (b c-a d) \left (4 a c d+b \left (c^2-5 d^2\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \cos (e+f x)}{15 d \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (a^2 d^2 \left (23 c^2+9 d^2\right )-a b \left (6 c^3 d+58 c d^3\right )-b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^2 \left (c^2-d^2\right )^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 (b c-a d) \left (b c^2+4 a c d-5 b d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^2 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 4.99, size = 424, normalized size = 0.92 \[ \frac {2 \left (\frac {d \cos (e+f x) \left (-2 \left (c^2-d^2\right ) \left (-4 a^2 c d^2+a b d \left (3 c^2+5 d^2\right )+b^2 \left (c^3-5 c d^2\right )\right ) (c+d \sin (e+f x))-\left (-a^2 d^2 \left (23 c^2+9 d^2\right )+a b \left (6 c^3 d+58 c d^3\right )+b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) (c+d \sin (e+f x))^2+3 \left (c^2-d^2\right )^2 (b c-a d)^2\right )}{\left (c^2-d^2\right )^3}-\frac {\left (\frac {c+d \sin (e+f x)}{c+d}\right )^{5/2} \left (d^2 \left (a^2 \left (15 c^3+17 c d^2\right )-2 a b d \left (27 c^2+5 d^2\right )+b^2 c \left (7 c^2+25 d^2\right )\right ) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-\left (-a^2 d^2 \left (23 c^2+9 d^2\right )+a b \left (6 c^3 d+58 c d^3\right )+b^2 \left (2 c^4-19 c^2 d^2-15 d^4\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )\right )}{(c-d)^3 (c+d)}\right )}{15 d^2 f (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*(-(((d^2*(-2*a*b*d*(27*c^2 + 5*d^2) + b^2*c*(7*c^2 + 25*d^2) + a^2*(15*c^3 + 17*c*d^2))*EllipticF[(-2*e + P
i - 2*f*x)/4, (2*d)/(c + d)] - (-(a^2*d^2*(23*c^2 + 9*d^2)) + a*b*(6*c^3*d + 58*c*d^3) + b^2*(2*c^4 - 19*c^2*d
^2 - 15*d^4))*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2
*d)/(c + d)]))*((c + d*Sin[e + f*x])/(c + d))^(5/2))/((c - d)^3*(c + d))) + (d*Cos[e + f*x]*(3*(b*c - a*d)^2*(
c^2 - d^2)^2 - 2*(c^2 - d^2)*(-4*a^2*c*d^2 + a*b*d*(3*c^2 + 5*d^2) + b^2*(c^3 - 5*c*d^2))*(c + d*Sin[e + f*x])
- (-(a^2*d^2*(23*c^2 + 9*d^2)) + a*b*(6*c^3*d + 58*c*d^3) + b^2*(2*c^4 - 19*c^2*d^2 - 15*d^4))*(c + d*Sin[e +
f*x])^2))/(c^2 - d^2)^3))/(15*d^2*f*(c + d*Sin[e + f*x])^(5/2))

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{d^{4} \cos \left (f x + e\right )^{4} + c^{4} + 6 \, c^{2} d^{2} + d^{4} - 2 \, {\left (3 \, c^{2} d^{2} + d^{4}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (c d^{3} \cos \left (f x + e\right )^{2} - c^{3} d - c d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)*sqrt(d*sin(f*x + e) + c)/(d^4*cos(f*x + e)^4 +
c^4 + 6*c^2*d^2 + d^4 - 2*(3*c^2*d^2 + d^4)*cos(f*x + e)^2 - 4*(c*d^3*cos(f*x + e)^2 - c^3*d - c*d^3)*sin(f*x
+ e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(7/2), x)

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maple [B]  time = 8.37, size = 1450, normalized size = 3.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((a^2*d^2-2*a*b*c*d+b^2*c^2)/d^2*(2/5/(c^2-d^2)/d^2*(-(-d*sin(f*x+e)-c
)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^3+16/15*c/(c^2-d^2)^2/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*
x+e)+c/d)^2+2/15*d*cos(f*x+e)^2/(c^2-d^2)^3*(23*c^2+9*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(15*c^3+1
7*c*d^2)/(15*c^6-45*c^4*d^2+45*c^2*d^4-15*d^6)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))
^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-
d))^(1/2),((c-d)/(c+d))^(1/2))+2/15*d*(23*c^2+9*d^2)/(c^2-d^2)^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-
sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*Ell
ipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+
d))^(1/2))))+2*b*(a*d-b*c)/d^2*(2/3/(c^2-d^2)/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3
*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(c
/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(
f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+8/3*c*d/(c^2-d^2)^
2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*
sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+Elli
pticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+b^2/d^2*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+
e)-c)*cos(f*x+e)^2)^(1/2)+2*c/(c^2-d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*
((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/
2),((c-d)/(c+d))^(1/2))+2/(c^2-d^2)*d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((
-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c
-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))))/cos(f*x+e)/(c
+d*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^2/(c + d*sin(e + f*x))^(7/2),x)

[Out]

int((a + b*sin(e + f*x))^2/(c + d*sin(e + f*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**2/(c+d*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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