\(\int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx\) [503]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=-\frac {c \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(a+b x)\right ) (b \sec (a+b x))^{-1+n} \sqrt {c \sin (a+b x)}}{(1-n) \sqrt [4]{\sin ^2(a+b x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2667, 2656} \[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=-\frac {c \sqrt {c \sin (a+b x)} (b \sec (a+b x))^{n-1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(a+b x)\right )}{(1-n) \sqrt [4]{\sin ^2(a+b x)}} \]

[In]

Int[(b*Sec[a + b*x])^n*(c*Sin[a + b*x])^(3/2),x]

[Out]

-((c*Hypergeometric2F1[-1/4, (1 - n)/2, (3 - n)/2, Cos[a + b*x]^2]*(b*Sec[a + b*x])^(-1 + n)*Sqrt[c*Sin[a + b*
x]])/((1 - n)*(Sin[a + b*x]^2)^(1/4)))

Rule 2656

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^(2*IntPar
t[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*
x]^2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 2667

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[b^2*(b*Cos[e
+ f*x])^(n - 1)*(b*Sec[e + f*x])^(n - 1), Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e,
f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \left (b^2 (b \cos (a+b x))^{-1+n} (b \sec (a+b x))^{-1+n}\right ) \int (b \cos (a+b x))^{-n} (c \sin (a+b x))^{3/2} \, dx \\ & = -\frac {c \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(a+b x)\right ) (b \sec (a+b x))^{-1+n} \sqrt {c \sin (a+b x)}}{(1-n) \sqrt [4]{\sin ^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 31.91 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.37 \[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=\frac {2 \cos ^2(a+b x)^{\frac {1}{2} (-1+n)} (b \sec (a+b x))^{-1+n} (c \sin (a+b x))^{5/2} \left (9 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {1}{2} (-1+n),\frac {9}{4},\sin ^2(a+b x)\right )+5 \operatorname {Hypergeometric2F1}\left (\frac {9}{4},\frac {1+n}{2},\frac {13}{4},\sin ^2(a+b x)\right ) \sin ^2(a+b x)\right )}{45 c} \]

[In]

Integrate[(b*Sec[a + b*x])^n*(c*Sin[a + b*x])^(3/2),x]

[Out]

(2*(Cos[a + b*x]^2)^((-1 + n)/2)*(b*Sec[a + b*x])^(-1 + n)*(c*Sin[a + b*x])^(5/2)*(9*Hypergeometric2F1[5/4, (-
1 + n)/2, 9/4, Sin[a + b*x]^2] + 5*Hypergeometric2F1[9/4, (1 + n)/2, 13/4, Sin[a + b*x]^2]*Sin[a + b*x]^2))/(4
5*c)

Maple [F]

\[\int \left (b \sec \left (b x +a \right )\right )^{n} \left (c \sin \left (b x +a \right )\right )^{\frac {3}{2}}d x\]

[In]

int((b*sec(b*x+a))^n*(c*sin(b*x+a))^(3/2),x)

[Out]

int((b*sec(b*x+a))^n*(c*sin(b*x+a))^(3/2),x)

Fricas [F]

\[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}} \left (b \sec \left (b x + a\right )\right )^{n} \,d x } \]

[In]

integrate((b*sec(b*x+a))^n*(c*sin(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*sin(b*x + a))*(b*sec(b*x + a))^n*c*sin(b*x + a), x)

Sympy [F(-1)]

Timed out. \[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate((b*sec(b*x+a))**n*(c*sin(b*x+a))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}} \left (b \sec \left (b x + a\right )\right )^{n} \,d x } \]

[In]

integrate((b*sec(b*x+a))^n*(c*sin(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(3/2)*(b*sec(b*x + a))^n, x)

Giac [F]

\[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}} \left (b \sec \left (b x + a\right )\right )^{n} \,d x } \]

[In]

integrate((b*sec(b*x+a))^n*(c*sin(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(3/2)*(b*sec(b*x + a))^n, x)

Mupad [F(-1)]

Timed out. \[ \int (b \sec (a+b x))^n (c \sin (a+b x))^{3/2} \, dx=\int {\left (c\,\sin \left (a+b\,x\right )\right )}^{3/2}\,{\left (\frac {b}{\cos \left (a+b\,x\right )}\right )}^n \,d x \]

[In]

int((c*sin(a + b*x))^(3/2)*(b/cos(a + b*x))^n,x)

[Out]

int((c*sin(a + b*x))^(3/2)*(b/cos(a + b*x))^n, x)