3.614 \(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=138 \[ \frac {\sqrt {2} (c-d)^2 \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {5}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]

[Out]

(c-d)^2*AppellF1(1/2+m,-5/2,1/2,3/2+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)/f/(1+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 = 0.19, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2788, 140, 139, 138} \[ \frac {\sqrt {2} (c-d)^2 \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {5}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(Sqrt[2]*(c - d)^2*AppellF1[1/2 + m, 1/2, -5/2, 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]])/(f*(1 + 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]

Rubi steps

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

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Mathematica [B]  time = 1.90, size = 365, normalized size = 2.64 \[ -\frac {3 \sqrt {2} (c+d) \sqrt {\sin (e+f x)+1} \tan \left (\frac {1}{4} (2 e+2 f x-\pi )\right ) (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^{5/2} F_1\left (\frac {1}{2};\frac {1}{2}-m,-\frac {5}{2};\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{f \sqrt {\cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )} \left (\sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \left (10 d F_1\left (\frac {3}{2};\frac {1}{2}-m,-\frac {3}{2};\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )+(2 m-1) (c+d) F_1\left (\frac {3}{2};\frac {3}{2}-m,-\frac {5}{2};\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )-3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,-\frac {5}{2};\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(-3*Sqrt[2]*(c + d)*AppellF1[1/2, 1/2 - m, -5/2, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)
/4]^2)/(c + d)]*Sqrt[1 + Sin[e + f*x]]*(a*(1 + Sin[e + f*x]))^m*(c + d*Sin[e + f*x])^(5/2)*Tan[(2*e - Pi + 2*f
*x)/4])/(f*Sqrt[Cos[(2*e - Pi + 2*f*x)/4]^2]*(-3*(c + d)*AppellF1[1/2, 1/2 - m, -5/2, 3/2, Cos[(2*e + Pi + 2*f
*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] + (10*d*AppellF1[3/2, 1/2 - m, -3/2, 5/2, Cos[(2*e + Pi +
2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] + (c + d)*(-1 + 2*m)*AppellF1[3/2, 3/2 - m, -5/2, 5/2
, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)])*Sin[(2*e - Pi + 2*f*x)/4]^2))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}\right )} \sqrt {d \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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))^(5/2),x, algorithm="fricas")

[Out]

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

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

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))^(5/2),x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + e) + a)^m, x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}\, dx \]

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))^(5/2),x)

[Out]

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

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

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + e) + a)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

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))^(5/2),x)

[Out]

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

[Out]

Timed out

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