\(\int (c \sin ^m(a+b x))^{5/2} \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 89 \[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\frac {2 c^2 \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2+5 m),\frac {1}{4} (6+5 m),\sin ^2(a+b x)\right ) \sin ^{1+2 m}(a+b x) \sqrt {c \sin ^m(a+b x)}}{b (2+5 m) \sqrt {\cos ^2(a+b x)}} \]

[Out]

2*c^2*cos(b*x+a)*hypergeom([1/2, 1/2+5/4*m],[3/2+5/4*m],sin(b*x+a)^2)*sin(b*x+a)^(1+2*m)*(c*sin(b*x+a)^m)^(1/2
)/b/(2+5*m)/(cos(b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3287, 2722} \[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\frac {2 c^2 \cos (a+b x) \sin ^{2 m+1}(a+b x) \sqrt {c \sin ^m(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5 m+2),\frac {1}{4} (5 m+6),\sin ^2(a+b x)\right )}{b (5 m+2) \sqrt {\cos ^2(a+b x)}} \]

[In]

Int[(c*Sin[a + b*x]^m)^(5/2),x]

[Out]

(2*c^2*Cos[a + b*x]*Hypergeometric2F1[1/2, (2 + 5*m)/4, (6 + 5*m)/4, Sin[a + b*x]^2]*Sin[a + b*x]^(1 + 2*m)*Sq
rt[c*Sin[a + b*x]^m])/(b*(2 + 5*m)*Sqrt[Cos[a + b*x]^2])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3287

Int[(u_.)*((b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Sin[e + f*x
])^n)^FracPart[p]/(c*Sin[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Sin[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps \begin{align*} \text {integral}& = \left (c^2 \sin ^{-\frac {m}{2}}(a+b x) \sqrt {c \sin ^m(a+b x)}\right ) \int \sin ^{\frac {5 m}{2}}(a+b x) \, dx \\ & = \frac {2 c^2 \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2+5 m),\frac {1}{4} (6+5 m),\sin ^2(a+b x)\right ) \sin ^{1+2 m}(a+b x) \sqrt {c \sin ^m(a+b x)}}{b (2+5 m) \sqrt {\cos ^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\frac {2 \sqrt {\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2+5 m),\frac {1}{4} (6+5 m),\sin ^2(a+b x)\right ) \left (c \sin ^m(a+b x)\right )^{5/2} \tan (a+b x)}{b (2+5 m)} \]

[In]

Integrate[(c*Sin[a + b*x]^m)^(5/2),x]

[Out]

(2*Sqrt[Cos[a + b*x]^2]*Hypergeometric2F1[1/2, (2 + 5*m)/4, (6 + 5*m)/4, Sin[a + b*x]^2]*(c*Sin[a + b*x]^m)^(5
/2)*Tan[a + b*x])/(b*(2 + 5*m))

Maple [F]

\[\int {\left (c \left (\sin ^{m}\left (b x +a \right )\right )\right )}^{\frac {5}{2}}d x\]

[In]

int((c*sin(b*x+a)^m)^(5/2),x)

[Out]

int((c*sin(b*x+a)^m)^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*sin(b*x+a)^m)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F(-1)]

Timed out. \[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((c*sin(b*x+a)**m)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )^{m}\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((c*sin(b*x+a)^m)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a)^m)^(5/2), x)

Giac [F]

\[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )^{m}\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((c*sin(b*x+a)^m)^(5/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a)^m)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \left (c \sin ^m(a+b x)\right )^{5/2} \, dx=\int {\left (c\,{\sin \left (a+b\,x\right )}^m\right )}^{5/2} \,d x \]

[In]

int((c*sin(a + b*x)^m)^(5/2),x)

[Out]

int((c*sin(a + b*x)^m)^(5/2), x)