3.7 \(\int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=336 \[ -\frac {2 a^3 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};1-\sin (e+f x)\right )}{f (2 n+3) (2 n+5) (2 n+7) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) (2 n+5) (2 n+7) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5) (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)} \]

[Out]

-2*a*B*cos(f*x+e)*(d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(3/2)/d/f/(7+2*n)-2*a^3*(2*B*(16*n^3+104*n^2+203*n+115
)+A*(32*n^3+224*n^2+478*n+301))*cos(f*x+e)*hypergeom([1/2, -n],[3/2],1-sin(f*x+e))*(d*sin(f*x+e))^n/f/(3+2*n)/
(5+2*n)/(7+2*n)/(sin(f*x+e)^n)/(a+a*sin(f*x+e))^(1/2)-2*a^3*(2*B*(4*n^2+23*n+35)+A*(8*n^2+50*n+77))*cos(f*x+e)
*(d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/(5+2*n)/(7+2*n)/(a+a*sin(f*x+e))^(1/2)-2*a^2*(2*B*(5+n)+A*(7+2*n))*cos(f*x+e
)*(d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(1/2)/d/f/(5+2*n)/(7+2*n)

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Rubi [A]  time = 0.87, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2976, 2981, 2776, 67, 65} \[ -\frac {2 a^3 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};1-\sin (e+f x)\right )}{f (2 n+3) (2 n+5) (2 n+7) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) (2 n+5) (2 n+7) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5) (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]),x]

[Out]

(-2*a^3*(2*B*(115 + 203*n + 104*n^2 + 16*n^3) + A*(301 + 478*n + 224*n^2 + 32*n^3))*Cos[e + f*x]*Hypergeometri
c2F1[1/2, -n, 3/2, 1 - Sin[e + f*x]]*(d*Sin[e + f*x])^n)/(f*(3 + 2*n)*(5 + 2*n)*(7 + 2*n)*Sin[e + f*x]^n*Sqrt[
a + a*Sin[e + f*x]]) - (2*a^3*(2*B*(35 + 23*n + 4*n^2) + A*(77 + 50*n + 8*n^2))*Cos[e + f*x]*(d*Sin[e + f*x])^
(1 + n))/(d*f*(3 + 2*n)*(5 + 2*n)*(7 + 2*n)*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*(2*B*(5 + n) + A*(7 + 2*n))*Cos
[e + f*x]*(d*Sin[e + f*x])^(1 + n)*Sqrt[a + a*Sin[e + f*x]])/(d*f*(5 + 2*n)*(7 + 2*n)) - (2*a*B*Cos[e + f*x]*(
d*Sin[e + f*x])^(1 + n)*(a + a*Sin[e + f*x])^(3/2))/(d*f*(7 + 2*n))

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 67

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

Rule 2776

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

Rule 2976

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx &=-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)}+\frac {2 \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} \left (\frac {1}{2} a d \left (2 B (1+n)+2 A \left (\frac {7}{2}+n\right )\right )+\frac {1}{2} a d (2 B (5+n)+A (7+2 n)) \sin (e+f x)\right ) \, dx}{d (7+2 n)}\\ &=-\frac {2 a^2 (2 B (5+n)+A (7+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n) (7+2 n)}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)}+\frac {4 \int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} \left (\frac {1}{4} a^2 d^2 \left (2 B \left (15+19 n+4 n^2\right )+A \left (49+42 n+8 n^2\right )\right )+\frac {1}{4} a^2 d^2 \left (2 B \left (35+23 n+4 n^2\right )+A \left (77+50 n+8 n^2\right )\right ) \sin (e+f x)\right ) \, dx}{d^2 (5+2 n) (7+2 n)}\\ &=-\frac {2 a^3 \left (2 B \left (35+23 n+4 n^2\right )+A \left (77+50 n+8 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (5+n)+A (7+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n) (7+2 n)}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)}+\frac {\left (a^2 \left (2 B \left (115+203 n+104 n^2+16 n^3\right )+A \left (301+478 n+224 n^2+32 n^3\right )\right )\right ) \int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} \, dx}{(3+2 n) (5+2 n) (7+2 n)}\\ &=-\frac {2 a^3 \left (2 B \left (35+23 n+4 n^2\right )+A \left (77+50 n+8 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (5+n)+A (7+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n) (7+2 n)}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)}+\frac {\left (a^4 \left (2 B \left (115+203 n+104 n^2+16 n^3\right )+A \left (301+478 n+224 n^2+32 n^3\right )\right ) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) (5+2 n) (7+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {2 a^3 \left (2 B \left (35+23 n+4 n^2\right )+A \left (77+50 n+8 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (5+n)+A (7+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n) (7+2 n)}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)}+\frac {\left (a^4 \left (2 B \left (115+203 n+104 n^2+16 n^3\right )+A \left (301+478 n+224 n^2+32 n^3\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n\right ) \operatorname {Subst}\left (\int \frac {x^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) (5+2 n) (7+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {2 a^3 \left (2 B \left (115+203 n+104 n^2+16 n^3\right )+A \left (301+478 n+224 n^2+32 n^3\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^3 \left (2 B \left (35+23 n+4 n^2\right )+A \left (77+50 n+8 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (5+n)+A (7+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n) (7+2 n)}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)}\\ \end {align*}

