Optimal. Leaf size=362 \[ \frac {3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac {a \left (18 a^2+49 b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{110 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {\left (9 a^4+23 a^2 b^2-32 b^4\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{55 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
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Rubi [A]
time = 0.48, antiderivative size = 362, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3950, 4167,
4087, 4092, 3919, 144, 143} \begin {gather*} \frac {a \left (18 a^2+49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{110 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 \left (9 a^2+32 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{220 b^2 d}-\frac {\left (9 a^4+23 a^2 b^2-32 b^4\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{55 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{44 b^2 d}+\frac {3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/3}}{11 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 3919
Rule 3950
Rule 4087
Rule 4092
Rule 4167
Rubi steps
\begin {align*} \int \sec ^4(c+d x) (a+b \sec (c+d x))^{2/3} \, dx &=\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac {3 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (a+\frac {8}{3} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{11 b}\\ &=-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac {9 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (-\frac {2 a b}{3}+\frac {2}{9} \left (9 a^2+32 b^2\right ) \sec (c+d x)\right ) \, dx}{88 b^2}\\ &=\frac {3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac {27 \int \frac {\sec (c+d x) \left (\frac {2}{27} b \left (3 a^2+64 b^2\right )+\frac {2}{27} a \left (18 a^2+49 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{440 b^2}\\ &=\frac {3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac {\left (a \left (18 a^2+49 b^2\right )\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{220 b^3}-\frac {\left (9 a^4+23 a^2 b^2-32 b^4\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{110 b^3}\\ &=\frac {3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}-\frac {\left (a \left (18 a^2+49 b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{220 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {\left (\left (9 a^4+23 a^2 b^2-32 b^4\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{110 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=\frac {3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}-\frac {\left (a \left (18 a^2+49 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{220 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac {\left (\left (9 a^4+23 a^2 b^2-32 b^4\right ) \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{110 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac {3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac {a \left (18 a^2+49 b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{110 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {\left (9 a^4+23 a^2 b^2-32 b^4\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{55 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(8052\) vs. \(2(362)=724\).
time = 38.01, size = 8052, normalized size = 22.24 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (\sec ^{4}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{{\cos \left (c+d\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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