3.377 \(\int (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=138 \[ \frac {64 a^3 (7 B+5 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (7 B+5 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

2/35*a*(7*B+5*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+64/105*a^3*(7*B
+5*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/105*a^2*(7*B+5*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.17, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4054, 12, 3793, 3792} \[ \frac {64 a^3 (7 B+5 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (7 B+5 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(64*a^3*(7*B + 5*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(7*B + 5*C)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*a*(7*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + a*Sec
[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3792

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 3793

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

Rule 4054

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

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {2 \int \frac {1}{2} a (7 B+5 C) \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx}{7 a}\\ &=\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{7} (7 B+5 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{35} (8 a (7 B+5 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (7 B+5 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{105} \left (32 a^2 (7 B+5 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {64 a^3 (7 B+5 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (7 B+5 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 79, normalized size = 0.57 \[ \frac {2 a^3 \tan (c+d x) \left (3 (7 B+20 C) \sec ^2(c+d x)+(98 B+115 C) \sec (c+d x)+301 B+15 C \sec ^3(c+d x)+230 C\right )}{105 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a^3*(301*B + 230*C + (98*B + 115*C)*Sec[c + d*x] + 3*(7*B + 20*C)*Sec[c + d*x]^2 + 15*C*Sec[c + d*x]^3)*Tan
[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.46, size = 115, normalized size = 0.83 \[ \frac {2 \, {\left ({\left (301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/105*((301*B + 230*C)*a^2*cos(d*x + c)^3 + (98*B + 115*C)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c
) + 15*C*a^2)*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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giac [A]  time = 1.86, size = 216, normalized size = 1.57 \[ \frac {8 \, {\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} {\left (7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {2} {\left (7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

8/105*((4*(2*sqrt(2)*(7*B*a^6*sgn(cos(d*x + c)) + 5*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 7*sqrt(2
)*(7*B*a^6*sgn(cos(d*x + c)) + 5*C*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(7*B*a^6*sgn(co
s(d*x + c)) + 5*C*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt(2)*(B*a^6*sgn(cos(d*x + c)) + C*a^
6*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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maple [A]  time = 1.44, size = 119, normalized size = 0.86 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (301 B \left (\cos ^{3}\left (d x +c \right )\right )+230 C \left (\cos ^{3}\left (d x +c \right )\right )+98 B \left (\cos ^{2}\left (d x +c \right )\right )+115 C \left (\cos ^{2}\left (d x +c \right )\right )+21 B \cos \left (d x +c \right )+60 C \cos \left (d x +c \right )+15 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{105 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(301*B*cos(d*x+c)^3+230*C*cos(d*x+c)^3+98*B*cos(d*x+c)^2+115*C*cos(d*x+c)^2+21*B*cos(
d*x+c)+60*C*cos(d*x+c)+15*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^3*a^2

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maxima [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((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 7.43, size = 590, normalized size = 4.28 \[ \frac {\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 (\frac {B\,a^2\,2{}\mathrm {i}}{d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (301\,B+230\,C\right )\,2{}\mathrm {i}}{105\,d}\right )}{{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1}-\frac {\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}}\,\left (\frac {B\,a^2\,2{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,10{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,2{}\mathrm {i}}{7\,d}\right )+\frac {B\,a^2\,2{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,10{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,2{}\mathrm {i}}{7\,d}\right )}{\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 {\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}}\,\left (\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (5\,B+9\,C\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (7\,B-8\,C\right )\,2{}\mathrm {i}}{35\,d}\right )-\frac {B\,a^2\,2{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (B+2\,C\right )\,2{}\mathrm {i}}{d}+\frac {a^2\,\left (B+C\right )\,4{}\mathrm {i}}{d}\right )}{\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 {\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}}\,\left (\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (63\,B+80\,C\right )\,2{}\mathrm {i}}{105\,d}\right )-\frac {B\,a^2\,2{}\mathrm {i}}{3\,d}+\frac {a^2\,\left (9\,B+10\,C\right )\,2{}\mathrm {i}}{3\,d}\right )}{\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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2),x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*2i)/d - (a^2*exp(c*1i + d*x*1i)*(301*B
+ 230*C)*2i)/(105*d)))/(exp(c*1i + d*x*1i) + 1) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/
2)*(exp(c*1i + d*x*1i)*((B*a^2*2i)/(7*d) + (a^2*(3*B + 4*C)*10i)/(7*d) - (a^2*(5*B + 2*C)*2i)/(7*d) - (a^2*(11
*B + 10*C)*2i)/(7*d)) + (B*a^2*2i)/(7*d) + (a^2*(3*B + 4*C)*10i)/(7*d) - (a^2*(5*B + 2*C)*2i)/(7*d) - (a^2*(11
*B + 10*C)*2i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - ((a + a/(exp(- c*1i - d*x*1i)/2
+ exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(5*B + 2*C)*2i)/(5*d) - (a^2*(5*B + 9*C)*4i)/(5*d) +
(a^2*(7*B - 8*C)*2i)/(35*d)) - (B*a^2*2i)/(5*d) - (a^2*(B + 2*C)*2i)/d + (a^2*(B + C)*4i)/d))/((exp(c*1i + d*
x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c
*1i + d*x*1i)*((a^2*(5*B + 2*C)*2i)/(3*d) - (a^2*(63*B + 80*C)*2i)/(105*d)) - (B*a^2*2i)/(3*d) + (a^2*(9*B + 1
0*C)*2i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(5/2)*(B + C*sec(c + d*x))*sec(c + d*x), x)

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