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Mathematica [A]  time = 18.38, size = 596, normalized size = 1.77 \[ \frac {2^{n+1} \tan \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right ) (a (\sin (e+f x)+1))^{5/2} \sin ^{-n}(e+f x) \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\tan ^2\left (\frac {1}{2} (e+f x)\right )+1}\right )^n \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right )^n (d \sin (e+f x))^n \left (\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(5 A+2 B) \, _2F_1\left (\frac {n+2}{2},n+\frac {9}{2};\frac {n+4}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+2}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(11 A+10 B) \, _2F_1\left (\frac {n+3}{2},n+\frac {9}{2};\frac {n+5}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+3}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {5 (3 A+4 B) \, _2F_1\left (\frac {n+4}{2},n+\frac {9}{2};\frac {n+6}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+4}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {5 (3 A+4 B) \, _2F_1\left (n+\frac {9}{2},\frac {n+5}{2};\frac {n+7}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+5}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(11 A+10 B) \, _2F_1\left (n+\frac {9}{2},\frac {n+6}{2};\frac {n+8}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+6}+\frac {(5 A+2 B) \tan \left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (n+\frac {9}{2},\frac {n+7}{2};\frac {n+9}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+7}\right )\right )\right )\right )\right )+\frac {A \, _2F_1\left (\frac {n+1}{2},n+\frac {9}{2};\frac {n+3}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+1}+\frac {A \tan ^7\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {n}{2}+4,n+\frac {9}{2};\frac {n}{2}+5;-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{n+8}\right )}{f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]),x]

[Out]

(2^(1 + n)*Sec[(e + f*x)/2]*(d*Sin[e + f*x])^n*(a*(1 + Sin[e + f*x]))^(5/2)*Tan[(e + f*x)/2]*(Tan[(e + f*x)/2]
/(1 + Tan[(e + f*x)/2]^2))^n*(1 + Tan[(e + f*x)/2]^2)^n*((A*Hypergeometric2F1[(1 + n)/2, 9/2 + n, (3 + n)/2, -
Tan[(e + f*x)/2]^2])/(1 + n) + (A*Hypergeometric2F1[4 + n/2, 9/2 + n, 5 + n/2, -Tan[(e + f*x)/2]^2]*Tan[(e + f
*x)/2]^7)/(8 + n) + Tan[(e + f*x)/2]*(((5*A + 2*B)*Hypergeometric2F1[(2 + n)/2, 9/2 + n, (4 + n)/2, -Tan[(e +
f*x)/2]^2])/(2 + n) + Tan[(e + f*x)/2]*(((11*A + 10*B)*Hypergeometric2F1[(3 + n)/2, 9/2 + n, (5 + n)/2, -Tan[(
e + f*x)/2]^2])/(3 + n) + Tan[(e + f*x)/2]*((5*(3*A + 4*B)*Hypergeometric2F1[(4 + n)/2, 9/2 + n, (6 + n)/2, -T
an[(e + f*x)/2]^2])/(4 + n) + Tan[(e + f*x)/2]*((5*(3*A + 4*B)*Hypergeometric2F1[9/2 + n, (5 + n)/2, (7 + n)/2
, -Tan[(e + f*x)/2]^2])/(5 + n) + Tan[(e + f*x)/2]*(((11*A + 10*B)*Hypergeometric2F1[9/2 + n, (6 + n)/2, (8 +
n)/2, -Tan[(e + f*x)/2]^2])/(6 + n) + ((5*A + 2*B)*Hypergeometric2F1[9/2 + n, (7 + n)/2, (9 + n)/2, -Tan[(e +
f*x)/2]^2]*Tan[(e + f*x)/2])/(7 + n))))))))/(f*Sqrt[Sec[(e + f*x)/2]^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^
5*Sin[e + f*x]^n)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a^{2} + {\left (B a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \left (d \sin \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.09, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (A +B \sin \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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