3.484 \(\int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx\)

Optimal. Leaf size=75 \[ \frac {d \sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac {5}{4},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2587, 2577} \[ \frac {d \sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac {5}{4},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[5/4, (1 + m)/2, (3 + m)/2, Sin[a + b*x]^2]*Sqrt[d*Sec[a + b*x]]*(c
*Sin[a + b*x])^(1 + m))/(b*c*(1 + m))

Rule 2577

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

Rule 2587

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*} \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx &=\left (d^2 \sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)}\right ) \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{3/2}} \, dx\\ &=\frac {d \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac {5}{4},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(a+b x)\right ) \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{1+m}}{b c (1+m)}\\ \end {align*}

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Mathematica [A]  time = 1.36, size = 96, normalized size = 1.28 \[ -\frac {2 \cot (a+b x) (d \sec (a+b x))^{3/2} \left (-\tan ^2(a+b x)\right )^{\frac {1-m}{2}} (c \sin (a+b x))^m \, _2F_1\left (\frac {1}{4} (3-2 m),\frac {1-m}{2};\frac {1}{4} (7-2 m);\sec ^2(a+b x)\right )}{b (2 m-3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cot[a + b*x]*Hypergeometric2F1[(3 - 2*m)/4, (1 - m)/2, (7 - 2*m)/4, Sec[a + b*x]^2]*(d*Sec[a + b*x])^(3/2)
*(c*Sin[a + b*x])^m*(-Tan[a + b*x]^2)^((1 - m)/2))/(b*(-3 + 2*m))

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m} d \sec \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}} \left (c \sin \left (b x +a \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,\sin \left (a+b\,x\right )\right )}^m\,{\left (\frac {d}{\cos \left (a+b\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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