3.1480 \(\int \frac{\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt{d \sin (e+f x)}} \, dx\)

Optimal. Leaf size=366 \[ \frac{\sec ^3(e+f x) \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}{3 d f}+\frac{5 a \sec (e+f x) (a \sin (e+f x)+b) \sqrt{a+b \sin (e+f x)}}{6 f \sqrt{d \sin (e+f x)}}-\frac{5 a (a+b)^{3/2} \tan (e+f x) \sqrt{-\frac{a (\csc (e+f x)-1)}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{6 \sqrt{d} f}-\frac{5 a b (a+b) (\sin (e+f x)+1) \tan (e+f x) \sqrt{-\frac{a (\csc (e+f x)-1)}{a+b}} \sqrt{\frac{a \csc (e+f x)+b}{b-a}} E\left (\sin ^{-1}\left (\sqrt{-\frac{b+a \csc (e+f x)}{a-b}}\right )|\frac{b-a}{a+b}\right )}{6 f \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}}} \]

[Out]

(5*a*Sec[e + f*x]*(b + a*Sin[e + f*x])*Sqrt[a + b*Sin[e + f*x]])/(6*f*Sqrt[d*Sin[e + f*x]]) + (Sec[e + f*x]^3*
Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(5/2))/(3*d*f) - (5*a*(a + b)^(3/2)*Sqrt[-((a*(-1 + Csc[e + f*x]))/(
a + b))]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[ArcSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]
*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a - b))]*Tan[e + f*x])/(6*Sqrt[d]*f) - (5*a*b*(a + b)*Sqrt[-((a*(-1 + Csc[
e + f*x]))/(a + b))]*Sqrt[(b + a*Csc[e + f*x])/(-a + b)]*EllipticE[ArcSin[Sqrt[-((b + a*Csc[e + f*x])/(a - b))
]], (-a + b)/(a + b)]*(1 + Sin[e + f*x])*Tan[e + f*x])/(6*f*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Sqrt[d*Sin[e
+ f*x]]*Sqrt[a + b*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [F]  time = 0.391059, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt{d \sin (e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(Sec[e + f*x]^4*(a + b*Sin[e + f*x])^(5/2))/Sqrt[d*Sin[e + f*x]],x]

[Out]

(Sec[e + f*x]^3*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(5/2))/(3*d*f) + (5*a*Defer[Int][(Sec[e + f*x]^2*(a
+ b*Sin[e + f*x])^(3/2))/Sqrt[d*Sin[e + f*x]], x])/6

Rubi steps

\begin{align*} \int \frac{\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt{d \sin (e+f x)}} \, dx &=\frac{\sec ^3(e+f x) \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}{3 d f}+\frac{1}{6} (5 a) \int \frac{\sec ^2(e+f x) (a+b \sin (e+f x))^{3/2}}{\sqrt{d \sin (e+f x)}} \, dx\\ \end{align*}

Mathematica [B]  time = 26.7563, size = 6142, normalized size = 16.78 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[e + f*x]^4*(a + b*Sin[e + f*x])^(5/2))/Sqrt[d*Sin[e + f*x]],x]

[Out]

Result too large to show

________________________________________________________________________________________

Maple [B]  time = 0.457, size = 2426, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(a+b*sin(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x)

[Out]

-1/12/f*2^(1/2)*(10*(-a^2+b^2)^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^
(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a^2+b^2)^(1/2)/sin(f*x+e
))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticE((((-a^2+b^2)^(1/2)*sin(f*x+e)+b*s
in(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^
(1/2))^(1/2))*cos(f*x+e)^4*b^2-5*(-a^2+b^2)^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(
b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a^2+b^2)^(1
/2)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticF((((-a^2+b^2)^(1/2)*s
in(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2)
)/(-a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^4*a^2-10*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(
-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a^2+b^2)^(1/2)
/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticE((((-a^2+b^2)^(1/2)*sin(
f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(
-a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^4*a^2*b+10*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-
a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a^2+b^2)^(1/2)/
sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticE((((-a^2+b^2)^(1/2)*sin(f
*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-
a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^4*b^3+10*(-a^2+b^2)^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x
+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a
^2+b^2)^(1/2)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticE((((-a^2+b^
2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+
b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^3*b^2-5*(-a^2+b^2)^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f
*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*
x+e)*a-a)/(-a^2+b^2)^(1/2)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*Ellipti
cF((((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/
2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^3*a^2-10*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+
e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e
)*a-a)/(-a^2+b^2)^(1/2)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticE(
(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*
((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^3*a^2*b+10*(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e
)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)
*a-a)/(-a^2+b^2)^(1/2)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticE((
((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*(
(b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*cos(f*x+e)^3*b^3+5*2^(1/2)*sin(f*x+e)*cos(f*x+e)^3*a*b^2+5*2^(1/
2)*cos(f*x+e)^4*a^2*b-2*2^(1/2)*cos(f*x+e)^4*b^3-5*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2*a^3+2^(1/2)*sin(f*x+e)*cos(
f*x+e)^2*a*b^2+5*2^(1/2)*cos(f*x+e)^3*a^2*b-4*cos(f*x+e)^2*2^(1/2)*a^2*b+4*2^(1/2)*cos(f*x+e)^2*b^3-2*sin(f*x+
e)*2^(1/2)*a^3-6*sin(f*x+e)*2^(1/2)*a*b^2-6*2^(1/2)*a^2*b-2*2^(1/2)*b^3)/cos(f*x+e)^3/(d*sin(f*x+e))^(1/2)/(a+
b*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^(5/2)*sec(f*x + e)^4/sqrt(d*sin(f*x + e)), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (2 \, a b \sec \left (f x + e\right )^{4} \sin \left (f x + e\right ) -{\left (b^{2} \cos \left (f x + e\right )^{2} - a^{2} - b^{2}\right )} \sec \left (f x + e\right )^{4}\right )} \sqrt{b \sin \left (f x + e\right ) + a} \sqrt{d \sin \left (f x + e\right )}}{d \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral((2*a*b*sec(f*x + e)^4*sin(f*x + e) - (b^2*cos(f*x + e)^2 - a^2 - b^2)*sec(f*x + e)^4)*sqrt(b*sin(f*x
+ e) + a)*sqrt(d*sin(f*x + e))/(d*sin(f*x + e)), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**4*(a+b*sin(f*x+e))**(5/2)/(d*sin(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)^(5/2)*sec(f*x + e)^4/sqrt(d*sin(f*x + e)), x)