3.1.90 \(\int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\) [90]

Optimal. Leaf size=122 \[ \frac {4 a \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {8 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{35 d}+\frac {12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d} \]

[Out]

12/35*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+4/5*a*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*a*sec(d*x+c)^3*tan(d
*x+c)/d/(a+a*sec(d*x+c))^(1/2)-8/35*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.14, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3888, 3885, 4086, 3877} \begin {gather*} \frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}+\frac {12 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac {8 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{35 d}+\frac {4 a \tan (c+d x)}{5 d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(4*a*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c +
d*x]]) - (8*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(35*d) + (12*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*a
*d)

Rule 3877

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(Cot[e + f*x]/(
f*Sqrt[a + b*Csc[e + f*x]])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3885

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3888

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*
Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[2*a*d*((n - 1)/(b*(2
*n - 1))), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 4086

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*B*m + A*b*(m + 1))/(b
*(m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B
, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx &=\frac {2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {6}{7} \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {12 \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx}{35 a}\\ &=\frac {2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {8 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{35 d}+\frac {12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {2}{5} \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {4 a \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {8 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{35 d}+\frac {12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 58, normalized size = 0.48 \begin {gather*} \frac {2 a \left (16+8 \sec (c+d x)+6 \sec ^2(c+d x)+5 \sec ^3(c+d x)\right ) \tan (c+d x)}{35 d \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*a*(16 + 8*Sec[c + d*x] + 6*Sec[c + d*x]^2 + 5*Sec[c + d*x]^3)*Tan[c + d*x])/(35*d*Sqrt[a*(1 + Sec[c + d*x])
])

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Maple [A]
time = 0.32, size = 82, normalized size = 0.67

method	result	size
default	\(-\frac {2 \left (16 \left (\cos ^{4}\left (d x +c \right )\right )-8 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-5\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{35 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/35/d*(16*cos(d*x+c)^4-8*cos(d*x+c)^3-2*cos(d*x+c)^2-cos(d*x+c)-5)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d
*x+c)^3/sin(d*x+c)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

16/35*(35*(d*cos(2*d*x + 2*c)^2 + d*sin(2*d*x + 2*c)^2 + 2*d*cos(2*d*x + 2*c) + d)*(cos(2*d*x + 2*c)^2 + sin(2
*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*sqrt(a)*integrate((((cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos
(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + c
os(2*d*x + 2*c)^2 + sin(10*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x +
6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(5/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(
2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*
cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))
*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)
) - ((cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x
+ 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2
*d*x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c))) - (cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) +
6*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(10*d*x +
10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*
d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))/(((2*(4*cos(8*d*x + 8*c) + 6*cos(6*d*x + 6*c) + 4*cos(4*d*
x + 4*c) + cos(2*d*x + 2*c))*cos(10*d*x + 10*c) + cos(10*d*x + 10*c)^2 + 8*(6*cos(6*d*x + 6*c) + 4*cos(4*d*x +
4*c) + cos(2*d*x + 2*c))*cos(8*d*x + 8*c) + 16*cos(8*d*x + 8*c)^2 + 12*(4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c)
)*cos(6*d*x + 6*c) + 36*cos(6*d*x + 6*c)^2 + 16*cos(4*d*x + 4*c)^2 + 8*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos
(2*d*x + 2*c)^2 + 2*(4*sin(8*d*x + 8*c) + 6*sin(6*d*x + 6*c) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(10*d
*x + 10*c) + sin(10*d*x + 10*c)^2 + 8*(6*sin(6*d*x + 6*c) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(8*d*x +
8*c) + 16*sin(8*d*x + 8*c)^2 + 12*(4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 36*sin(6*d*x + 6
*c)^2 + 16*sin(4*d*x + 4*c)^2 + 8*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (2*(4*cos(8*d*x + 8*c) + 6*cos(6*d*x + 6*c) + 4*cos(4*d*x + 4*c) + co
s(2*d*x + 2*c))*cos(10*d*x + 10*c) + cos(10*d*x + 10*c)^2 + 8*(6*cos(6*d*x + 6*c) + 4*cos(4*d*x + 4*c) + cos(2
*d*x + 2*c))*cos(8*d*x + 8*c) + 16*cos(8*d*x + 8*c)^2 + 12*(4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(6*d*x +
6*c) + 36*cos(6*d*x + 6*c)^2 + 16*cos(4*d*x + 4*c)^2 + 8*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)
^2 + 2*(4*sin(8*d*x + 8*c) + 6*sin(6*d*x + 6*c) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(10*d*x + 10*c) +
sin(10*d*x + 10*c)^2 + 8*(6*sin(6*d*x + 6*c) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*si
n(8*d*x + 8*c)^2 + 12*(4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 36*sin(6*d*x + 6*c)^2 + 16*si
n(4*d*x + 4*c)^2 + 8*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1))^2)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)), x) - (7*
cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(2*d*x + 2*c) - (7*cos(2*d*x + 2*c) + 2)*sin(7/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a))/((d*cos(2*d*x + 2*c)^2 + d*sin(2*d*x + 2*c)^2 + 2*d*
cos(2*d*x + 2*c) + d)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4))

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Fricas [A]
time = 2.67, size = 82, normalized size = 0.67 \begin {gather*} \frac {2 \, {\left (16 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 5\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{35 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*cos(d*x + c)^3 + 8*cos(d*x + c)^2 + 6*cos(d*x + c) + 5)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d
*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*sec(c + d*x)**4, x)

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Giac [A]
time = 0.83, size = 120, normalized size = 0.98 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (35 \, a^{4} - {\left (35 \, a^{4} + {\left (9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 49 \, a^{4}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{35 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/35*sqrt(2)*(35*a^4 - (35*a^4 + (9*a^4*tan(1/2*d*x + 1/2*c)^2 - 49*a^4)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x
+ 1/2*c)^2)*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2
*c)^2 + a)*d)

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Mupad [B]
time = 5.93, size = 331, normalized size = 2.71 \begin {gather*} -\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,32{}\mathrm {i}}{35\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {\left (\frac {16{}\mathrm {i}}{7\,d}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,16{}\mathrm {i}}{7\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {16{}\mathrm {i}}{5\,d}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,128{}\mathrm {i}}{35\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,16{}\mathrm {i}}{35\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(1/2)/cos(c + d*x)^4,x)

[Out]

((16i/(5*d) + (exp(c*1i + d*x*1i)*128i)/(35*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))
/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - ((16i/(7*d) + (exp(c*1i + d*x*1i)*16i)/(7*d))*(a + a/
(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3)
- (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*32i)/(35*d*(exp(c*1i + d*x
*1i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*16i)/(35*d*(exp
(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))

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