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

Optimal. Leaf size=406 \[ \frac {\sqrt {2} (c-d) \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+7)+2 B m+7 B+2 C m-5 C)) \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {3}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+7) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}-\frac {\sqrt {2} (c-d) \cos (e+f x) (2 c C (m+1)-d (B (2 m+7)+2 C m)) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {3}{2};\frac {1}{2},-\frac {3}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+7) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{5/2}}{d f (2 m+7)} \]

[Out]

-2*C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(5/2)/d/f/(7+2*m)+(c-d)*(2*c*(2*C*m+C)-d*(7*B-5*C+2*B*m+2*
C*m-A*(7+2*m)))*AppellF1(1/2+m,-3/2,1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin(
f*x+e))^m*2^(1/2)*(c+d*sin(f*x+e))^(1/2)/d/f/(1+2*m)/(7+2*m)/(1-sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/
2)-(c-d)*(2*c*C*(1+m)-d*(2*C*m+B*(7+2*m)))*AppellF1(3/2+m,-3/2,1/2,5/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f
*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*2^(1/2)*(c+d*sin(f*x+e))^(1/2)/a/d/f/(3+2*m)/(7+2*m)/(1-sin(f*x+e))^(
1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)

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Rubi [A]  time = 1.03, antiderivative size = 403, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 6, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {3045, 2987, 2788, 140, 139, 138} \[ \frac {\sqrt {2} (c-d) \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+7)+2 B m+7 B+2 C m-5 C)) \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {3}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+7) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} (c-d) \cos (e+f x) (B d (2 m+7)-2 c C (m+1)+2 C d m) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {3}{2};\frac {1}{2},-\frac {3}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+7) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{5/2}}{d f (2 m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(5/2))/(d*f*(7 + 2*m)) + (Sqrt[2]*(c - d)*(2*c*
(C + 2*C*m) - d*(7*B - 5*C + 2*B*m + 2*C*m - A*(7 + 2*m)))*AppellF1[1/2 + m, 1/2, -3/2, 3/2 + m, (1 + Sin[e +
f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]])/(d*f
*(1 + 2*m)*(7 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)]) + (Sqrt[2]*(c - d)*(2*C*d*m -
2*c*C*(1 + m) + B*d*(7 + 2*m))*AppellF1[3/2 + m, 1/2, -3/2, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f
*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c + d*Sin[e + f*x]])/(a*d*f*(3 + 2*m)*(7 + 2*m)
*Sqrt[1 - Sin[e + f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)])

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dis
t[(a^2*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[a - b*x], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] &
& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !IntegerQ[m]

Rule 2987

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

Rule 3045

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac {2 \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \left (\frac {1}{2} a \left (2 A d \left (\frac {7}{2}+m\right )+2 C \left (\frac {5 d}{2}+c m\right )\right )+\frac {1}{2} a (2 C d m-2 c C (1+m)+B d (7+2 m)) \sin (e+f x)\right ) \, dx}{a d (7+2 m)}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac {(2 C d m-2 c C (1+m)+B d (7+2 m)) \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{3/2} \, dx}{a d (7+2 m)}+\frac {(2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx}{d (7+2 m)}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac {(a (2 C d m-2 c C (1+m)+B d (7+2 m)) \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} (c+d x)^{3/2}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (7+2 m) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^2 (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} (c+d x)^{3/2}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (7+2 m) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac {\left (a (2 C d m-2 c C (1+m)+B d (7+2 m)) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} (c+d x)^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^2 (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} (c+d x)^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac {\left ((a c-a d) (2 C d m-2 c C (1+m)+B d (7+2 m)) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {c+d \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)} \sqrt {\frac {a (c+d \sin (e+f x))}{a c-a d}}}+\frac {\left (a (a c-a d) (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {c+d \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)} \sqrt {\frac {a (c+d \sin (e+f x))}{a c-a d}}}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac {\sqrt {2} (c-d) (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) F_1\left (\frac {1}{2}+m;\frac {1}{2},-\frac {3}{2};\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)}}{d f (1+2 m) (7+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} (c-d) (2 C d m-2 c C (1+m)+B d (7+2 m)) F_1\left (\frac {3}{2}+m;\frac {1}{2},-\frac {3}{2};\frac {5}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} \sqrt {c+d \sin (e+f x)}}{d f (3+2 m) (7+2 m) (a-a \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\\ \end {align*}

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Mathematica [B]  time = 9.63, size = 6591, normalized size = 16.23 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left ({\left (C c + B d\right )} \cos \left (f x + e\right )^{2} - {\left (A + C\right )} c - B d + {\left (C d \cos \left (f x + e\right )^{2} - B c - {\left (A + C\right )} d\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral(-((C*c + B*d)*cos(f*x + e)^2 - (A + C)*c - B*d + (C*d*cos(f*x + e)^2 - B*c - (A + C)*d)*sin(f*x + e))
*sqrt(d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, x)

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giac [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+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 4.47, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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