\(\int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx\) [261]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}+\frac {4 (c \sin (a+b x))^{3/2}}{5 b c d^3 \sqrt {d \cos (a+b x)}}-\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b d^4 \sqrt {\sin (2 a+2 b x)}} \]

[Out]

2/5*(c*sin(b*x+a))^(3/2)/b/c/d/(d*cos(b*x+a))^(5/2)+4/5*(c*sin(b*x+a))^(3/2)/b/c/d^3/(d*cos(b*x+a))^(1/2)+4/5*
(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)*(c*sin
(b*x+a))^(1/2)/b/d^4/sin(2*b*x+2*a)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2651, 2652, 2719} \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {4 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{5 b d^4 \sqrt {\sin (2 a+2 b x)}}+\frac {4 (c \sin (a+b x))^{3/2}}{5 b c d^3 \sqrt {d \cos (a+b x)}}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}} \]

[In]

Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(7/2),x]

[Out]

(2*(c*Sin[a + b*x])^(3/2))/(5*b*c*d*(d*Cos[a + b*x])^(5/2)) + (4*(c*Sin[a + b*x])^(3/2))/(5*b*c*d^3*Sqrt[d*Cos
[a + b*x]]) - (4*Sqrt[d*Cos[a + b*x]]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(5*b*d^4*Sqrt[Sin[2*a
+ 2*b*x]])

Rule 2651

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]

Rule 2652

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a*Sin[e +
f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]), Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}+\frac {2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{3/2}} \, dx}{5 d^2} \\ & = \frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}+\frac {4 (c \sin (a+b x))^{3/2}}{5 b c d^3 \sqrt {d \cos (a+b x)}}-\frac {4 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)} \, dx}{5 d^4} \\ & = \frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}+\frac {4 (c \sin (a+b x))^{3/2}}{5 b c d^3 \sqrt {d \cos (a+b x)}}-\frac {\left (4 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{5 d^4 \sqrt {\sin (2 a+2 b x)}} \\ & = \frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}+\frac {4 (c \sin (a+b x))^{3/2}}{5 b c d^3 \sqrt {d \cos (a+b x)}}-\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b d^4 \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)} \tan (a+b x)}{3 b d^4} \]

[In]

Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(7/2),x]

[Out]

(2*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[3/4, 9/4, 7/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a +
b*x]]*Tan[a + b*x])/(3*b*d^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(411\) vs. \(2(139)=278\).

Time = 0.31 (sec) , antiderivative size = 412, normalized size of antiderivative = 3.07

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {c \sin \left (b x +a \right )}\, \left (4 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \cot \left (b x +a \right )-2 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \cot \left (b x +a \right )+4 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \csc \left (b x +a \right )-2 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \csc \left (b x +a \right )-2 \sqrt {2}\, \cot \left (b x +a \right )+\sqrt {2}\, \csc \left (b x +a \right )+\sqrt {2}\, \left (\sec ^{2}\left (b x +a \right )\right ) \csc \left (b x +a \right )\right )}{5 b \,d^{3} \sqrt {d \cos \left (b x +a \right )}}\)	\(412\)

[In]

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/5/b*2^(1/2)*(c*sin(b*x+a))^(1/2)/d^3/(d*cos(b*x+a))^(1/2)*(4*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-cs
c(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*cot(b*
x+a)-2*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*Elliptic
F((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*cot(b*x+a)+4*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(
b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*csc(b*x+
a)-2*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF(
(-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*csc(b*x+a)-2*2^(1/2)*cot(b*x+a)+2^(1/2)*csc(b*x+a)+2^(1/2)*sec(b
*x+a)^2*csc(b*x+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {2 \, {\left (i \, \sqrt {i \, c d} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - i \, \sqrt {-i \, c d} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - i \, \sqrt {i \, c d} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + i \, \sqrt {-i \, c d} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {d \cos \left (b x + a\right )} {\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{5 \, b d^{4} \cos \left (b x + a\right )^{3}} \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

-2/5*(I*sqrt(I*c*d)*cos(b*x + a)^3*elliptic_e(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) - I*sqrt(-I*c*d)*cos(
b*x + a)^3*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - I*sqrt(I*c*d)*cos(b*x + a)^3*elliptic_f(arc
sin(cos(b*x + a) + I*sin(b*x + a)), -1) + I*sqrt(-I*c*d)*cos(b*x + a)^3*elliptic_f(arcsin(cos(b*x + a) - I*sin
(b*x + a)), -1) - sqrt(d*cos(b*x + a))*(2*cos(b*x + a)^2 + 1)*sqrt(c*sin(b*x + a))*sin(b*x + a))/(b*d^4*cos(b*
x + a)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(7/2), x)

Giac [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\int \frac {\sqrt {c\,\sin \left (a+b\,x\right )}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \]

[In]

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(7/2),x)

[Out]

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(7/2), x)