3.266 \(\int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx\)

Optimal. Leaf size=112 \[ \frac {64 (c \sin (a+b x))^{3/2}}{231 b c d^5 (d \cos (a+b x))^{3/2}}+\frac {16 (c \sin (a+b x))^{3/2}}{77 b c d^3 (d \cos (a+b x))^{7/2}}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}} \]

[Out]

2/11*(c*sin(b*x+a))^(3/2)/b/c/d/(d*cos(b*x+a))^(11/2)+16/77*(c*sin(b*x+a))^(3/2)/b/c/d^3/(d*cos(b*x+a))^(7/2)+
64/231*(c*sin(b*x+a))^(3/2)/b/c/d^5/(d*cos(b*x+a))^(3/2)

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Rubi [A]  time = 0.17, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2571, 2563} \[ \frac {64 (c \sin (a+b x))^{3/2}}{231 b c d^5 (d \cos (a+b x))^{3/2}}+\frac {16 (c \sin (a+b x))^{3/2}}{77 b c d^3 (d \cos (a+b x))^{7/2}}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(13/2),x]

[Out]

(2*(c*Sin[a + b*x])^(3/2))/(11*b*c*d*(d*Cos[a + b*x])^(11/2)) + (16*(c*Sin[a + b*x])^(3/2))/(77*b*c*d^3*(d*Cos
[a + b*x])^(7/2)) + (64*(c*Sin[a + b*x])^(3/2))/(231*b*c*d^5*(d*Cos[a + b*x])^(3/2))

Rule 2563

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

Rule 2571

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

Rubi steps

\begin {align*} \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx &=\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}+\frac {8 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx}{11 d^2}\\ &=\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}+\frac {16 (c \sin (a+b x))^{3/2}}{77 b c d^3 (d \cos (a+b x))^{7/2}}+\frac {32 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx}{77 d^4}\\ &=\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}+\frac {16 (c \sin (a+b x))^{3/2}}{77 b c d^3 (d \cos (a+b x))^{7/2}}+\frac {64 (c \sin (a+b x))^{3/2}}{231 b c d^5 (d \cos (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 67, normalized size = 0.60 \[ \frac {2 (28 \cos (2 (a+b x))+4 \cos (4 (a+b x))+45) \sec ^6(a+b x) (c \sin (a+b x))^{3/2} \sqrt {d \cos (a+b x)}}{231 b c d^7} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(13/2),x]

[Out]

(2*Sqrt[d*Cos[a + b*x]]*(45 + 28*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Sec[a + b*x]^6*(c*Sin[a + b*x])^(3/2))
/(231*b*c*d^7)

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fricas [A]  time = 0.57, size = 64, normalized size = 0.57 \[ \frac {2 \, {\left (32 \, \cos \left (b x + a\right )^{4} + 24 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{231 \, b d^{7} \cos \left (b x + a\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x, algorithm="fricas")

[Out]

2/231*(32*cos(b*x + a)^4 + 24*cos(b*x + a)^2 + 21)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d
^7*cos(b*x + a)^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(13/2), x)

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maple [A]  time = 0.21, size = 60, normalized size = 0.54 \[ \frac {2 \left (32 \left (\cos ^{4}\left (b x +a \right )\right )+24 \left (\cos ^{2}\left (b x +a \right )\right )+21\right ) \sqrt {c \sin \left (b x +a \right )}\, \cos \left (b x +a \right ) \sin \left (b x +a \right )}{231 b \left (d \cos \left (b x +a \right )\right )^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x)

[Out]

2/231/b*(32*cos(b*x+a)^4+24*cos(b*x+a)^2+21)*(c*sin(b*x+a))^(1/2)*cos(b*x+a)*sin(b*x+a)/(d*cos(b*x+a))^(13/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(13/2), x)

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mupad [B]  time = 6.21, size = 216, normalized size = 1.93 \[ -\frac {\sqrt {c\,\sin \left (a+b\,x\right )}\,\left (2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )\,\left (2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {1984\,\sin \left (a+b\,x\right )\,\left (-2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}+1\right )}{231\,b\,d^6}+\frac {256\,\sin \left (3\,a+3\,b\,x\right )\,\left (-2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}+1\right )}{77\,b\,d^6}+\frac {128\,\sin \left (5\,a+5\,b\,x\right )\,\left (-2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}+1\right )}{231\,b\,d^6}\right )}{32\,{\left ({\sin \left (a+b\,x\right )}^2-1\right )}^3\,\sqrt {-d\,\left (2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(13/2),x)

[Out]

-((c*sin(a + b*x))^(1/2)*(2*sin(a/2 + (b*x)/2)^2 - 1)*(sin(5*a + 5*b*x)*1i + 2*sin((5*a)/2 + (5*b*x)/2)^2 - 1)
*((1984*sin(a + b*x)*(sin(5*a + 5*b*x)*1i - 2*sin((5*a)/2 + (5*b*x)/2)^2 + 1))/(231*b*d^6) + (256*sin(3*a + 3*
b*x)*(sin(5*a + 5*b*x)*1i - 2*sin((5*a)/2 + (5*b*x)/2)^2 + 1))/(77*b*d^6) + (128*sin(5*a + 5*b*x)*(sin(5*a + 5
*b*x)*1i - 2*sin((5*a)/2 + (5*b*x)/2)^2 + 1))/(231*b*d^6)))/(32*(sin(a + b*x)^2 - 1)^3*(-d*(2*sin(a/2 + (b*x)/
2)^2 - 1))^(1/2))

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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((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(13/2),x)

[Out]

Timed out

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