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]

1/3*sec(f*x+e)^3*(a+b*sin(f*x+e))^(5/2)*(d*sin(f*x+e))^(1/2)/d/f+5/6*a*sec(f*x+e)*(b+a*sin(f*x+e))*(a+b*sin(f*
x+e))^(1/2)/f/(d*sin(f*x+e))^(1/2)-5/6*a*(a+b)^(3/2)*EllipticF(d^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(d*s
in(f*x+e))^(1/2),((-a-b)/(a-b))^(1/2))*(-a*(-1+csc(f*x+e))/(a+b))^(1/2)*(a*(1+csc(f*x+e))/(a-b))^(1/2)*tan(f*x
+e)/f/d^(1/2)-5/6*a*b*(a+b)*EllipticE(((-b-a*csc(f*x+e))/(a-b))^(1/2),((-a+b)/(a+b))^(1/2))*(1+sin(f*x+e))*(-a
*(-1+csc(f*x+e))/(a+b))^(1/2)*((b+a*csc(f*x+e))/(-a+b))^(1/2)*tan(f*x+e)/f/(a*(1+csc(f*x+e))/(a-b))^(1/2)/(d*s
in(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(1/2)

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Rubi [F]  time = 0.39, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [A]  time = 22.95, size = 667, normalized size = 1.82 \[ \frac {\sin (e+f x) \sqrt {a+b \sin (e+f x)} \left (\frac {1}{3} \sec ^3(e+f x) \left (a^2+2 a b \sin (e+f x)+b^2\right )+\frac {1}{6} \sec (e+f x) \left (5 a^2+5 a b \sin (e+f x)-2 b^2\right )\right )}{f \sqrt {d \sin (e+f x)}}+\frac {5 a \sqrt {\sin (e+f x)} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right ) \sqrt {\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{2 \tan ^2\left (\frac {1}{2} (e+f x)\right )+2}} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (e+f x)\right )+a+2 b \tan \left (\frac {1}{2} (e+f x)\right )}{\tan ^2\left (\frac {1}{2} (e+f x)\right )+1}} \left (\frac {2 \sqrt {b^2-a^2} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right ) \sqrt {\frac {a \left (a \tan ^2\left (\frac {1}{2} (e+f x)\right )+a+2 b \tan \left (\frac {1}{2} (e+f x)\right )\right )}{a^2-b^2}} \left (a \sqrt {\frac {a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b^2-a^2}-b}} \sqrt {-\frac {a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b^2-a^2}+b}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {b^2-a^2}}{\sqrt {b^2-a^2}}}}{\sqrt {2}}\right )|\frac {2 \sqrt {b^2-a^2}}{b+\sqrt {b^2-a^2}}\right )-b \tan \left (\frac {1}{2} (e+f x)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {-b-a \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {b^2-a^2}}{\sqrt {b^2-a^2}}}}{\sqrt {2}}\right )|\frac {2 \sqrt {b^2-a^2}}{\sqrt {b^2-a^2}-b}\right )\right )}{\sqrt {\frac {a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b^2-a^2}-b}} \left (a \tan ^2\left (\frac {1}{2} (e+f x)\right )+a+2 b \tan \left (\frac {1}{2} (e+f x)\right )\right )}-2 b \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{6 f \left (\tan ^3\left (\frac {1}{2} (e+f x)\right )+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {d \sin (e+f x)}} \]

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]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, a b \sec \left (f x + e\right )^{4} \sin \left (f x + e\right ) - {\left (b^{2} \cos \left (f x + e\right )^{2} - a^{2} - b^{2}\right )} \sec \left (f x + e\right )^{4}\right )} \sqrt {b \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right )}}{d \sin \left (f x + e\right )}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt {d \sin \left (f x + e\right )}}\,{d x} \]

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)

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maple [B]  time = 0.72, size = 2490, normalized size = 6.80 \[ \text {result too large to display} \]

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*(10*cos(f*x+e)^4*(-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)/si
n(f*x+e))^(1/2)*(a*(-1+cos(f*x+e))/(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))*b^2-5*cos(f*x+e)^4*(-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)*(a*(-1+cos(f*x+e))/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticF((-(-(-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))*a^2-10*cos(f*x+e)^4*(-(-(-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)*(a*(-1+cos(f*x+e))/(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))*a^2*b+10*cos(f*x+e)^4*(-(-(-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+co
s(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)*(a*(-1+cos(f*x+e))/(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))*b^3+10*cos(f*x+e)^3*(-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)*(a*(-1+cos(f*x+e))/(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))*b^2-5*cos(f*x+e)^3*(-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)*(a*(-1+cos(f*x+e))/(b+(-a^2+b^2)^(1/2)
)/sin(f*x+e))^(1/2)*EllipticF((-(-(-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))*a^2-10*cos(f*x+e)^3*(-(-(-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)*si
n(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)*(a*(-1+cos(f*x+e))/(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))*a^2*b+10*cos(f*x+e)^3*(-(-(-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)*(a*(-1+cos(f*x+e))/(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))*b^3+5*2^(1/2)*sin(f*x+e)*co
s(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*2^(1/2)*cos(f*x+e)^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)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt {d \sin \left (f x + e\right )}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\cos \left (e+f\,x\right )}^4\,\sqrt {d\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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